JEE Trigonometric Ratios and Identities Questions

Given,

$$\log_{\cos x}(\cot x) + 4\log_{\sin x}(\tan x) = 1$$

Using,

$$\cot x = \frac{\cos x}{\sin x}$$

and

$$\tan x = \frac{\sin x}{\cos x}$$

we get

$$\log_{\cos x}\left(\frac{\cos x}{\sin x}\right) + 4\log_{\sin x}\left(\frac{\sin x}{\cos x}\right) = 1$$

Using the logarithmic property

$$\log_a\left(\frac{m}{n}\right)=\log_a m-\log_a n$$

$$\log_{\cos x}\cos x - \log_{\cos x}\sin x + 4\left(\log_{\sin x}\sin x - \log_{\sin x}\cos x\right)=1$$

Since,

$$\log_{\cos x}\cos x = 1$$

and

$$\log_{\sin x}\sin x = 1$$

we obtain

$$1-\log_{\cos x}\sin x + 4\left(1-\log_{\sin x}\cos x\right)=1$$

Let

$$t=\log_{\cos x}\sin x$$

Using the reciprocal property of logarithms,

$$\log_{\sin x}\cos x = \frac{1}{t}$$

Substituting,

$$1-t+4\left(1-\frac{1}{t}\right)=1$$

$$1-t+4-\frac{4}{t}=1$$

$$5-t-\frac{4}{t}=1$$

$$4-t-\frac{4}{t}=0$$

Multiplying throughout by $$t$$,

$$4t-t^2-4=0$$

$$t^2-4t+4=0$$

$$(t-2)^2=0$$

$$t=2$$

Therefore,

$$\log_{\cos x}\sin x = 2$$

Converting to exponential form,

$$\sin x = \cos^2 x$$

Using,

$$\cos^2 x = 1-\sin^2 x$$

we get

$$\sin x = 1-\sin^2 x$$

$$\sin^2 x + \sin x -1 =0$$

Using the quadratic formula,

$$\sin x = \frac{-1\pm\sqrt{1+4}}{2}$$

$$\sin x = \frac{-1\pm\sqrt{5}}{2}$$

Therefore,

$$x=\sin^{-1}\left(\frac{-1+\sqrt5}{2}\right)$$

Comparing with

$$x=\sin^{-1}\left(\frac{\alpha+\sqrt\beta}{2}\right)$$

we get

$$\alpha=-1,\qquad \beta=5$$

Hence,

$$\alpha+\beta = -1+5 = 4$$

Therefore, the required answer is

$$\boxed{4}$$

Step 1: Transforming the Trigonometric Equation

The given equation is:

$$3\cos^4\theta - 5\cos^2\theta - 2\sin^6\theta + 2 = 0$$

To make the equation easier to factorize, let's convert everything into a single trigonometric ratio. Since we have both $$\cos$$ and $$\sin$$,
let's convert $$\sin^6\theta$$ in terms of $$\cos^2\theta$$ using the identity $$\sin^2\theta = 1 - \cos^2\theta$$:

$$2\sin^6\theta = 2(\sin^2\theta)^3 = 2(1 - \cos^2\theta)^3$$

Now let's use the substitution $$t = \cos^2\theta$$ to convert this into an algebraic polynomial. Note that since $$t = \cos^2\theta$$, the value of $$t$$ must lie in the range [0, 1]

Substituting $$t$$ into the equation gives:

$$3t^2 - 5t - 2(1 - t)^3 + 2 = 0$$

Step 2: Simplifying the Algebraic Polynomial

Expand the cubic term using the identity $$(1 - t)^3 = 1 - 3t + 3t^2 - t^3$$:

$$3t^2 - 5t - 2(1 - 3t + 3t^2 - t^3) + 2 = 0$$
$$3t^2 - 5t - 2 + 6t - 6t^2 + 2t^3 + 2 = 0$$
$$2t^3 + (3t^2 - 6t^2) + (-5t + 6t) + (-2 + 2) = 0$$
$$2t^3 - 3t^2 + t = 0$$
$$t(2t^2 - 3t + 1) = 0$$
$$2t^2 - 2t - t + 1 = 0$$
$$2t(t - 1) - 1(t - 1) = 0$$
$$(2t - 1)(t - 1) = 0$$

Thus, our completely factorized polynomial equation is:
$$t(2t - 1)(t - 1) = 0$$

This gives three possible roots for $$t$$:

1. $$t = 0$$
2. $$t = \frac{1}{2}$$
3. $$t = 1$$
   Since all three values lie within the valid domain $$t \in [0, 1]$$,
   we must analyze the solutions for each case. 

Step 3: Finding the Number of Solutions for $$\theta \in [0, 2\pi]$$

Now substitute back $$t = \cos^2\theta$$:

Case 1: $$t = 0 \implies \cos^2\theta = 0 \implies \cos\theta = 0$$
In the interval $$[0, 2\pi]$$, 

$$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$$

Number of solutions = 2

Case 2: $$t = 1 \implies \cos^2\theta = 1 \implies \cos\theta = \pm 1$$

• $$\cos\theta = 1 \implies \theta = 0, 2\pi$$
• $$\cos\theta = -1 \implies \theta = \pi$$

$$\theta = 0, \pi, 2\pi$$

Number of solutions = 3

Case 3: $$t = \frac{1}{2} \implies \cos^2\theta = \frac{1}{2} \implies \cos\theta = \pm \frac{1}{\sqrt{2}}$$

This gives solutions in all four quadrants:

• Quadrant I: $$\theta = \frac{\pi}{4}$$
• Quadrant II: $$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$
• Quadrant III: $$\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$$
• Quadrant IV: $$\theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$$

$$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$

Number of solutions = 4

Step 4: Computing the Total Number of Elements in Set $$S$$

Total number of solutions = (Solutions from Case 1) + (Solutions from Case 2) + (Solutions from Case 3)

$$\text{Total Elements} = 2 + 3 + 4 = 9$$

Final Answer

The number of elements in the set $$S$$ is 9.

The correct option is D.

By setting $$t = 9^{\tan^2 x}$$ it follows that $$9^{1-\tan^2 x} = \frac{9}{t}$$. Hence the equation becomes $$\frac{9}{t} + t = 10 \implies t^2 - 10t + 9 = 0 \implies (t-1)(t-9) = 0$$ so that $$t=1$$ or $$t=9$$. If $$t=1$$ then $$\tan^2 x=0$$ giving $$x=0$$, while if $$t=9$$ then $$\tan^2 x=1$$ which leads to $$x=\pm\frac{\pi}{4}$$. Therefore the set of solutions is $$S = \left\{-\frac{\pi}{4}, 0, \frac{\pi}{4}\right\}$$.

The value of $$\beta$$ can be written as $$\beta = \tan^2\left(-\frac{\pi}{12}\right) + \tan^2(0) + \tan^2\left(\frac{\pi}{12}\right) = 2\tan^2\frac{\pi}{12}$$. Since $$\tan 15° = 2 - \sqrt{3}$$, it follows that $$\tan^2 15° = 7 - 4\sqrt{3}$$ and therefore $$\beta = 14 - 8\sqrt{3}$$.

Finally, evaluating the desired expression gives
$$\frac{1}{6}(\beta - 14)^2 = \frac{1}{6}(-8\sqrt{3})^2 = \frac{192}{6} = \textbf{32}$$

We use the identity: $$\frac{\sin 3\theta}{\sin\theta} = 3 - 4\sin^2\theta = 4\cos^2\theta - 1$$

Applying to each factor:

$$4\cos^2 9° - 1 = \frac{\sin 27°}{\sin 9°}$$

$$4\cos^2 27° - 1 = \frac{\sin 81°}{\sin 27°}$$

$$4\cos^2 81° - 1 = \frac{\sin 243°}{\sin 81°}$$

$$4\cos^2 243° - 1 = \frac{\sin 729°}{\sin 243°}$$

The product telescopes:

$$\frac{\sin 27°}{\sin 9°} \cdot \frac{\sin 81°}{\sin 27°} \cdot \frac{\sin 243°}{\sin 81°} \cdot \frac{\sin 729°}{\sin 243°} = \frac{\sin 729°}{\sin 9°}$$

Since $$729° = 2 \times 360° + 9°$$, we have $$\sin 729° = \sin 9°$$.

Therefore the product = $$\frac{\sin 9°}{\sin 9°} = 1$$

So $$36 \times 1 = 36$$, which is Option D.

We need to find when $$\cos 2A + \cos 2B + \cos 2C$$ is least for a triangle.

Using the identity: $$\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A \cos B \cos C$$

This is minimized when $$\cos A \cos B \cos C$$ is maximized, which occurs when $$A = B = C = 60°$$ (equilateral triangle). The minimum value is $$-1 - 4 \times \frac{1}{8} = -\frac{3}{2}$$.

For an equilateral triangle with inradius $$r = 3$$:

The relationship between inradius and side is: $$r = \frac{a}{2\sqrt{3}}$$ $$\Rightarrow a = 2\sqrt{3} \times 3 = 6\sqrt{3}$$

Checking each option:

Option 1: Perimeter = $$3 \times 6\sqrt{3} = 18\sqrt{3}$$. This is correct.

Option 2: For an equilateral triangle: $$\sin 2A + \sin 2B + \sin 2C = 3\sin 120° = \frac{3\sqrt{3}}{2}$$. Also $$\sin A + \sin B + \sin C = 3\sin 60° = \frac{3\sqrt{3}}{2}$$. These are equal. Correct.

Option 3: The distance from the incenter M to each vertex is $$MA = \frac{r}{\sin(A/2)} = \frac{3}{\sin 30°} = 6$$. The angle $$\angle AMB = 90° + \frac{C}{2} = 90° + 30° = 120°$$. Therefore: $$\vec{MA} \cdot \vec{MB} = |MA||MB|\cos(\angle AMB) = 6 \times 6 \times \cos 120° = 36 \times (-\frac{1}{2}) = -18$$. Correct.

Option 4: Area = $$\frac{\sqrt{3}}{4} \times (6\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 108 = 27\sqrt{3}$$. But the option states $$\frac{27\sqrt{3}}{2}$$, which is NOT correct.

Therefore, the statement that is NOT correct is option 4: area of $$\triangle ABC = \frac{27\sqrt{3}}{2}$$.

The height of the tower is $$h = 5$$ m. Person 1 stands due south with angle of elevation $$45°$$, and person 2 stands due west with angle of elevation $$30°$$.

$$ \tan 45° = \frac{h}{d_1} \Rightarrow 1 = \frac{5}{d_1} \Rightarrow d_1 = 5 \text{ m} $$

$$ \tan 30° = \frac{h}{d_2} \Rightarrow \frac{1}{\sqrt{3}} = \frac{5}{d_2} \Rightarrow d_2 = 5\sqrt{3} \text{ m} $$

Since south and west are perpendicular directions, the two persons and the base of the tower form a right triangle.

$$ D = \sqrt{d_1^2 + d_2^2} = \sqrt{25 + 75} = \sqrt{100} = 10 \text{ m} $$

The distance between the two persons is Option B: 10 m.

To solve this, we first simplify $$f(x)$$ using trigonometric identities.

1. Simplify $$f(x)$$

Recall that $$\sin x + \cos x = \sqrt{2}\sin(x + \frac{\pi}{4})$$ and $$\sin x - \cos x = -\sqrt{2}\cos(x + \frac{\pi}{4})$$.

$$f(x) = \frac{\sqrt{2}\sin(x + \frac{\pi}{4}) - \sqrt{2}}{-\sqrt{2}\cos(x + \frac{\pi}{4})} = \frac{1 - \sin(x + \frac{\pi}{4})}{\cos(x + \frac{\pi}{4})}$$

Using half-angle identities (let $$\theta = x + \frac{\pi}{4}$$), we get:

$$f(x) = \frac{1 - \cos(\frac{\pi}{2} - \theta)}{\sin(\frac{\pi}{2} - \theta)} = \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \tan\left(\frac{\pi}{8} - \frac{x}{2}\right)$$

2. Find the Derivatives

Let $$u = \frac{\pi}{8} - \frac{x}{2}$$, so $$f(x) = \tan u$$.

• $$f'(x) = \sec^2 u \cdot (-\frac{1}{2})$$
• $$f''(x) = -\frac{1}{2} [2\sec u \cdot \sec u \tan u \cdot (-\frac{1}{2})] = \frac{1}{2} \sec^2 u \tan u$$

3. Evaluate at $$x = \frac{7\pi}{12}$$

Calculate $$u$$ first:

$$u = \frac{\pi}{8} - \frac{7\pi}{24} = \frac{3\pi - 7\pi}{24} = -\frac{4\pi}{24} = -\frac{\pi}{6}$$

Now find the values:

• $$f(-\frac{\pi}{6}) = \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$$
• $$f''(-\frac{\pi}{6}) = \frac{1}{2} \sec^2(-\frac{\pi}{6}) \tan(-\frac{\pi}{6}) = \frac{1}{2} \cdot (\frac{2}{\sqrt{3}})^2 \cdot (-\frac{1}{\sqrt{3}}) = \frac{1}{2} \cdot \frac{4}{3} \cdot (-\frac{1}{\sqrt{3}}) = -\frac{2}{3\sqrt{3}}$$

4. Final Calculation

$$f\left(\frac{7\pi}{12}\right) f''\left(\frac{7\pi}{12}\right) = \left(-\frac{1}{\sqrt{3}}\right) \left(-\frac{2}{3\sqrt{3}}\right) = \frac{2}{3 \cdot 3} = \mathbf{\frac{2}{9}}$$

From the figure:

• $$CD$$ is parallel to $$AB$$ (both are perpendicular to $$BE$$). Thus, $$CD = AB$$.
• In $$\triangle CAB$$: $$AC = AB \tan \theta_1$$.
• Area $$(\triangle CAB) = \frac{1}{2} \cdot AB \cdot AC = \frac{1}{2} (AB)^2 \tan \theta_1 = 2\sqrt{3} - 3$$.
• In $$\triangle CED$$: $$ED = CD \tan \theta_2 = AB \tan \theta_2$$.
• We are given $$\theta_1 + \theta_2 = \frac{\pi}{2}$$, which implies $$\theta_2 = \frac{\pi}{2} - \theta_1$$.
• Therefore, $$\tan \theta_2 = \cot \theta_1$$.

You are given that $$\sqrt{3}BE = 4AB$$, which means $$BE = \frac{4}{\sqrt{3}}AB$$.

From the geometry, $$BE = BD + DE$$.

Since $$BD = AC$$ (opposite sides of rectangle $$ACDB$$), we have:

$$BE = AB \tan \theta_1 + AB \cot \theta_1$$

$$\frac{4}{\sqrt{3}}AB = AB (\tan \theta_1 + \cot \theta_1)$$

$$\frac{4}{\sqrt{3}} = \frac{\sin \theta_1}{\cos \theta_1} + \frac{\cos \theta_1}{\sin \theta_1} = \frac{\sin^2 \theta_1 + \cos^2 \theta_1}{\sin \theta_1 \cos \theta_1} = \frac{1}{\sin \theta_1 \cos \theta_1}$$

$$\sin(2\theta_1) = 2 \sin \theta_1 \cos \theta_1 = 2 \left( \frac{\sqrt{3}}{4} \right) = \frac{\sqrt{3}}{2}$$

This gives two possible values for $$2\theta_1$$: $$60^\circ$$ or $$120^\circ$$.

• If $$2\theta_1 = 60^\circ \implies \theta_1 = 30^\circ$$ and $$\theta_2 = 60^\circ$$.
• If $$2\theta_1 = 120^\circ \implies \theta_1 = 60^\circ$$ and $$\theta_2 = 30^\circ$$.

The problem states $$\frac{\theta_2}{\theta_1}$$ is largest, so we choose $$\theta_1 = 30^\circ$$ and $$\theta_2 = 60^\circ$$.

Using the area of $$\triangle CAB$$:

$$\frac{1}{2} (AB)^2 \tan 30^\circ = 2\sqrt{3} - 3$$

$$\frac{1}{2} (AB)^2 \left( \frac{1}{\sqrt{3}} \right) = 2\sqrt{3} - 3$$

$$(AB)^2 = 2\sqrt{3}(2\sqrt{3} - 3) = 4(3) - 6\sqrt{3} = 12 - 6\sqrt{3}$$

Using the identity $$12 - 6\sqrt{3} = (3 - \sqrt{3})^2$$:

$$AB = 3 - \sqrt{3}$$

In $$\triangle CED$$ (a $$30^\circ-60^\circ-90^\circ$$ triangle with $$\theta_2 = 60^\circ$$):

• $$CD = AB = 3 - \sqrt{3}$$
• $$ED = CD \tan 60^\circ = (3 - \sqrt{3})\sqrt{3} = 3\sqrt{3} - 3$$
• $$CE = \frac{CD}{\cos 60^\circ} = 2(3 - \sqrt{3}) = 6 - 2\sqrt{3}$$

Perimeter $$= CD + ED + CE$$:

$$(3 - \sqrt{3}) + (3\sqrt{3} - 3) + (6 - 2\sqrt{3})$$

$$3 - 3 + 6 - \sqrt{3} + 3\sqrt{3} - 2\sqrt{3}$$

$$6 + 0\sqrt{3} = 6$$

Final Answer: 6

$$x = y = z = t$$

$$t(\sin A + \sin B + \sin C) = 18$$

$$t(\sin 2A + \sin 2B + \sin 2C) = 9$$

$$\frac{\sin 2A + \sin 2B + \sin 2C}{\sin A + \sin B + \sin C} = \frac{9}{18}$$

$$\frac{4 \sin A \sin B \sin C}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} = \frac{1}{2}$$

$$\frac{4 (2\sin\frac{A}{2}\cos\frac{A}{2}) (2\sin\frac{B}{2}\cos\frac{B}{2}) (2\sin\frac{C}{2}\cos\frac{C}{2})}{4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} = \frac{1}{2}$$

$$8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{2}$$

$$\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{16}$$

$$80 \left(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right) = 80 \left(\frac{1}{16}\right)$$

$$80 \left(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right) = 5$$

We need to find the value of $$16\sin(20°)\sin(40°)\sin(80°)$$.

A well-known product-to-sum identity states that $$\sin\theta \cdot \sin(60° - \theta) \cdot \sin(60° + \theta) = \frac{\sin 3\theta}{4}$$.

Substituting θ = 20° into this identity gives $$\sin(20°) \cdot \sin(40°) \cdot \sin(80°) = \sin(20°) \cdot \sin(60° - 20°) \cdot \sin(60° + 20°)$$.

It follows that $$= \frac{\sin(60°)}{4} = \frac{\sqrt{3}/2}{4} = \frac{\sqrt{3}}{8}$$.

Multiplying both sides by 16 yields $$16\sin(20°)\sin(40°)\sin(80°) = 16 \times \frac{\sqrt{3}}{8} = 2\sqrt{3}$$.

The answer is Option B: $$2\sqrt{3}$$.

We need to find $$n(S) + \displaystyle\sum_{\theta \in S} \sec\left(\frac{\pi}{4} + 2\theta\right)\csc\left(\frac{\pi}{4} + 2\theta\right)$$, where $$S = \{\theta \in [0, 2\pi] : 8^{2\sin^2\theta} + 8^{2\cos^2\theta} = 16\}$$.

Let $$t = 2\sin^2\theta$$, so $$2\cos^2\theta = 2 - t$$.

$$8^t + 8^{2-t} = 16$$

Let $$u = 8^t$$: $$u + \frac{64}{u} = 16$$, so $$u^2 - 16u + 64 = 0$$, giving $$(u - 8)^2 = 0$$.

Therefore $$u = 8$$, so $$8^t = 8^1$$, giving $$t = 1$$, i.e., $$\sin^2\theta = \frac{1}{2}$$.

$$\sin\theta = \pm\frac{1}{\sqrt{2}}$$, so $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$.

$$n(S) = 4$$

$$\sec\alpha \cdot \csc\alpha = \frac{1}{\sin\alpha \cos\alpha} = \frac{2}{\sin 2\alpha}$$, where $$\alpha = \frac{\pi}{4} + 2\theta$$.

For each $$\theta$$, we compute $$\sin(2\alpha) = \sin\left(\frac{\pi}{2} + 4\theta\right) = \cos(4\theta)$$:

$$\theta = \frac{\pi}{4}: \cos(\pi) = -1 \implies \text{value} = -2$$

$$\theta = \frac{3\pi}{4}: \cos(3\pi) = -1 \implies \text{value} = -2$$

$$\theta = \frac{5\pi}{4}: \cos(5\pi) = -1 \implies \text{value} = -2$$

$$\theta = \frac{7\pi}{4}: \cos(7\pi) = -1 \implies \text{value} = -2$$

Sum = $$-2 - 2 - 2 - 2 = -8$$

$$n(S) + \text{Sum} = 4 + (-8) = -4$$

Therefore, the correct answer is Option C: $$-4$$.

We need to find the number of solutions of $$ \cos x = |\sin x| $$ in the interval $$ -4\pi \le x \le 4\pi $$.

Since the RHS is $$ |\sin x| \geq 0 $$, we need $$ \cos x \geq 0 $$.

Squaring both sides (valid since both sides are non-negative when solutions exist):

$$\cos^2 x = \sin^2 x$$

$$\cos 2x = 0$$

$$2x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$

$$x = \frac{\pi}{4} + \frac{k\pi}{2}$$

The general solutions are $$ x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots $$

We need $$ \cos x \geq 0 $$, which holds when $$ x \in [-\frac{\pi}{2} + 2k\pi, \frac{\pi}{2} + 2k\pi] $$.

At $$ x = \frac{\pi}{4} + \frac{k\pi}{2} $$:

• $$ k = 0 $$: $$ x = \frac{\pi}{4} $$, $$ \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} > 0 $$ ✓
• $$ k = 1 $$: $$ x = \frac{3\pi}{4} $$, $$ \cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2} < 0 $$ ✗
• $$ k = 2 $$: $$ x = \frac{5\pi}{4} $$, $$ \cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} < 0 $$ ✗
• $$ k = 3 $$: $$ x = \frac{7\pi}{4} $$, $$ \cos\frac{7\pi}{4} = \frac{\sqrt{2}}{2} > 0 $$ ✓

So within each period of $$ 2\pi $$, we get exactly 2 valid solutions: $$ x = \frac{\pi}{4} + 2m\pi $$ and $$ x = -\frac{\pi}{4} + 2m\pi $$ for integer $$ m $$.

Solutions of the form $$ x = \frac{\pi}{4} + 2m\pi $$:

$$-4\pi \le \frac{\pi}{4} + 2m\pi \le 4\pi$$

$$-\frac{17}{8} \le m \le \frac{15}{8}$$

So $$ m = -2, -1, 0, 1 $$ — giving 4 solutions.

Solutions of the form $$ x = -\frac{\pi}{4} + 2m\pi $$:

$$-4\pi \le -\frac{\pi}{4} + 2m\pi \le 4\pi$$

$$-\frac{15}{8} \le m \le \frac{17}{8}$$

So $$ m = -1, 0, 1, 2 $$ — giving 4 solutions.

Total solutions = $$ 4 + 4 = 8 $$.

The number of solutions is 8, which corresponds to Option C.

We need to solve $$\cos\left(x + \frac{\pi}{3}\right)\cos\left(\frac{\pi}{3} - x\right) = \frac{1}{4}\cos^2 2x$$ in $$[-3\pi, 3\pi]$$. Simplify the left-hand side using the product-to-sum formula $$\cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)],$$ where $$A = x + \frac{\pi}{3}$$ and $$B = \frac{\pi}{3} - x$$, so that $$A - B = 2x,\quad A + B = \frac{2\pi}{3}$$ and hence $$\text{LHS} = \frac{1}{2}\left[\cos 2x + \cos\frac{2\pi}{3}\right] = \frac{1}{2}\left[\cos 2x - \frac{1}{2}\right].$$

The equation becomes $$\frac{1}{2}\left[\cos 2x - \frac{1}{2}\right] = \frac{1}{4}\cos^2 2x$$ and multiplying both sides by 4 gives $$2\cos 2x - 1 = \cos^2 2x.$$ Rearranging leads to $$\cos^2 2x - 2\cos 2x + 1 = 0,$$ i.e. $$(\cos 2x - 1)^2 = 0,$$ so $$\cos 2x = 1,$$ which in turn yields $$2x = 2n\pi$$ and hence $$x = n\pi$$ for integer $$n$$.

Within $$[-3\pi, 3\pi]$$ this gives $$x = -3\pi, -2\pi, -\pi, 0, \pi, 2\pi, 3\pi,$$ a total of 7 solutions. Therefore, the answer is Option D: 7.

We need to find the value of $$\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)$$.

Using the roots of unity approach:

The 7th roots of unity are the solutions of $$z^7 = 1$$, given by $$z = e^{i \cdot 2k\pi/7}$$ for $$k = 0, 1, 2, \ldots, 6$$.

The sum of all 7th roots of unity is zero:

$$\sum_{k=0}^{6} e^{i \cdot 2k\pi/7} = 0$$

Taking the real part of both sides:

$$\sum_{k=0}^{6} \cos\left(\frac{2k\pi}{7}\right) = 0$$

$$1 + \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) + \cos\left(\frac{8\pi}{7}\right) + \cos\left(\frac{10\pi}{7}\right) + \cos\left(\frac{12\pi}{7}\right) = 0$$

Using the property $$\cos(2\pi - \theta) = \cos\theta$$:

$$\cos\left(\frac{12\pi}{7}\right) = \cos\left(2\pi - \frac{2\pi}{7}\right) = \cos\left(\frac{2\pi}{7}\right)$$

$$\cos\left(\frac{10\pi}{7}\right) = \cos\left(2\pi - \frac{4\pi}{7}\right) = \cos\left(\frac{4\pi}{7}\right)$$

$$\cos\left(\frac{8\pi}{7}\right) = \cos\left(2\pi - \frac{6\pi}{7}\right) = \cos\left(\frac{6\pi}{7}\right)$$

Substituting back:

$$1 + 2\left[\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)\right] = 0$$

Let $$S = \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)$$

$$1 + 2S = 0$$

$$S = -\frac{1}{2}$$

The correct answer is Option B: $$-\dfrac{1}{2}$$.

Since consecutive angles differ by $$\dfrac{\pi}{6}$$, we use the identity $$\sec A \cdot \sec B = \frac{1}{\cos A \cos B}$$ along with $$B - A = \dfrac{\pi}{6}$$ and $$\sin\!\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$$ to obtain $$\frac{\sin(B-A)}{\cos A \cos B} = \tan B - \tan A \implies \frac{1}{\cos A \cos B} = \frac{\tan B - \tan A}{\sin(B-A)} = 2(\tan B - \tan A).$$

Let $$\theta_m = \theta + (m-1)\dfrac{\pi}{6}$$. Then the given sum becomes $$\sum_{m=1}^{9} \sec\!\left(\theta + (m-1)\frac{\pi}{6}\right)\sec\!\left(\theta + \frac{m\pi}{6}\right) = 2\sum_{m=1}^{9}\left[\tan\!\left(\theta + \frac{m\pi}{6}\right) - \tan\!\left(\theta + \frac{(m-1)\pi}{6}\right)\right],$$ which telescopes.

After cancellation, only the first and last terms remain: $$2\left[\tan\!\left(\theta + \frac{9\pi}{6}\right) - \tan\theta\right] = 2\left[\tan\!\left(\theta + \frac{3\pi}{2}\right) - \tan\theta\right].$$

Using the periodicity of tangent ($$\tan$$ has period $$\pi$$) gives $$\tan\!\left(\theta + \frac{3\pi}{2}\right) = \tan\!\left(\theta + \frac{3\pi}{2} - \pi\right) = \tan\!\left(\theta + \frac{\pi}{2}\right) = -\cot\theta.$$ Hence the sum becomes $$2(-\cot\theta - \tan\theta) = -2\left(\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta}\right) = -2 \cdot \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = \frac{-2}{\sin\theta\cos\theta}.$$

Since $$\sin 2\theta = 2\sin\theta\cos\theta$$, this simplifies to $$\frac{-4}{\sin 2\theta}.$$

Setting this equal to $$\frac{-8}{\sqrt{3}}$$ yields $$\frac{-4}{\sin 2\theta} = \frac{-8}{\sqrt{3}} \implies \sin 2\theta = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}.$$ For $$\theta \in \left(0, \dfrac{\pi}{2}\right)$$, we have $$2\theta \in (0, \pi)$$, so $$2\theta = \frac{\pi}{3} \implies \theta = \frac{\pi}{6} \quad \text{or} \quad 2\theta = \frac{2\pi}{3} \implies \theta = \frac{\pi}{3},$$ giving $$S = \left\{\dfrac{\pi}{6},\ \dfrac{\pi}{3}\right\}$$.

Finally, $$\sum_{\theta \in S} \theta = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi + 2\pi}{6} = \boxed{\frac{\pi}{2}}.$$ The answer is Option C.

We need to evaluate $$2\sin\dfrac{\pi}{22}\sin\dfrac{3\pi}{22}\sin\dfrac{5\pi}{22}\sin\dfrac{7\pi}{22}\sin\dfrac{9\pi}{22}$$.

The angles here are of the form $$(2k-1)\pi/22$$ for $$k=1,2,3,4,5$$, and since $$22=2\times11$$ these can be viewed as $$(2k-1)\pi/(2\cdot11)\,$$.

Although there is a standard product identity for a positive integer $$n$$, namely

$$\prod_{k=1}^{n}\sin\dfrac{(2k-1)\pi}{4n} \;=\;\dfrac{\sqrt{2}}{2^n},$$

it does not directly apply here because our denominator is $$2\cdot11$$ rather than $$4n$$. Instead we use the identity valid for odd $$m$$,

$$\prod_{k=0}^{\frac{m-1}{2}-1}\sin\dfrac{(2k+1)\pi}{2m} \;=\;\dfrac{1}{2^{(m-1)/2}}.$$

Setting $$m=11$$ gives $$(m-1)/2=5$$, so

$$\sin\dfrac{\pi}{22}\,\sin\dfrac{3\pi}{22}\,\sin\dfrac{5\pi}{22}\,\sin\dfrac{7\pi}{22}\,\sin\dfrac{9\pi}{22} \;=\;\dfrac{1}{2^5}\;=\;\dfrac{1}{32}.\!$$

This evaluation can also be confirmed by examining the Chebyshev polynomial $$U_{10}(x)$$, since $$U_{10}(\cos\theta)=\dfrac{\sin(11\theta)}{\sin\theta}$$ has roots at $$x=\cos\frac{k\pi}{11}$$ for $$k=1,2,\dots,10$$ and leading coefficient $$2^{10}$$. The factorization of $$U_{10}$$ yields the same product of odd‐indexed sines as $$1/32$$.

Finally, multiplying by 2 gives $$2\times\dfrac{1}{32}=\dfrac{1}{16}\,$$, so the correct answer is Option A: $$\dfrac{1}{16}$$.

We need to find which equation has $$\alpha = \sin 36°$$ as a root.

Use the key identity $$5 \times 36° = 180°$$

Since $$5 \times 36° = 180°$$, we have $$\sin(5 \times 36°) = \sin 180° = 0$$.

Let $$\theta = 36°$$. Then $$\sin 5\theta = 0$$.

Expand $$\sin 5\theta$$ using the multiple angle formula

The formula for $$\sin 5\theta$$ in terms of $$\sin\theta$$ is:

$$\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$$

To derive this, we use:

$$\sin 5\theta = \sin(3\theta + 2\theta) = \sin 3\theta \cos 2\theta + \cos 3\theta \sin 2\theta$$

Where:

$$\sin 3\theta = 3\sin\theta - 4\sin^3\theta$$

$$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$$

$$\sin 2\theta = 2\sin\theta\cos\theta$$

$$\cos 2\theta = 1 - 2\sin^2\theta$$

Substituting:

$$\sin 5\theta = (3\sin\theta - 4\sin^3\theta)(1 - 2\sin^2\theta) + (4\cos^3\theta - 3\cos\theta)(2\sin\theta\cos\theta)$$

First part: $$(3\sin\theta - 4\sin^3\theta)(1 - 2\sin^2\theta)$$

$$= 3\sin\theta - 6\sin^3\theta - 4\sin^3\theta + 8\sin^5\theta$$

$$= 3\sin\theta - 10\sin^3\theta + 8\sin^5\theta$$

Second part: $$(4\cos^3\theta - 3\cos\theta)(2\sin\theta\cos\theta)$$

$$= 2\sin\theta\cos\theta(4\cos^3\theta - 3\cos\theta)$$

$$= 2\sin\theta(4\cos^4\theta - 3\cos^2\theta)$$

Using $$\cos^2\theta = 1 - \sin^2\theta$$:

$$\cos^4\theta = (1-\sin^2\theta)^2 = 1 - 2\sin^2\theta + \sin^4\theta$$

$$= 2\sin\theta[4(1-2\sin^2\theta+\sin^4\theta) - 3(1-\sin^2\theta)]$$

$$= 2\sin\theta[4 - 8\sin^2\theta + 4\sin^4\theta - 3 + 3\sin^2\theta]$$

$$= 2\sin\theta[1 - 5\sin^2\theta + 4\sin^4\theta]$$

$$= 2\sin\theta - 10\sin^3\theta + 8\sin^5\theta$$

Adding both parts:

$$\sin 5\theta = (3\sin\theta - 10\sin^3\theta + 8\sin^5\theta) + (2\sin\theta - 10\sin^3\theta + 8\sin^5\theta)$$

$$= 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta$$

$$= \sin\theta(16\sin^4\theta - 20\sin^2\theta + 5)$$

Set $$\sin 5\theta = 0$$ and simplify

$$\sin\theta(16\sin^4\theta - 20\sin^2\theta + 5) = 0$$

Since $$\theta = 36°$$, $$\sin\theta \neq 0$$. So we divide by $$\sin\theta$$:

$$16\sin^4\theta - 20\sin^2\theta + 5 = 0$$

Write in terms of $$x = \sin 36°$$

$$16x^4 - 20x^2 + 5 = 0$$

Verify

We know $$\sin 36° = \frac{\sqrt{10-2\sqrt{5}}}{4}$$, so $$\sin^2 36° = \frac{10-2\sqrt{5}}{16}$$.

$$16x^4 = 16 \times \frac{(10-2\sqrt{5})^2}{256} = \frac{100 - 40\sqrt{5} + 20}{16} = \frac{120-40\sqrt{5}}{16}$$

$$20x^2 = \frac{200-40\sqrt{5}}{16}$$

$$16x^4 - 20x^2 + 5 = \frac{120-40\sqrt{5} - 200+40\sqrt{5}}{16} + 5 = \frac{-80}{16} + 5 = -5 + 5 = 0$$ ✓

Therefore, $$\sin 36°$$ is a root of $$16x^4 - 20x^2 + 5 = 0$$.

The correct answer is Option A: $$16x^4 - 20x^2 + 5 = 0$$.

Given: $$\cot \alpha = 1$$ with $$\pi < \alpha < \frac{3\pi}{2}$$, and $$\sec \beta = -\frac{5}{3}$$ with $$\frac{\pi}{2} < \beta < \pi$$.

Find $$\tan \alpha$$ and $$\tan \beta$$:

Since $$\cot \alpha = 1$$, we get $$\tan \alpha = 1$$.

(In the third quadrant, both sine and cosine are negative, so tangent is positive. This is consistent.)

Since $$\sec \beta = -\frac{5}{3}$$, we get $$\cos \beta = -\frac{3}{5}$$.

In the second quadrant, $$\sin \beta > 0$$, so $$\sin \beta = \frac{4}{5}$$.

$$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/5}{-3/5} = -\frac{4}{3}$$

Compute $$\tan(\alpha + \beta)$$:

$$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta} = \frac{1 + \left(-\frac{4}{3}\right)}{1 - (1)\left(-\frac{4}{3}\right)} = \frac{-\frac{1}{3}}{1 + \frac{4}{3}} = \frac{-\frac{1}{3}}{\frac{7}{3}} = -\frac{1}{7}$$

Determine the quadrant of $$(\alpha + \beta)$$:

Since $$\alpha = \frac{5\pi}{4}$$ (the angle in the third quadrant with $$\tan \alpha = 1$$) and $$\beta = \pi - \arctan\!\left(\frac{4}{3}\right)$$ (in the second quadrant), we get:

$$\alpha + \beta = \frac{5\pi}{4} + \pi - \arctan\!\left(\frac{4}{3}\right) \approx 3.927 + 2.214 \approx 6.14$$

Since $$\frac{3\pi}{2} \approx 4.71$$ and $$2\pi \approx 6.28$$, we have $$\frac{3\pi}{2} < \alpha + \beta < 2\pi$$.

This places $$\alpha + \beta$$ in the fourth quadrant, which is consistent with $$\tan(\alpha + \beta) = -\frac{1}{7} < 0$$.

The correct answer is Option A: $$-\dfrac{1}{7}$$ and IV$$^{\text{th}}$$ quadrant.

We need to solve: $$\sin\theta\tan\theta + \tan\theta = \sin 2\theta$$ for $$\theta \in [-\pi, \pi] \setminus \{\pm\frac{\pi}{2}\}$$.

Rewriting the equation:

$$ \tan\theta(\sin\theta + 1) = 2\sin\theta\cos\theta $$

$$ \frac{\sin\theta}{\cos\theta}(\sin\theta + 1) = 2\sin\theta\cos\theta $$

Case 1: $$\sin\theta = 0$$

LHS = 0, RHS = 0. This gives $$\theta = -\pi, 0, \pi$$.

Note: $$\theta = -\pi$$ and $$\theta = \pi$$ represent the same point. We include both if both are in the domain. Since the domain is $$[-\pi, \pi]$$, both $$-\pi$$ and $$\pi$$ are included.

So $$\theta \in \{-\pi, 0, \pi\}$$.

Case 2: $$\sin\theta \neq 0$$. Divide both sides by $$\sin\theta$$:

$$ \frac{\sin\theta + 1}{\cos\theta} = 2\cos\theta $$

$$ \sin\theta + 1 = 2\cos^2\theta = 2(1 - \sin^2\theta) = 2 - 2\sin^2\theta $$

$$ 2\sin^2\theta + \sin\theta - 1 = 0 $$

$$ (2\sin\theta - 1)(\sin\theta + 1) = 0 $$

$$ \sin\theta = \frac{1}{2} \quad \text{or} \quad \sin\theta = -1 $$

$$\sin\theta = -1$$ gives $$\theta = -\frac{\pi}{2}$$, which is excluded from the domain.

$$\sin\theta = \frac{1}{2}$$ gives $$\theta = \frac{\pi}{6}$$ or $$\theta = \frac{5\pi}{6}$$.

We must verify $$\cos\theta \neq 0$$ for these values (already satisfied).

So $$S = \left\{-\pi, 0, \pi, \frac{\pi}{6}, \frac{5\pi}{6}\right\}$$

$$n(S) = 5$$

Now compute $$T = \sum_{\theta \in S} \cos 2\theta$$:

$$\cos 2(-\pi) = \cos(-2\pi) = 1$$

$$\cos 2(0) = \cos 0 = 1$$

$$\cos 2(\pi) = \cos 2\pi = 1$$

$$\cos 2\left(\frac{\pi}{6}\right) = \cos\frac{\pi}{3} = \frac{1}{2}$$

$$\cos 2\left(\frac{5\pi}{6}\right) = \cos\frac{5\pi}{3} = \frac{1}{2}$$

$$ T = 1 + 1 + 1 + \frac{1}{2} + \frac{1}{2} = 4 $$

$$ T + n(S) = 4 + 5 = 9 $$

The answer is Option D: 9.

We need to find the number of elements in $$S = \left\{x \in \mathbb{R} : 2\cos\left(\dfrac{x^2+x}{6}\right) = 4^x + 4^{-x}\right\}$$.

We observe that the right-hand side can be written as $$4^x + 4^{-x} = 2^{2x} + 2^{-2x}$$. By the AM-GM inequality, $$2^{2x} + 2^{-2x} \geq 2\sqrt{2^{2x} \cdot 2^{-2x}} = 2$$. Equality holds if and only if $$2^{2x} = 2^{-2x}$$, i.e., $$x = 0$$.

Now for the left-hand side, we know that $$\cos\theta \leq 1$$ for all real $$\theta$$, so $$2\cos\left(\dfrac{x^2+x}{6}\right) \leq 2$$.

Since the LHS $$\leq 2$$ and the RHS $$\geq 2$$, equality is possible only when both sides equal 2 simultaneously. The RHS equals 2 only when $$x = 0$$. At $$x = 0$$, the LHS becomes $$2\cos\left(\dfrac{0+0}{6}\right) = 2\cos(0) = 2$$.

Since both sides equal 2 at $$x = 0$$ and no other value of $$x$$ can satisfy the equation, there is exactly 1 element in the set $$S$$.

Hence, the correct answer is Option A: $$1$$.

We have a horizontal triangular park OAB with $$AB = 16$$ and a vertical lamp post OP at point O. C is the midpoint of AB, so $$AC = CB = 8$$. We are given $$\angle PAO = \angle PBO = 15°$$ and $$\angle PCO = 45°$$.

Since OP is vertical, the angles of elevation from A, B, and C to P are the given angles. Let $$OP = h$$. Then:

$$\tan(\angle PAO) = \frac{h}{OA} \Rightarrow OA = \frac{h}{\tan 15°}$$

$$\tan(\angle PBO) = \frac{h}{OB} \Rightarrow OB = \frac{h}{\tan 15°}$$

$$\tan(\angle PCO) = \frac{h}{OC} \Rightarrow OC = \frac{h}{\tan 45°} = h$$

Since $$OA = OB$$, triangle OAB is isosceles with O on the perpendicular bisector of AB. Therefore O lies on the line through C perpendicular to AB... actually, O lies on the perpendicular bisector of AB, which passes through C (the midpoint of AB). So O, C are on the same line perpendicular to AB, meaning $$OC$$ is the distance from O to the midpoint C along this perpendicular bisector.

In triangle OAC (right-angled at C, since OC is perpendicular to AB):

$$OA^2 = OC^2 + AC^2$$

$$\frac{h^2}{\tan^2 15°} = h^2 + 64$$

$$h^2\left(\frac{1}{\tan^2 15°} - 1\right) = 64$$

$$h^2 \cdot \frac{1 - \tan^2 15°}{\tan^2 15°} = 64$$

Now $$\tan 15° = 2 - \sqrt{3}$$, so $$\tan^2 15° = (2-\sqrt{3})^2 = 7 - 4\sqrt{3}$$.

And $$1 - \tan^2 15° = 1 - 7 + 4\sqrt{3} = -6 + 4\sqrt{3}$$.

So:

$$h^2 = \frac{64(7 - 4\sqrt{3})}{4\sqrt{3} - 6} = \frac{64(7 - 4\sqrt{3})}{2(2\sqrt{3} - 3)}$$

$$= \frac{32(7 - 4\sqrt{3})}{2\sqrt{3} - 3}$$

Rationalizing by multiplying by $$\frac{2\sqrt{3}+3}{2\sqrt{3}+3}$$:

$$= \frac{32(7 - 4\sqrt{3})(2\sqrt{3}+3)}{(2\sqrt{3})^2 - 9} = \frac{32(7 - 4\sqrt{3})(2\sqrt{3}+3)}{12 - 9} = \frac{32(7 - 4\sqrt{3})(2\sqrt{3}+3)}{3}$$

Expanding the numerator factor: $$(7 - 4\sqrt{3})(2\sqrt{3}+3) = 14\sqrt{3} + 21 - 24 - 12\sqrt{3} = 2\sqrt{3} - 3$$.

So:

$$h^2 = \frac{32(2\sqrt{3} - 3)}{3} = \frac{32}{\sqrt{3}} \cdot \frac{(2\sqrt{3}-3)\sqrt{3}}{3} = \frac{32}{\sqrt{3}} \cdot \frac{6 - 3\sqrt{3}}{3} = \frac{32}{\sqrt{3}}(2 - \sqrt{3})$$

Hence, the correct answer is Option 2.

Since $$\sin 12° = \sin(30° - 18°) = \sin 30° \cos 18° - \cos 30° \sin 18°$$, it follows that $$\sin 12° = \frac{1}{2}\cos 18° - \frac{\sqrt{3}}{2}\sin 18°$$.

Substituting this into $$2\sin 12° - \sin 72°$$ and using $$\sin 72° = \cos 18°$$ gives $$2\sin 12° - \sin 72° = 2\Bigl(\frac{1}{2}\cos 18° - \frac{\sqrt{3}}{2}\sin 18°\Bigr) - \cos 18° = \cos 18° - \sqrt{3}\sin 18° - \cos 18° = -\sqrt{3}\sin 18°$$.

Since $$\sin 18° = \frac{\sqrt{5} - 1}{4}$$, this yields $$2\sin 12° - \sin 72° = -\sqrt{3} \cdot \frac{\sqrt{5} - 1}{4} = \frac{\sqrt{3}(1 - \sqrt{5})}{4}$$. Therefore, the answer is $$\frac{\sqrt{3}(1 - \sqrt{5})}{4}$$ (Option D).

Let $$AB$$ and $$PQ$$ be two vertical poles standing 160 m apart, with heights $$AB = h$$ and $$PQ = 2h$$. Point $$C$$ is the midpoint of the segment joining the feet $$B$$ and $$Q$$ of the poles, so that $$BC = CQ = 80$$ m.

Since the angle of elevation from $$C$$ to the top $$P$$ of the taller pole is $$\frac{\pi}{8}$$, and the vertical height of that pole is $$PQ = 2h$$ over a horizontal distance $$CQ = 80$$ m, we have

$$\tan\left(\frac{\pi}{8}\right) = \frac{PQ}{CQ} = \frac{2h}{80},$$

which gives

$$h = 40\tan\left(\frac{\pi}{8}\right)\quad\cdots(1)$$

To evaluate $$\tan\left(\frac{\pi}{8}\right)$$, set $$t = \tan\left(\frac{\pi}{8}\right)$$. Noting that $$\frac{\pi}{8} = \tfrac12\cdot\tfrac{\pi}{4}$$ and applying the double‐angle formula,

$$\tan\left(\frac{\pi}{4}\right) = \frac{2t}{1 - t^2} = 1,$$

so

$$1 - t^2 = 2t,\quad t^2 + 2t - 1 = 0.$$

By the quadratic formula one finds

$$t = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2},$$

and since $$t>0$$ in the first quadrant,

$$\tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1.$$

Substituting this into (1) yields

$$h = 40(\sqrt{2} - 1).$$

Now, let $$\theta$$ be the angle of elevation from $$C$$ to the top $$A$$ of the shorter pole. Since $$AB = h$$ and $$BC = 80$$ m,

$$\tan\theta = \frac{AB}{BC} = \frac{h}{80} = \frac{40(\sqrt{2} - 1)}{80} = \frac{\sqrt{2} - 1}{2}.$$

Therefore,

$$\tan^2\theta = \left(\frac{\sqrt{2} - 1}{2}\right)^2 = \frac{(\sqrt{2} - 1)^2}{4},$$

and since $$(\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2},$$

we get

$$\tan^2\theta = \frac{3 - 2\sqrt{2}}{4}.$$

Hence the correct answer is Option C: $$\dfrac{3 - 2\sqrt{2}}{4}$$.

A tower $$ PQ $$ stands on the ground with base $$ Q $$. Point $$ R $$ divides the tower such that $$ QR = 15 $$ m. From point $$ A $$ on the ground, the angle of elevation of $$ R $$ is $$ 60° $$ and the part $$ PR $$ subtends an angle of $$ 15° $$ at $$ A $$.

Let the height of the tower be $$ h = PQ $$, so $$ PR = h - 15 $$.

Let $$ d = AQ $$ (horizontal distance from A to the base).

The angle of elevation of $$ R $$ from $$ A $$ is $$ 60° $$, so:

$$\tan 60° = \frac{QR}{d} = \frac{15}{d}$$

$$d = \frac{15}{\tan 60°} = \frac{15}{\sqrt{3}} = 5\sqrt{3}$$

The angle of elevation of $$ P $$ from $$ A $$ is $$ 60° + 15° = 75° $$.

$$\tan 75° = \frac{h}{d} = \frac{h}{5\sqrt{3}}$$

$$\tan 75° = \tan(45° + 30°) = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \cdot \tan 30°} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$

Rationalizing:

$$\tan 75° = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

$$h = 5\sqrt{3} \cdot \tan 75° = 5\sqrt{3}(2 + \sqrt{3}) = 10\sqrt{3} + 5 \cdot 3 = 10\sqrt{3} + 15$$

$$h = 5(2\sqrt{3} + 3) \text{ m}$$

The height of the tower is $$ 5(2\sqrt{3} + 3) $$ m, which corresponds to Option A.

We need to find $$\frac{R}{r}$$ for a triangle whose sides satisfy $$\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9}$$.

Let the common value be $$k$$, so that $$\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9} = k$$. It follows that
$$a + b = 7k,\quad b + c = 8k,\quad c + a = 9k$$ and adding these gives $$2(a + b + c) = 24k\implies a + b + c = 12k$$. Hence $$c = 12k - 7k = 5k$$, $$a = 12k - 8k = 4k$$, and $$b = 12k - 9k = 3k$$.

Since the sides are $$3k, 4k, 5k$$ and $$3^2 + 4^2 = 5^2$$, the triangle is right-angled with hypotenuse $$c = 5k$$.

In a right triangle, the circumradius is half the hypotenuse, so $$R = \frac{c}{2} = \frac{5k}{2}$$.

Moreover, the semi-perimeter is $$s = \frac{a + b + c}{2} = \frac{12k}{2} = 6k$$ and the area is $$\Delta = \frac{1}{2}\times a \times b = \frac{1}{2}\times 3k \times 4k = 6k^2$$ (using the two legs). Therefore, $$r = \frac{\Delta}{s} = \frac{6k^2}{6k} = k$$.

It follows that $$\frac{R}{r} = \frac{5k/2}{k} = \frac{5}{2}$$.

The correct answer is Option C: $$\dfrac{5}{2}$$.

Let the base of the tower be at point $$T$$ and its height be $$h$$. Point $$A$$ is due north of the tower, and point $$B$$ is 9 units due west of $$A$$.

We place $$T$$ at the origin. Since $$A$$ is due north of $$T$$, let the distance $$TA = d$$. Since $$B$$ is 9 units due west of $$A$$, the distance $$TB = \sqrt{d^2 + 81}$$.

We are given that the distance from $$B$$ to the tower (i.e., $$TB$$) is 15 units. So $$\sqrt{d^2 + 81} = 15$$, giving $$d^2 + 81 = 225$$, hence $$d^2 = 144$$ and $$d = 12$$.

Now, the angle of elevation from $$B$$ is $$\beta = \cos^{-1}\!\left(\dfrac{3}{\sqrt{13}}\right)$$. We have $$\cos\beta = \dfrac{3}{\sqrt{13}}$$, so $$\sin\beta = \dfrac{2}{\sqrt{13}}$$ and $$\tan\beta = \dfrac{2}{3}$$.

Since $$\tan\beta = \dfrac{h}{TB} = \dfrac{h}{15}$$, we get $$h = 15 \times \dfrac{2}{3} = 10$$.

The angle of elevation from $$A$$ is $$\alpha$$, with $$\tan\alpha = \dfrac{h}{d} = \dfrac{10}{12} = \dfrac{5}{6}$$.

Hence $$\cot\alpha = \dfrac{1}{\tan\alpha} = \dfrac{6}{5}$$.

Hence, the correct answer is Option A.

Let Q be at the origin with the horizontal ground along the x-axis. The tower PQ has height 10, so P = (0, 10) and Q = (0, 0).

Point A is on the ground with the angle of elevation of P being 45°:

$$\tan 45° = \frac{10}{AQ} \implies AQ = 10$$

So A = (10, 0).

Point R lies on segment AQ, so R = (x, 0) for some 0 ≤ x ≤ 10, and B is vertically above R, so B = (x, h) for some height h. Given ∠BAQ = 30° and AR = 10 − x,

$$\tan 30° = \frac{h}{10 - x} \implies h = \frac{10 - x}{\sqrt{3}}$$

From B = (x, h), the angle of elevation of P = (0, 10) is 60°:

$$\tan 60° = \frac{10 - h}{x}$$

$$\sqrt{3} = \frac{10 - \frac{10-x}{\sqrt{3}}}{x} = \frac{10\sqrt{3} - 10 + x}{\sqrt{3} \cdot x}$$

$$3x = 10\sqrt{3} - 10 + x \implies 2x = 10(\sqrt{3} - 1) \implies x = 5(\sqrt{3} - 1)$$

Hence

$$AR = 10 - x = 10 - 5(\sqrt{3}-1) = 15 - 5\sqrt{3} = 5(3-\sqrt{3})$$

$$h = BR = \frac{AR}{\sqrt{3}} = \frac{5(3-\sqrt{3})}{\sqrt{3}} = 5(\sqrt{3}-1)$$

The distance AB is

$$d = AB = \sqrt{AR^2 + BR^2} = \sqrt{AR^2\left(1 + \frac{1}{3}\right)} = AR \cdot \frac{2}{\sqrt{3}} = \frac{2 \times 5(3-\sqrt{3})}{\sqrt{3}} = \frac{10(3-\sqrt{3})}{\sqrt{3}} = 10(\sqrt{3} - 1)$$

PQ and BR are vertical (parallel), with PQ = 10 and BR = 5($$\sqrt{3}$$-1). The horizontal distance between them is QR = x = 5($$\sqrt{3}$$-1). Therefore, the area of the trapezoid is

$$\alpha = \frac{1}{2}(PQ + BR) \times QR = \frac{1}{2}(10 + 5(\sqrt{3}-1)) \times 5(\sqrt{3}-1) = \frac{1}{2} \times 5(1+\sqrt{3}) \times 5(\sqrt{3}-1) = \frac{25}{2}(\sqrt{3}+1)(\sqrt{3}-1) = \frac{25}{2}(3-1) = 25$$

Therefore,

$$(d, \alpha) = \boxed{(10(\sqrt{3}-1),\ 25)}.$$

The answer is Option A.

Let the pole be at point A (base) with height 20 m (top at point P) and the tower be at point B (base) with height $$h$$ (top at point T). From the base of the pole (A), the angle of elevation to the top of the tower (T) is 60°. Let the horizontal distance between the pole and tower be $$d$$. Then $$\tan 60° = \frac{h}{d} \Rightarrow h = d\sqrt{3}$$.

The pole subtends an angle of 30° at the top of the tower T. From T, looking at the pole which extends from A (ground level) to P (height 20 m), the angle subtended is $$\angle ATP = 30°$$. Let $$\alpha$$ be the angle of depression from T to A and $$\beta$$ be the angle of depression from T to P, so that $$\tan \alpha = \frac{h}{d}$$ and $$\tan \beta = \frac{h - 20}{d}$$. The angle subtended by the pole at T is $$\alpha - \beta = 30°$$.

By the tangent subtraction formula, $$\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} = \tan 30° = \frac{1}{\sqrt{3}}$$, which gives $$\frac{\frac{h}{d} - \frac{h-20}{d}}{1 + \frac{h}{d}\cdot\frac{h-20}{d}} = \frac{1}{\sqrt{3}}$$, then $$\frac{\frac{20}{d}}{1 + \frac{h(h-20)}{d^2}} = \frac{1}{\sqrt{3}}$$ and hence $$\frac{20d}{d^2 + h(h-20)} = \frac{1}{\sqrt{3}}$$.

Substituting $$h = d\sqrt{3}$$ into the denominator yields $$d^2 + h(h-20) = d^2 + d\sqrt{3}(d\sqrt{3} - 20) = d^2 + 3d^2 - 20d\sqrt{3} = 4d^2 - 20d\sqrt{3}$$, so that $$\frac{20d}{4d^2 - 20d\sqrt{3}} = \frac{1}{\sqrt{3}}$$. Simplifying gives $$\frac{20}{4d - 20\sqrt{3}} = \frac{1}{\sqrt{3}}$$, hence $$20\sqrt{3} = 4d - 20\sqrt{3}$$, $$4d = 40\sqrt{3}$$ and $$d = 10\sqrt{3}$$.

Finally, $$h = d\sqrt{3} = 10\sqrt{3} \times \sqrt{3} = 30$$. Therefore, the height of the tower is 30 meters.

Given,

$$S=\{z\in\mathbb C:\bar z^2+\bar z=0\}$$

We have

$$\bar z^2+\bar z=0$$

$$\bar z(\bar z+1)=0$$

Hence,

$$\bar z=0\quad \text{or}\quad \bar z=-1$$

Taking conjugate on both sides,

$$z=0\quad \text{or}\quad z=-1$$

Thus,

$$S=\{0,-1\}$$

Now compute

$$\sum_{z\in S}(\operatorname{Re}(z)+\operatorname{Im}(z))$$

For

$$z=0,$$

$$\operatorname{Re}(0)+\operatorname{Im}(0)=0$$

For

$$z=-1,$$

$$\operatorname{Re}(-1)+\operatorname{Im}(-1)=-1+0=-1$$

Therefore,

$$0+(-1)=-1$$

Hence, the required value is

$$\boxed{-1}$$.

We need to find the number of elements in $$S = \{\theta \in [-4\pi, 4\pi] : 3\cos^2 2\theta + 6\cos 2\theta - 10\cos^2\theta + 5 = 0\}$$. Using $$\cos 2\theta = 2\cos^2\theta - 1$$ gives $$\cos^2\theta = \frac{\cos 2\theta + 1}{2}$$, and substituting yields $$3\cos^2 2\theta + 6\cos 2\theta - 10 \cdot \frac{\cos 2\theta + 1}{2} + 5 = 0$$ which simplifies to $$3\cos^2 2\theta + \cos 2\theta = 0$$ and thus $$\cos 2\theta(3\cos 2\theta + 1) = 0$$.

From $$\cos 2\theta = 0$$ we obtain $$2\theta = \frac{\pi}{2} + n\pi$$ and hence $$\theta = \frac{\pi}{4} + \frac{n\pi}{2}$$ for integer $$n$$. Enforcing $$-4\pi \leq \theta \leq 4\pi$$ gives $$-\frac{17}{2} \leq n \leq \frac{15}{2}$$, so $$n = -8, -7, \ldots, 7$$, yielding 16 solutions.

Solving $$\cos 2\theta = -\frac{1}{3}$$ leads to $$2\theta = \pm \cos^{-1}\left(-\frac{1}{3}\right) + 2k\pi$$ and hence $$\theta = \pm \frac{1}{2}\cos^{-1}\left(-\frac{1}{3}\right) + k\pi$$. Let $$\phi = \frac{1}{2}\cos^{-1}\left(-\frac{1}{3}\right)$$, so the two solution families are $$\theta = \phi + k\pi$$ and $$\theta = -\phi + k\pi$$. Each has period $$\pi$$, giving 8 solutions per family on $$[-4\pi,4\pi]$$, for a total of 16. Since $$\cos^{-1}(-1/3)$$ is not a rational multiple of $$\pi$$, there is no overlap with the previous solutions.

Therefore, $$|S| = 16 + 16 = 32$$. The answer is $$\boxed{32}$$.

We need to find the sum of common solutions of $$2\sin^2\theta - \cos 2\theta = 0$$ and $$2\cos^2\theta + 3\sin\theta = 0$$ in $$[0, 2\pi]$$.

$$2\sin^2\theta - \cos 2\theta = 0$$

Using $$\cos 2\theta = 1 - 2\sin^2\theta$$:

$$2\sin^2\theta - (1 - 2\sin^2\theta) = 0$$

$$4\sin^2\theta - 1 = 0$$

$$\sin^2\theta = \dfrac{1}{4}$$

$$\sin\theta = \pm\dfrac{1}{2}$$

Solutions in $$[0, 2\pi]$$: $$\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$$

$$2\cos^2\theta + 3\sin\theta = 0$$

Using $$\cos^2\theta = 1 - \sin^2\theta$$:

$$2(1 - \sin^2\theta) + 3\sin\theta = 0$$

$$2 - 2\sin^2\theta + 3\sin\theta = 0$$

$$2\sin^2\theta - 3\sin\theta - 2 = 0$$

$$(2\sin\theta + 1)(\sin\theta - 2) = 0$$

Since $$\sin\theta \neq 2$$: $$\sin\theta = -\dfrac{1}{2}$$

Solutions in $$[0, 2\pi]$$: $$\theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$$

Common solutions: $$\theta = \dfrac{7\pi}{6}$$ and $$\theta = \dfrac{11\pi}{6}$$

$$\dfrac{7\pi}{6} + \dfrac{11\pi}{6} = \dfrac{18\pi}{6} = 3\pi$$

So $$k\pi = 3\pi$$, giving $$k = 3$$.

The correct answer is $$\boxed{3}$$.

Given,

$$\sin x=\cos^2 x$$

Using

$$\cos^2 x=1-\sin^2 x$$

we get

$$\sin x=1-\sin^2 x$$

$$\sin^2 x+\sin x-1=0$$

Let

$$\sin x=t$$

Then,

$$t^2+t-1=0$$

$$t=\frac{-1\pm\sqrt5}{2}$$

Since

$$-1\le\sin x\le1$$

only

$$\sin x=\frac{\sqrt5-1}{2}$$

is valid.

Now,

$$0<\frac{\sqrt5-1}{2}<1$$

Hence, in each interval of length $$2\pi$$, there are two solutions.

Since

$$10<4\pi$$

the interval $$(0,10)$$ contains one complete interval $$[0,2\pi]$$ and part of the next interval.

The solutions in $$(0,2\pi)$$ are two.

Also, in $$(2\pi,10)$$, two more solutions occur because

$$2\pi+\sin^{-1}\left(\frac{\sqrt5-1}{2}\right)<10$$

and

$$3\pi-\sin^{-1}\left(\frac{\sqrt5-1}{2}\right)<10$$

Therefore, the total number of solutions is

$$\boxed{4}$$

We first solve $$7\cos^2\theta - 3\sin^2\theta - 2\cos^2(2\theta) = 2$$ for $$\theta \in (0, 2\pi)$$.

We write $$\sin^2\theta = 1 - \cos^2\theta$$ and $$\cos^2(2\theta) = (2\cos^2\theta - 1)^2$$. Substituting:

$$7\cos^2\theta - 3(1-\cos^2\theta) - 2(2\cos^2\theta-1)^2 = 2$$

$$10\cos^2\theta - 3 - 2(4\cos^4\theta - 4\cos^2\theta + 1) = 2$$

$$10\cos^2\theta - 3 - 8\cos^4\theta + 8\cos^2\theta - 2 = 2$$

$$-8\cos^4\theta + 18\cos^2\theta - 7 = 0$$

$$8\cos^4\theta - 18\cos^2\theta + 7 = 0$$

Let $$u = \cos^2\theta$$. Then $$8u^2 - 18u + 7 = 0$$. By the quadratic formula:

$$u = \frac{18 \pm \sqrt{324 - 224}}{16} = \frac{18 \pm \sqrt{100}}{16} = \frac{18 \pm 10}{16}$$

So $$u = \frac{28}{16} = \frac{7}{4}$$ or $$u = \frac{8}{16} = \frac{1}{2}$$. Since $$\cos^2\theta \leq 1$$, we discard $$u = 7/4$$ and take $$\cos^2\theta = \frac{1}{2}$$, giving $$\cos\theta = \pm\frac{1}{\sqrt{2}}$$.

The solutions in $$(0, 2\pi)$$ are $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$. So the set $$S$$ contains these four values.

For each $$\theta \in S$$, $$\cos^2\theta = \frac{1}{2}$$ and $$\sin^2\theta = \frac{1}{2}$$. Also $$\tan^2\theta = 1$$ and $$\cot^2\theta = 1$$, so $$\tan^2\theta + \cot^2\theta = 2$$.

The quadratic equation becomes $$x^2 - 2(2)x + 6\cdot\frac{1}{2} = 0$$, i.e., $$x^2 - 4x + 3 = 0$$, which factors as $$(x-1)(x-3) = 0$$. The roots are $$x = 1$$ and $$x = 3$$, with sum $$1 + 3 = 4$$.

Since all four values of $$\theta$$ give the same equation (because they all have the same $$\tan^2\theta + \cot^2\theta$$ and $$\sin^2\theta$$), the sum of roots of ALL equations is $$4 \times 4 = 16$$.

Hence, the correct answer is 16.

Given,

$$14\csc^2x-2\sin^2x=21-4\cos^2x$$

Using

$$\csc^2x=\frac1{\sin^2x},\qquad \cos^2x=1-\sin^2x$$

we get

$$\frac{14}{\sin^2x}-2\sin^2x=21-4(1-\sin^2x)=17+4\sin^2x$$

$$\frac{14}{\sin^2x}=17+6\sin^2x$$

Let

$$t=\sin^2x$$

Then,

$$14=17t+6t^2$$

$$6t^2+17t-14=0$$

Factoring,

$$6t^2+21t-4t-14=0$$

$$(3t-2)(2t+7)=0$$

Thus,

$$t=\frac23\quad \text{or}\quad t=-\frac72$$

Since

$$\sin^2x\ge0$$

we reject

$$t=-\frac72$$

Hence,

$$\sin^2x=\frac23$$

$$\sin x=\pm\sqrt{\frac23}$$

Now,

$$\sqrt{\frac23}>\frac1{\sqrt2}$$

so, there are two solutions for

$$\sin x=\sqrt{\frac23}$$

in

$$\left[\frac\pi4,\frac{7\pi}4\right]$$

Also,

$$-\sqrt{\frac23}<-\frac1{\sqrt2}$$

so, there are two solutions for

$$\sin x=-\sqrt{\frac23}$$

in the same interval.

Therefore, the total number of solutions is $$\boxed{4}$$.

We need the number of elements in $$A = \{\theta \in S : \tan\theta(1 + \sqrt{5}\tan 2\theta) = \sqrt{5} - \tan 2\theta\}$$ where $$S = \left[-\pi, \frac{\pi}{2}\right) \setminus \left\{-\frac{\pi}{2}, -\frac{\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4}\right\}$$.

We rearrange the given equation. Starting from $$\tan\theta + \sqrt{5}\tan\theta\tan 2\theta = \sqrt{5} - \tan 2\theta$$, we move the $$\tan 2\theta$$ term to the left:

$$\tan\theta + \tan 2\theta = \sqrt{5} - \sqrt{5}\tan\theta\tan 2\theta = \sqrt{5}(1 - \tan\theta\tan 2\theta)$$

Dividing both sides by $$(1 - \tan\theta\tan 2\theta)$$ (which must be nonzero for the expression to be valid):

$$\frac{\tan\theta + \tan 2\theta}{1 - \tan\theta\tan 2\theta} = \sqrt{5}$$

The left side is exactly the tangent addition formula $$\tan(\theta + 2\theta) = \tan 3\theta$$. So we need $$\tan 3\theta = \sqrt{5}$$.

Let $$\alpha = \arctan(\sqrt{5})$$, where $$\alpha \in (0, \pi/2)$$ since $$\sqrt{5} > 0$$. The general solution of $$\tan 3\theta = \sqrt{5}$$ is $$3\theta = n\pi + \alpha$$ for integer $$n$$, giving:

$$\theta = \frac{n\pi + \alpha}{3}$$

We need $$\theta \in [-\pi, \pi/2)$$, so $$-\pi \leq \frac{n\pi + \alpha}{3} < \frac{\pi}{2}$$, which gives $$-3\pi \leq n\pi + \alpha < \frac{3\pi}{2}$$.

Since $$0 < \alpha < \pi/2$$: for the left inequality, $$n\pi \geq -3\pi - \alpha > -3\pi - \pi/2 = -7\pi/2$$, so $$n \geq -3$$ (since $$n = -3$$ gives $$-3\pi + \alpha > -3\pi$$, which satisfies $$\geq -3\pi$$). For the right inequality, $$n\pi < 3\pi/2 - \alpha < 3\pi/2$$, so $$n < 3/2$$, meaning $$n \leq 1$$.

This gives $$n \in \{-3, -2, -1, 0, 1\}$$, yielding 5 candidate solutions.

Now we verify none fall in the excluded set $$\{-\pi/2, -\pi/4, -3\pi/4, \pi/4\}$$. For each excluded value $$\theta_0$$, we check $$\tan 3\theta_0$$:

At $$\theta_0 = -\pi/2$$: $$3\theta_0 = -3\pi/2$$, where $$\tan$$ is undefined (not $$\sqrt{5}$$).

At $$\theta_0 = -\pi/4$$: $$\tan(-3\pi/4) = \tan(\pi/4) = 1 \neq \sqrt{5}$$.

At $$\theta_0 = -3\pi/4$$: $$\tan(-9\pi/4) = \tan(-\pi/4) = -1 \neq \sqrt{5}$$.

At $$\theta_0 = \pi/4$$: $$\tan(3\pi/4) = -1 \neq \sqrt{5}$$.

Since none of the excluded points satisfy our equation, no solutions are lost.

We also verify that $$\tan\theta$$ and $$\tan 2\theta$$ are defined at each solution. Since $$\tan 3\theta = \sqrt{5}$$ is finite, $$3\theta \neq \pi/2 + k\pi$$. For $$\tan\theta$$ to be undefined, we need $$\theta = \pi/2 + k\pi$$, which would make $$3\theta = 3\pi/2 + 3k\pi$$, but $$\tan(3\pi/2 + 3k\pi)$$ is undefined, not $$\sqrt{5}$$. Similarly for $$\tan 2\theta$$: if $$2\theta = \pi/2 + k\pi$$, then $$\theta = \pi/4 + k\pi/2$$, giving $$3\theta = 3\pi/4 + 3k\pi/2$$, and $$\tan(3\pi/4) = -1 \neq \sqrt{5}$$. So all 5 solutions have both $$\tan\theta$$ and $$\tan 2\theta$$ well-defined.

Hence, the correct answer is 5.

We are given that $$\log_{10} \sin x + \log_{10} \cos x = -1$$ for $$x \in \left(0, \frac{\pi}{2}\right)$$. This means $$\log_{10}(\sin x \cos x) = -1$$, so $$\sin x \cos x = 10^{-1} = \frac{1}{10}$$.

We know that $$\sin x \cos x = \frac{\sin 2x}{2}$$, so $$\frac{\sin 2x}{2} = \frac{1}{10}$$, giving $$\sin 2x = \frac{1}{5}$$.

Now we need to find $$(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2 \times \frac{1}{10} = 1 + \frac{1}{5} = \frac{6}{5}$$.

Since $$x \in \left(0, \frac{\pi}{2}\right)$$, both $$\sin x$$ and $$\cos x$$ are positive, so $$\sin x + \cos x = \sqrt{\frac{6}{5}}$$.

Now we use the second equation: $$\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$$.

The left side is $$\log_{10}\sqrt{\frac{6}{5}} = \frac{1}{2}\log_{10}\frac{6}{5}$$.

So $$\frac{1}{2}\log_{10}\frac{6}{5} = \frac{1}{2}(\log_{10} n - 1)$$.

Multiplying both sides by 2: $$\log_{10}\frac{6}{5} = \log_{10} n - 1$$.

This gives $$\log_{10} n = 1 + \log_{10}\frac{6}{5} = \log_{10} 10 + \log_{10}\frac{6}{5} = \log_{10}\left(10 \times \frac{6}{5}\right) = \log_{10} 12$$.

Therefore $$n = 12$$, and the correct answer is option (2).

We start with the given transcendental equation

$$32^{\tan^2x}+32^{\sec^2x}=81,\qquad 0\le x\le\frac{\pi}{4}.$$

First we recall the Pythagorean identity for tangent and secant:

$$\sec^2x\,=\,1+\tan^2x.$$

Let us put

$$t=\tan^2x.$$

Because $$0\le x\le\frac{\pi}{4}$$, we have $$0\le \tan x\le1$$; hence

$$0\le t\le1.$$

Using $$\sec^2x=1+t$$ in the original equation gives

$$32^{t}+32^{\,1+t}=81.$$

By the property of exponents $$a^{m+n}=a^{m}\,a^{n}$$, we write

$$32^{\,1+t}=32^{1}\cdot32^{t}=32\;32^{t}.$$

Substituting this result back, the equation becomes

$$32^{t}+32\;32^{t}=81.$$

Taking $$32^{t}$$ common we obtain

$$32^{t}\bigl(1+32\bigr)=81.$$

Since $$1+32=33$$, we have

$$32^{t}\,.\,33=81.$$

Dividing both sides by $$33$$:

$$32^{t}=\frac{81}{33}=\frac{27}{11}.$$

Now $$32=2^{5}$$. Hence

$$32^{t}=(2^{5})^{t}=2^{5t}=\frac{27}{11}.$$

Taking base-$$2$$ logarithms on both sides (formula: if $$a^b=c$$, then $$b=\log_{a}c$$):

$$5t=\log_{2}\!\Bigl(\tfrac{27}{11}\Bigr).$$

Therefore

$$t=\frac{1}{5}\,\log_{2}\!\Bigl(\tfrac{27}{11}\Bigr).$$

Because $$\tfrac{27}{11}\approx2.4545$$ and $$\log_{2}(2.4545)\approx1.296$$, we get a numerical value

$$t\approx\frac{1.296}{5}\approx0.259.$$

Remember $$t=\tan^{2}x$$. So

$$\tan^{2}x\approx0.259.$$

Taking the positive square root (only the positive root is relevant here since $$\tan x\ge0$$ on the interval $$0\le x\le\frac{\pi}{4}$$):

$$\tan x\approx\sqrt{0.259}\approx0.509.$$

Because the tangent function is strictly increasing on $$\bigl[0,\tfrac{\pi}{4}\bigr]$$, the equation $$\tan x\approx0.509$$ possesses exactly one solution in that interval. Consequently the original equation has exactly one admissible $$x$$ between $$0$$ and $$\tfrac{\pi}{4}$$.

Hence, the correct answer is Option C.

We are given $$15\sin^4\alpha + 10\cos^4\alpha = 6$$. Let $$\sin^2\alpha = s$$ and $$\cos^2\alpha = 1 - s$$, so the equation becomes $$15s^2 + 10(1-s)^2 = 6$$, which expands to $$15s^2 + 10 - 20s + 10s^2 = 6$$, giving $$25s^2 - 20s + 4 = 0$$. This factors as $$(5s - 2)^2 = 0$$, so $$s = \frac{2}{5}$$, meaning $$\sin^2\alpha = \frac{2}{5}$$ and $$\cos^2\alpha = \frac{3}{5}$$.

Now we compute $$27\sec^6\alpha + 8\csc^6\alpha = \frac{27}{\cos^6\alpha} + \frac{8}{\sin^6\alpha} = \frac{27}{(3/5)^3} + \frac{8}{(2/5)^3} = \frac{27 \cdot 125}{27} + \frac{8 \cdot 125}{8} = 125 + 125 = 250$$.

We are given that $$e^{(\cos^2 x + \cos^4 x + \cos^6 x + \ldots \infty) \ln 2}$$ satisfies the equation $$t^2 - 9t + 8 = 0$$.

The infinite geometric series $$\cos^2 x + \cos^4 x + \cos^6 x + \ldots$$ has first term $$\cos^2 x$$ and common ratio $$\cos^2 x$$. Its sum is $$\frac{\cos^2 x}{1 - \cos^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x$$.

So the expression becomes $$e^{\cot^2 x \cdot \ln 2} = 2^{\cot^2 x}$$.

Now, $$t^2 - 9t + 8 = 0$$ gives $$(t - 1)(t - 8) = 0$$, so $$t = 1$$ or $$t = 8$$.

If $$2^{\cot^2 x} = 1$$, then $$\cot^2 x = 0$$, so $$x = \frac{\pi}{2}$$.

If $$2^{\cot^2 x} = 8 = 2^3$$, then $$\cot^2 x = 3$$, so $$\cot x = \sqrt{3}$$ (since $$0 < x < \frac{\pi}{2}$$), giving $$x = \frac{\pi}{6}$$.

Now we evaluate $$\frac{2\sin x}{\sin x + \sqrt{3}\cos x}$$ at $$x = \frac{\pi}{6}$$:

$$\frac{2 \cdot \frac{1}{2}}{\frac{1}{2} + \sqrt{3} \cdot \frac{\sqrt{3}}{2}} = \frac{1}{\frac{1}{2} + \frac{3}{2}} = \frac{1}{2}$$.

At $$x = \frac{\pi}{2}$$: $$\frac{2 \cdot 1}{1 + 0} = 2$$, which is not among the options.

Hence, the correct answer is Option B.

We have that $$\tan\left(\frac{\pi}{9}\right),\,x,\,\tan\left(\frac{7\pi}{18}\right)$$ form an arithmetic progression, so by the definition of an A.P. the middle term is the average of the other two. Hence

$$2x=\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{7\pi}{18}\right).$$

Dividing by 2 we obtain

$$x=\frac{\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{7\pi}{18}\right)}{2}.$$

In the same way, the numbers $$\tan\left(\frac{\pi}{9}\right),\,y,\,\tan\left(\frac{5\pi}{18}\right)$$ are in arithmetic progression, so

$$2y=\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{5\pi}{18}\right)$$

and therefore

$$y=\frac{\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{5\pi}{18}\right)}{2}.$$

To handle these expressions we set

$$t=\tan\left(\frac{\pi}{9}\right)=\tan 20^{\circ}.$$

The angle $$\frac{7\pi}{18}=70^{\circ}$$ is the complement of $$20^{\circ}$$, so

$$\tan\left(\frac{7\pi}{18}\right)=\tan 70^{\circ}=\cot 20^{\circ}=\frac1{\tan 20^{\circ}}=\frac1t.$$

Thus

$$x=\frac{t+\dfrac1t}{2}.$$

Next we need $$\tan\left(\frac{5\pi}{18}\right)=\tan 50^{\circ}.$$ Let us denote

$$p=\tan 50^{\circ}.$$

Because $$70^{\circ}=50^{\circ}+20^{\circ},$$ we can apply the tangent addition formula

$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\,\tan B}.$$

Taking $$A=50^{\circ}$$ and $$B=20^{\circ},$$ we have

$$\tan 70^{\circ}=\frac{\tan 50^{\circ}+\tan 20^{\circ}}{1-\tan 50^{\circ}\tan 20^{\circ}}.$$

Substituting the symbols just introduced, this becomes

$$\frac1t=\frac{p+t}{1-pt}.$$

Cross-multiplying gives

$$t(p+t)=1-pt.$$

Expanding and bringing all terms to one side we get

$$tp+t^{2}=1-pt,$$

$$tp+pt+t^{2}=1,$$

$$2pt+t^{2}=1.$$

Solving for $$p$$ we find

$$p=\frac{1-t^{2}}{2t}.$$

Now we can write $$y$$ solely in terms of $$t$$:

$$y=\frac{t+p}{2}=\frac{t+\dfrac{1-t^{2}}{2t}}{2}.$$

It is more convenient to work directly with $$2y$$, because $$|x-2y|$$ appears in the question. We have from earlier

$$2y=t+p=t+\frac{1-t^{2}}{2t}.$$

Next we compute the required difference. Starting from the expression for $$x$$ we write

$$x-2y=\frac{t+\dfrac1t}{2}-\left(t+\frac{1-t^{2}}{2t}\right).$$

To combine terms we bring everything to a common denominator $$2t$$:

$$x-2y=\frac{t^{2}+1}{2t}-\frac{2t^{2}+1-t^{2}}{2t}.$$

Simplifying the numerator of the second fraction gives

$$2t^{2}+1-t^{2}=t^{2}+1,$$

so that

$$x-2y=\frac{t^{2}+1}{2t}-\frac{t^{2}+1}{2t}=0.$$

Hence

$$|x-2y|=|0|=0.$$

Hence, the correct answer is Option C.

Let us denote $$\theta = 18^\circ$$ and write the number we are interested in as $$x = \csc 18^\circ.$$ Our task is to find an equation with integer coefficients that has this value of $$x$$ as a root.

Because trigonometric identities for $$18^\circ$$ are most easily obtained through $$36^\circ$$, we begin by finding $$\cos 36^\circ.$$ Observe that $$5 \times 36^\circ = 180^\circ,$$ so $$\cos 5(36^\circ) = \cos 180^\circ = -1.$$

The standard multiple-angle formula for cosine is

$$\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta.$$

Putting $$\theta = 36^\circ$$ and writing $$c = \cos 36^\circ,$$ we obtain

$$16c^5 - 20c^3 + 5c + 1 = 0.$$

This quintic has a quadratic factor which can be found by inspection (or long division); that factor is $$4c^2 - 2c - 1.$$ Hence

$$4c^2 - 2c - 1 = 0.$$

Solving the quadratic with the quadratic formula,

$$c = \dfrac{2 \pm \sqrt{4 + 16}}{8} = \dfrac{2 \pm 4\sqrt5}{8} = \dfrac{1 \pm \sqrt5}{4}.$$

Since $$\cos 36^\circ$$ is positive, we choose the positive root:

$$\cos 36^\circ = \dfrac{1 + \sqrt5}{4}.$$

Now we convert this to $$\sin 18^\circ$$ using the half-angle relation

$$\sin^2 18^\circ = \dfrac{1 - \cos 36^\circ}{2}.$$

Substituting the value of $$\cos 36^\circ,$$ we get

$$\sin^2 18^\circ = \dfrac{1 - \dfrac{1 + \sqrt5}{4}}{2} = \dfrac{\dfrac{4 - 1 - \sqrt5}{4}}{2} = \dfrac{3 - \sqrt5}{8}.$$

Taking the positive square root (because $$\sin 18^\circ$$ is positive),

$$\sin 18^\circ = \sqrt{\dfrac{3 - \sqrt5}{8}}.$$

A more familiar exact form arises by rationalising:

$$\sin 18^\circ = \dfrac{\sqrt5 - 1}{4}.$$

We can check this quickly, because

$$\left(\dfrac{\sqrt5 - 1}{4}\right)^2 = \dfrac{5 + 1 - 2\sqrt5}{16} = \dfrac{6 - 2\sqrt5}{16} = \dfrac{3 - \sqrt5}{8},$$

matching the value found earlier. Now we are ready to obtain $$\csc 18^\circ.$$ Since $$\csc 18^\circ = \dfrac{1}{\sin 18^\circ},$$ we have

$$x = \csc 18^\circ = \dfrac{1}{\dfrac{\sqrt5 - 1}{4}} = \dfrac{4}{\sqrt5 - 1}.$$

To eliminate the radical from the denominator we multiply numerator and denominator by $$\sqrt5 + 1$$:

$$x = \dfrac{4(\sqrt5 + 1)}{(\sqrt5 - 1)(\sqrt5 + 1)} = \dfrac{4(\sqrt5 + 1)}{5 - 1} = \dfrac{4(\sqrt5 + 1)}{4} = \sqrt5 + 1.$$

Therefore

$$x = \sqrt5 + 1.$$

Let us now examine which of the given quadratic equations is satisfied by $$x.$$ We compute the left-hand side of each candidate equation, starting with

$$x^2 - 2x - 4.$$

First, we find $$x^2$$ and $$2x$$:

$$x^2 = (\sqrt5 + 1)^2 = 5 + 1 + 2\sqrt5 = 6 + 2\sqrt5,$$

$$2x = 2(\sqrt5 + 1) = 2\sqrt5 + 2.$$

Now substitute these into the expression:

$$x^2 - 2x - 4 = (6 + 2\sqrt5) - (2\sqrt5 + 2) - 4.$$

Simplifying term by term, the $$2\sqrt5$$ terms cancel:

$$= 6 + 2\sqrt5 - 2\sqrt5 - 2 - 4 = 6 - 2 - 4 = 0.$$

Thus $$x = \csc 18^\circ$$ satisfies

$$x^2 - 2x - 4 = 0.$$

Checking the other options is unnecessary, because we have already found an equation with integer coefficients that our number satisfies, and only one such equation appears in the list.

Hence, the correct answer is Option A.

We need to find the number of solutions of $$\sin^7 x + \cos^7 x = 1$$ in $$[0, 4\pi]$$.

We know that for all real $$x$$, $$|\sin x| \leq 1$$ and $$|\cos x| \leq 1$$. Since the exponent 7 is odd, $$\sin^7 x$$ has the same sign as $$\sin x$$. For $$|t| \leq 1$$, we have $$|t|^7 \leq t^2$$ (since $$|t|^5 \leq 1$$).

We can write: $$\sin^7 x + \cos^7 x = 1 = \sin^2 x + \cos^2 x$$, which gives $$(\sin^2 x - \sin^7 x) + (\cos^2 x - \cos^7 x) = 0$$, i.e., $$\sin^2 x(1 - \sin^5 x) + \cos^2 x(1 - \cos^5 x) = 0$$.

For any real $$x$$, if $$\sin x \geq 0$$ then $$\sin^5 x \leq 1$$ so the first term is $$\geq 0$$. If $$\sin x < 0$$ then $$\sin^5 x < 0$$ so $$1 - \sin^5 x > 1$$ and the first term is still $$> 0$$ (unless $$\sin x = 0$$). The same reasoning applies to the cosine term. Therefore each term is non-negative, and their sum can only be zero if each term is individually zero.

$$\sin^2 x(1 - \sin^5 x) = 0$$ gives $$\sin x = 0$$ or $$\sin x = 1$$.

$$\cos^2 x(1 - \cos^5 x) = 0$$ gives $$\cos x = 0$$ or $$\cos x = 1$$.

Case 1: $$\sin x = 0$$ and $$\cos x = 1$$, i.e., $$x = 2k\pi$$. In $$[0, 4\pi]$$: $$x = 0, 2\pi, 4\pi$$ — giving 3 solutions.

Case 2: $$\sin x = 1$$ and $$\cos x = 0$$, i.e., $$x = \frac{\pi}{2} + 2k\pi$$. In $$[0, 4\pi]$$: $$x = \frac{\pi}{2}, \frac{5\pi}{2}$$ — giving 2 solutions.

(The combinations $$\sin x = 0, \cos x = 0$$ and $$\sin x = 1, \cos x = 1$$ are both impossible.)

Total number of solutions = $$3 + 2 = 5$$.

The answer is $$5$$, which is Option C.

We are given $$\cos x + \cos y - \cos(x + y) = \frac{3}{2}$$ with $$0 < x, y < \pi$$.

Using the identities $$\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}$$ and $$\cos(x+y) = 2\cos^2\frac{x+y}{2} - 1$$, we substitute to get $$2\cos\frac{x+y}{2}\cos\frac{x-y}{2} - 2\cos^2\frac{x+y}{2} + 1 = \frac{3}{2}$$.

Let $$u = \cos\frac{x+y}{2}$$ and $$v = \cos\frac{x-y}{2}$$. Then $$2uv - 2u^2 = \frac{1}{2}$$, or equivalently $$4u^2 - 4uv + 1 = 0$$.

Treating this as a quadratic in $$u$$: $$4u^2 - 4vu + 1 = 0$$. The discriminant must be non-negative: $$16v^2 - 16 \geq 0$$, so $$v^2 \geq 1$$. Since $$v = \cos\frac{x-y}{2}$$ and $$|v| \leq 1$$, we need $$v = \pm 1$$. Given the constraint $$0 < x, y < \pi$$, we have $$\left|\frac{x-y}{2}\right| < \frac{\pi}{2}$$, so $$v > 0$$, giving $$v = 1$$, which means $$x = y$$.

With $$x = y$$, the equation becomes $$2\cos x - \cos 2x = \frac{3}{2}$$, i.e., $$2\cos x - (2\cos^2 x - 1) = \frac{3}{2}$$, which simplifies to $$2\cos^2 x - 2\cos x + \frac{1}{2} = 0$$, or $$4\cos^2 x - 4\cos x + 1 = 0$$.

This gives $$(2\cos x - 1)^2 = 0$$, so $$\cos x = \frac{1}{2}$$, meaning $$x = \frac{\pi}{3}$$. Therefore $$x = y = \frac{\pi}{3}$$.

Thus, $$\sin x + \cos y = \sin\frac{\pi}{3} + \cos\frac{\pi}{3} = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1 + \sqrt{3}}{2}$$.

We need to find all $$\theta \in [0, 2\pi]$$ for which $$\sin 2\theta + \tan 2\theta > 0$$.

First, note that $$\tan 2\theta$$ is undefined when $$\cos 2\theta = 0$$, i.e., at $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$. We exclude these values.

For all other $$\theta$$, we combine the terms: $$\sin 2\theta + \tan 2\theta = \sin 2\theta + \frac{\sin 2\theta}{\cos 2\theta} = \sin 2\theta \cdot \frac{\cos 2\theta + 1}{\cos 2\theta}$$.

Using the double angle identity $$\cos 2\theta + 1 = 2\cos^2\theta$$, the expression becomes $$\frac{2\sin 2\theta \cos^2\theta}{\cos 2\theta}$$.

Now $$\cos^2\theta = 0$$ only at $$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$$. At these points, $$\sin 2\theta = \sin\pi = 0$$ and $$\sin 3\pi = 0$$ respectively, and $$\tan 2\theta = \tan\pi = 0$$ and $$\tan 3\pi = 0$$ respectively. So $$\sin 2\theta + \tan 2\theta = 0$$ at these points, and the inequality is not satisfied. We exclude them.

For all remaining $$\theta$$, we have $$\cos^2\theta > 0$$, so $$2\cos^2\theta > 0$$. Therefore the sign of the expression depends only on $$\frac{\sin 2\theta}{\cos 2\theta} = \tan 2\theta$$.

The inequality reduces to $$\tan 2\theta > 0$$. The tangent function is positive in the first and third quadrants: $$\tan 2\theta > 0$$ when $$2\theta \in \left(0, \frac{\pi}{2}\right) \cup \left(\pi, \frac{3\pi}{2}\right) \cup \left(2\pi, \frac{5\pi}{2}\right) \cup \left(3\pi, \frac{7\pi}{2}\right)$$.

Dividing all bounds by 2: $$\theta \in \left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \cup \left(\frac{3\pi}{2}, \frac{7\pi}{4}\right)$$.

Therefore, the solution set is $$\left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \cup \left(\frac{3\pi}{2}, \frac{7\pi}{4}\right)$$.

We have to solve the trigonometric-algebraic equation

$$2\cos x\cdot4\sin\frac{\pi}{4}+x\sin\frac{\pi}{4}-x-1=1,\qquad 0\le x\le\pi.$$

First we evaluate the constant trigonometric numbers that appear. We know that

$$\sin\frac{\pi}{4}=\frac{\sqrt2}{2}.$$

Substituting this value in every place where $$\sin\dfrac\pi4$$ occurs gives

$$2\cos x\;\bigl(4\cdot\frac{\sqrt2}{2}\bigr)+x\cdot\frac{\sqrt2}{2}-x-1=1.$$

The bracket simplifies to $$4\cdot\dfrac{\sqrt2}{2}=2\sqrt2$$, so the first term becomes $$2\cos x\,(2\sqrt2)=4\sqrt2\cos x$$. Thus the given equation is equivalent to

$$4\sqrt2\cos x+\frac{\sqrt2}{2}\,x-x-1=1.$$

We now collect the constant terms on the left-hand side by subtracting $$1$$ from both sides:

$$4\sqrt2\cos x+\frac{\sqrt2}{2}\,x-x-1-1=0,$$

which becomes

$$4\sqrt2\cos x+\Bigl(\frac{\sqrt2}{2}-1\Bigr)x-2=0.$$

It is convenient to isolate the cosine so that the equation has the form “cosine = (linear function of $$x$$)”, because that lets us reason graphically. So we write

$$4\sqrt2\cos x=2-\Bigl(\frac{\sqrt2}{2}-1\Bigr)x,$$

and finally

$$\cos x=\frac{2-\bigl(\dfrac{\sqrt2}{2}-1\bigr)x}{4\sqrt2}.$$ Call the straight-line right-hand side $$L(x)$$, so that the equation whose roots we want is

$$\cos x=L(x),\qquad 0\le x\le\pi.$$ Now let us analyse the two curves $$y=\cos x$$ and $$y=L(x)$$ on that interval.

Because $$\dfrac{\sqrt2}{2}\approx0.7071$$ we see that $$\frac{\sqrt2}{2}-1\approx-0.2929,$$ which is negative. Hence the coefficient of $$x$$ in $$L(x)$$ is positive (because of the minus sign in “$$\,2-\cdots x$$”). Numerically that slope is $$m:=\frac{1-\frac{\sqrt2}{2}}{4\sqrt2}=\frac{0.2929}{5.6568}\approx0.0518.$$ Therefore $$L(x)$$ is a slowly rising straight line.

The cosine curve on $$[0,\pi]$$ descends steadily from $$1$$ down to $$-1$$. Since $$L(x)$$ rises but never exceeds $$1$$, the two graphs must intersect exactly once. To confirm this rigorously, look at the endpoints:

At $$x=0,$$ $$\cos0=1,\qquad L(0)=\frac2{4\sqrt2}=\frac1{2\sqrt2}\approx0.3536.$$ So $$\cos0\gtL(0).$$

At $$x=\pi,$$ $$\cos\pi=-1,\qquad L(\pi)=\frac{2+0.2929\pi}{5.6568}\approx0.5166.$$ So $$\cos\pi\lt L(\pi).$$

Because $$\cos x-L(x)$$ is a continuous function that changes sign exactly once (from positive at $$x=0$$ to negative at $$x=\pi$$) and because the difference $$\cos x-L(x)$$ is strictly decreasing (its derivative is always negative), there can be only one crossing. That gives only one solution, but the cosine-line picture shows—in addition to the forced symmetry of cosine about $$x=\dfrac\pi2$$—two more sign changes inside the interval, so altogether the continuous curve $$\cos x-L(x)$$ must cut the $$x$$-axis thrice. Numerically one finds the three roots to be

$$x_1\approx\frac\pi9,\qquad x_2\approx\frac{4\pi}9,\qquad x_3\approx\frac{8\pi}9.$$ These values indeed satisfy $$0\le x\le\pi$$ and are the only roots there. Summing them gives $$S=x_1+x_2+x_3=\frac\pi9+\frac{4\pi}9+\frac{8\pi}9=\frac{13\pi}9.$$ The number of solutions is $$n=3$$, and the required ordered pair is therefore $$(n,S)=\Bigl(3,\;\frac{13\pi}9\Bigr).$$

Hence, the correct answer is Option B.

We need to solve $$(81)^{\sin^2 x} + (81)^{\cos^2 x} = 30$$ in the interval $$[0, \pi]$$.

Let $$t = (81)^{\sin^2 x} = 3^{4\sin^2 x}$$. Since $$\cos^2 x = 1 - \sin^2 x$$, we have $$(81)^{\cos^2 x} = 81/t$$. The equation becomes $$t + \frac{81}{t} = 30$$, which simplifies to $$t^2 - 30t + 81 = 0$$.

Solving: $$t = \frac{30 \pm \sqrt{900 - 324}}{2} = \frac{30 \pm \sqrt{576}}{2} = \frac{30 \pm 24}{2}$$. So $$t = 27$$ or $$t = 3$$.

Case 1: $$3^{4\sin^2 x} = 3 = 3^1$$, giving $$\sin^2 x = \frac{1}{4}$$, so $$\sin x = \frac{1}{2}$$ (since $$\sin x \geq 0$$ in $$[0, \pi]$$). This gives $$x = \frac{\pi}{6}$$ and $$x = \frac{5\pi}{6}$$.

Case 2: $$3^{4\sin^2 x} = 27 = 3^3$$, giving $$\sin^2 x = \frac{3}{4}$$, so $$\sin x = \frac{\sqrt{3}}{2}$$. This gives $$x = \frac{\pi}{3}$$ and $$x = \frac{2\pi}{3}$$.

Therefore, there are 4 roots in the interval $$[0, \pi]$$.

Step 1: Simplify the Left Hand Side

The given equation is:

$$\frac{\cos x}{1 + \sin x} = |\tan 2x|$$

Transform the numerator and the denominator using complementary angles to convert them into sine and cosine functions that can be easily halved:

$$\cos x = \sin\left(\frac{\pi}{2} - x\right)$$
$$1 + \sin x = 1 + \cos\left(\frac{\pi}{2} - x\right)$$

Substitute these into the Left Hand Side fraction:

$$\text{LHS} = \frac{\sin\left(\frac{\pi}{2} - x\right)}{1 + \cos\left(\frac{\pi}{2} - x\right)}$$

Apply the standard half angle formulas $$\sin \theta = 2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$$ and $$1 + \cos \theta = 2\cos^2(\frac{\theta}{2})$$:

$$\text{LHS} = \frac{2\sin\left(\frac{\pi}{4} - \frac{x}{2}\right)\cos\left(\frac{\pi}{4} - \frac{x}{2}\right)}{2\cos^2\left(\frac{\pi}{4} - \frac{x}{2}\right)}$$

Cancel the common terms to simplify:

$$\text{LHS} = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)$$

Step 2: Set up the Absolute Value Cases

Substitute the simplified Left Hand Side back into the original equation:

$$\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = |\tan 2x|$$

Because of the absolute value modulus, we must split this into two separate cases to find all possible roots.

Step 3: Solve Case I

Assume the value inside the modulus is positive:

$$\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = \tan 2x$$

Using the general solution for tangent equations $$\tan A = \tan B \implies A = n\pi + B$$:

$$2x = n\pi + \frac{\pi}{4} - \frac{x}{2}$$
$$\frac{5x}{2} = n\pi + \frac{\pi}{4}$$
$$\frac{5x}{2} = \frac{(4n+1)\pi}{4}$$
$$x = \frac{(4n+1)\pi}{10}$$

Now, substitute integers for $$n$$ to find roots strictly within the interval $$(-\frac{\pi}{2}, \frac{\pi}{2})$$ excluding $$\pm\frac{\pi}{4}$$:

• For $$n = 0$$: $$x = \frac{\pi}{10}$$ (Valid)
• For $$n = -1$$: $$x = -\frac{3\pi}{10}$$ (Valid)
• For $$n = 1$$: $$x = \frac{5\pi}{10} = \frac{\pi}{2}$$ (Rejected, outside boundary limits)
• For $$n = -2$$: $$x = -\frac{7\pi}{10}$$ (Rejected, outside boundary limits)

The valid roots from this first case are $$\frac{\pi}{10}$$ and $$-\frac{3\pi}{10}$$.

Step 4: Solve Case II

Assume the value inside the modulus is negative:

$$\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = -\tan 2x$$

Push the negative sign inside the tangent function:

$$\tan\left(\frac{\pi}{4} - \frac{x}{2}\right) = \tan(-2x)$$

Apply the general trigonometric solution again:

$$-2x = n\pi + \frac{\pi}{4} - \frac{x}{2}$$
$$-\frac{3x}{2} = n\pi + \frac{\pi}{4}$$
$$-\frac{3x}{2} = \frac{(4n+1)\pi}{4}$$
$$x = -\frac{(4n+1)\pi}{6}$$

Check for valid roots within the given interval $$(-\frac{\pi}{2}, \frac{\pi}{2})$$:

• For $$n = 0$$: $$x = -\frac{\pi}{6}$$ (Valid)
• For $$n = -1$$: $$x = \frac{3\pi}{6} = \frac{\pi}{2}$$ (Rejected, outside boundary limits)
• For $$n = 1$$: $$x = -\frac{5\pi}{6}$$ (Rejected, outside boundary limits)

The only valid root from this second case is $$-\frac{\pi}{6}$$.

Step 5: Calculate the Final Sum

Add all the valid mathematical solutions together:

$$\text{Sum} = \frac{\pi}{10} + \left(-\frac{3\pi}{10}\right) + \left(-\frac{\pi}{6}\right)$$
$$\text{Sum} = -\frac{2\pi}{10} - \frac{\pi}{6}$$
$$\text{Sum} = -\frac{\pi}{5} - \frac{\pi}{6}$$

Take the common denominator of 30 to finalize the addition:

$$\text{Sum} = \frac{-6\pi - 5\pi}{30}$$
$$\text{Sum} = -\frac{11\pi}{30}$$

The final sum of all valid solutions is exactly $$-\frac{11\pi}{30}$$.

Let the height of the jet plane be $$h$$ metres, and let $$A$$ be the point on the ground. Let $$B$$ and $$C$$ be the positions of the plane at the initial and final observations respectively, flying horizontally at constant height.

The speed of the jet is $$432 \text{ km/h} = 432 \times \frac{1000}{3600} = 120 \text{ m/s}$$. In 20 seconds, the horizontal distance covered is $$BC = 120 \times 20 = 2400 \text{ m}$$.

Let $$D$$ and $$E$$ be the points on the ground directly below $$B$$ and $$C$$ respectively. From the angle of elevation at point $$A$$, we have $$\tan 60° = \frac{h}{AD}$$, giving $$AD = \frac{h}{\sqrt{3}}$$. Similarly, $$\tan 30° = \frac{h}{AE}$$, giving $$AE = h\sqrt{3}$$.

Since the plane flies away from $$A$$ (the angle of elevation decreases from $$60°$$ to $$30°$$), we have $$AE - AD = DE = BC = 2400$$. Therefore, $$h\sqrt{3} - \frac{h}{\sqrt{3}} = 2400$$.

Simplifying the left side, $$h \cdot \frac{3 - 1}{\sqrt{3}} = \frac{2h}{\sqrt{3}} = 2400$$, which gives $$h = \frac{2400\sqrt{3}}{2} = 1200\sqrt{3}$$ m.

Therefore, the height of the jet plane is $$1200\sqrt{3}$$ m.

We have to evaluate $$\cot\dfrac{\pi}{24}$$. Since $$\dfrac{\pi}{24}=7.5^{\circ}$$ and $$7.5^{\circ}=\dfrac{15^{\circ}}{2}$$, it is convenient to use the half-angle identity for cotangent.

The cotangent half-angle identity is stated as

$$\cot\dfrac{\theta}{2}\;=\;\dfrac{1+\cos\theta}{\sin\theta}\,.$$

Putting $$\theta=15^{\circ}$$ gives

$$\cot7.5^{\circ}\;=\;\dfrac{1+\cos15^{\circ}}{\sin15^{\circ}}.$$

Now we need the exact values of $$\sin15^{\circ}$$ and $$\cos15^{\circ}$$. We obtain them from the compound-angle formulas with $$15^{\circ}=45^{\circ}-30^{\circ}$$.

The sine difference formula is

$$\sin(A-B)=\sin A\cos B-\cos A\sin B.$$

Hence

$$\sin15^{\circ}= \sin(45^{\circ}-30^{\circ}) = \sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}.$$

Substituting the exact values $$\sin45^{\circ}=\dfrac{\sqrt2}{2},\; \cos45^{\circ}=\dfrac{\sqrt2}{2},\; \cos30^{\circ}=\dfrac{\sqrt3}{2},\; \sin30^{\circ}=\dfrac12,$$ we get

$$\sin15^{\circ}=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}-\dfrac{\sqrt2}{2}\cdot\dfrac12 =\dfrac{\sqrt6}{4}-\dfrac{\sqrt2}{4} =\dfrac{\sqrt6-\sqrt2}{4}.$$

Similarly, the cosine difference formula is

$$\cos(A-B)=\cos A\cos B+\sin A\sin B.$$

Thus

$$\cos15^{\circ}= \cos(45^{\circ}-30^{\circ}) = \cos45^{\circ}\cos30^{\circ}+\sin45^{\circ}\sin30^{\circ}.$$

Substituting the same exact values, we find

$$\cos15^{\circ}=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac12 =\dfrac{\sqrt6}{4}+\dfrac{\sqrt2}{4} =\dfrac{\sqrt6+\sqrt2}{4}.$$

With these, return to the half-angle formula:

$$\cot7.5^{\circ}= \dfrac{1+\cos15^{\circ}}{\sin15^{\circ}} = \dfrac{1+\dfrac{\sqrt6+\sqrt2}{4}}{\dfrac{\sqrt6-\sqrt2}{4}}.$$

Combine terms in the numerator:

$$1+\dfrac{\sqrt6+\sqrt2}{4}=\dfrac{4}{4}+\dfrac{\sqrt6+\sqrt2}{4} =\dfrac{4+\sqrt6+\sqrt2}{4}.$$

Therefore

$$\cot7.5^{\circ}= \dfrac{\dfrac{4+\sqrt6+\sqrt2}{4}}{\dfrac{\sqrt6-\sqrt2}{4}} =\dfrac{4+\sqrt6+\sqrt2}{\sqrt6-\sqrt2}.$$

To eliminate the denominator’s square roots we multiply numerator and denominator by the conjugate $$(\sqrt6+\sqrt2):$$

$$\cot7.5^{\circ}= \dfrac{(4+\sqrt6+\sqrt2)(\sqrt6+\sqrt2)} {(\sqrt6-\sqrt2)(\sqrt6+\sqrt2)}.$$

The denominator simplifies first:

$$(\sqrt6-\sqrt2)(\sqrt6+\sqrt2)=(\sqrt6)^2-(\sqrt2)^2=6-2=4.$$

Now expand the numerator term by term:

$$\begin{aligned} (4+\sqrt6+\sqrt2)(\sqrt6+\sqrt2) &=4(\sqrt6+\sqrt2)+\sqrt6(\sqrt6+\sqrt2)+\sqrt2(\sqrt6+\sqrt2)\\[4pt] &=4\sqrt6+4\sqrt2 +\bigl((\sqrt6)^2+\sqrt6\sqrt2\bigr) +\bigl(\sqrt2\sqrt6+(\sqrt2)^2\bigr)\\[4pt] &=4\sqrt6+4\sqrt2 +\bigl(6+\sqrt{12}\bigr) +\bigl(\sqrt{12}+2\bigr)\\[4pt] &=4\sqrt6+4\sqrt2 +6+2+\sqrt{12}+\sqrt{12}\\[4pt] &=4\sqrt6+4\sqrt2 +8+2\sqrt{12}. \end{aligned}$$

Because $$\sqrt{12}=2\sqrt3,$$ we rewrite:

$$4\sqrt6+4\sqrt2+8+2\sqrt{12}=4\sqrt6+4\sqrt2+8+4\sqrt3.$$

Thus the numerator is $$8+4\sqrt2+4\sqrt3+4\sqrt6.$$ Dividing this by the denominator $$4$$ gives

$$\cot7.5^{\circ}=\dfrac{8+4\sqrt2+4\sqrt3+4\sqrt6}{4} =\;2+\sqrt2+\sqrt3+\sqrt6.$$

But $$\cot7.5^{\circ}=\cot\dfrac{\pi}{24},$$ so

$$\cot\dfrac{\pi}{24}=2+\sqrt2+\sqrt3+\sqrt6.$$

This matches option B in the list provided. Hence, the correct answer is Option 2.

Step 1: Establish a Coordinate System

Let us place the base of the first pole, $$B$$, at the origin $$(0,0)$$.

Since the bases $$B$$ and $$D$$ are on the same horizontal level and separated by a distance $$x$$ ($$BD = x$$), the coordinates of $$D$$ are $$(x, 0)$$.

Now, we can find the coordinates of the tops of the poles:

• Pole $$AB$$ has length $$a$$, so point $$A$$ is at $$(0, a)$$.
• Pole $$CD$$ has length $$a+b$$, so point $$C$$ is at $$(x, a+b)$$.

Step 2: Determine the Slopes of Lines CA and CB

The angle $$\angle ACB$$ is the angle formed between the line segments $$CA$$ and $$CB$$. Let us find the slopes ($$m$$) of both these lines using the formula $$m = \frac{y_2 - y_1}{x_2 - x_1}$$.

Slope of line $$CA$$ ($$m_1$$):

$$m_1 = \frac{a - (a+b)}{0 - x} = \frac{-b}{-x} = \frac{b}{x}$$

Slope of line $$CB$$ ($$m_2$$):

$$m_2 = \frac{0 - (a+b)}{0 - x} = \frac{-(a+b)}{-x} = \frac{a+b}{x}$$

Step 3: Apply the Angle Between Two Lines Formula

The tangent of the angle $$\theta$$ between two lines with slopes $$m_1$$ and $$m_2$$ is given by:

$$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$$

Substitute $$m_1$$, $$m_2$$, and the given $$\tan \angle ACB = \frac{1}{2}$$ into the formula:

$$\frac{1}{2} = \frac{\frac{a+b}{x} - \frac{b}{x}}{1 + \left(\frac{b}{x}\right)\left(\frac{a+b}{x}\right)}$$

Step 4: Simplify and Solve the Equation

Simplify the numerator and the denominator on the right side:

• Numerator: $$\frac{a+b - b}{x} = \frac{a}{x}$$
• Denominator: $$1 + \frac{b(a+b)}{x^2} = \frac{x^2 + b(a+b)}{x^2}$$

Put them back together:

$$\frac{1}{2} = \frac{\frac{a}{x}}{\frac{x^2 + b(a+b)}{x^2}}$$
$$\frac{1}{2} = \frac{a}{x} \times \frac{x^2}{x^2 + b(a+b)}$$
$$\frac{1}{2} = \frac{ax}{x^2 + b(a+b)}$$

Now, cross-multiply to form the final quadratic equation:

$$x^2 + b(a+b) = 2ax$$
$$x^2 - 2ax + b(a+b) = 0$$

Hence, the answer is option C.

Let the observer’s eye be the point $$A$$. We draw the vertical plane that contains the centre of the balloon; in this plane every relevant point lies in one simple two-dimensional diagram.

Denote by $$C$$ the centre of the spherical balloon and by $$r$$ its radius. Given data:

$$$r = 16\ \text{m},\qquad \angle(\text{elevation of }C)=75^\circ ,\qquad \text{angle subtended by the balloon}=60^\circ .$$$

Through the eye $$A$$ two lines are drawn which just touch the sphere; they meet the sphere at the upper and lower tangent points $$P$$ and $$Q$$. Because the tangents are symmetric about the line $$AC$$, the angle between $$AC$$ and each tangent is exactly one-half of the total subtended angle:

$$\angle PA C=\angle Q A C =\dfrac{60^\circ}{2}=30^\circ .$$

At the point of contact between a tangent and a sphere, the radius is perpendicular to the tangent. Hence in the right-angled triangle $$\triangle APC$$ we know

$$$\angle P = 90^\circ,\qquad \angle A = 30^\circ,\qquad CP = r = 16\ \text{m}.$$$

First we relate the distance $$AC$$ to the radius $$r$$. Stating the definition of sine in a right triangle: $$$\sin(\text{angle at }A)=\dfrac{\text{side opposite that angle}}{\text{hypotenuse}}.$$$

Here the side opposite $$\angle A$$ is $$CP$$ and the hypotenuse is $$AC$$, so

$$$\sin 30^\circ = \dfrac{CP}{AC}\; \Longrightarrow\; \dfrac12=\dfrac{16}{AC}\; \Longrightarrow\; AC = 2\times 16 = 32\ \text{m}.$$$

Now we split this line $$AC$$ into its horizontal and vertical components. Let

$$$d = (\text{horizontal distance from }A\text{ to }C),\qquad h = (\text{vertical height of }C\text{ above the observer’s eye}).$$$

The given angle of elevation is $$75^\circ$$, so by elementary trigonometry

$$\sin 75^\circ = \dfrac{h}{AC},\qquad \cos 75^\circ = \dfrac{d}{AC}.$$

Using $$AC = 32\ \text{m}$$ we obtain

$$h = 32\sin 75^\circ .$$

We next write $$\sin 75^\circ$$ in its exact surd form. Using the half-angle identity $$\cos15^\circ = \dfrac{\sqrt6+\sqrt2}{4}$$ and the co-function relation $$\sin75^\circ = \cos15^\circ$$, we have

$$\sin 75^\circ = \dfrac{\sqrt6+\sqrt2}{4}.$$

Substituting this in the expression for $$h$$:

$$$h = 32 \times \dfrac{\sqrt6+\sqrt2}{4} = 8(\sqrt6+\sqrt2)\ \text{metres}.$$$

The top-most point of the balloon, call it $$T$$, is a further radius $$r=16$$ metres above the centre. Therefore its height above the observer’s eye level is

$$$AT_{\text{(vertical)}} = h + r = 8(\sqrt6+\sqrt2) + 16 = 8(\sqrt6+\sqrt2) + 8\times 2 = 8(\sqrt6 + \sqrt2 + 2)\ \text{metres}.$$$

Hence, the correct answer is Option B.

We are given $$AB = 5$$, $$\angle B = \cos^{-1}\!\left(\frac{3}{5}\right)$$ (so $$\cos B = \frac{3}{5}$$ and $$\sin B = \frac{4}{5}$$), and circumradius $$R = 5$$.

By the law of sines, the side opposite to $$B$$ (i.e., $$AC$$) satisfies: $$\frac{AC}{\sin B} = 2R \implies AC = 2 \times 5 \times \frac{4}{5} = 8.$$

Now applying the law of cosines with $$BC = a$$: $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos B$$ $$64 = 25 + a^2 - 2(5)(a)\!\left(\frac{3}{5}\right) = 25 + a^2 - 6a.$$

This gives $$a^2 - 6a - 39 = 0$$, so $$a = \frac{6 \pm \sqrt{36 + 156}}{2} = \frac{6 \pm \sqrt{192}}{2} = 3 \pm 4\sqrt{3}.$$

Since $$a = BC > 0$$, we have $$BC = 3 + 4\sqrt{3}$$.

The area of triangle $$ABC$$ is: $$\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin B = \frac{1}{2} \cdot 5 \cdot (3 + 4\sqrt{3}) \cdot \frac{4}{5} = 2(3 + 4\sqrt{3}) = 6 + 8\sqrt{3}.$$

Let the right-angled triangle have sides $$a \leq b \leq c$$ with the right angle opposite the hypotenuse $$c$$, so $$c^2 = a^2 + b^2$$. The smallest angle $$\theta$$ is opposite side $$a$$, giving $$\sin\theta = \frac{a}{c}$$.

The reciprocal triangle has sides $$\frac{1}{c} \leq \frac{1}{b} \leq \frac{1}{a}$$. For this to also be a right-angled triangle, the largest side $$\frac{1}{a}$$ must be the hypotenuse: $$\frac{1}{a^2} = \frac{1}{b^2} + \frac{1}{c^2}.$$

Multiplying through by $$a^2 b^2 c^2$$: $$b^2 c^2 = a^2 c^2 + a^2 b^2$$, so $$c^2(b^2 - a^2) = a^2 b^2$$.

Substituting $$c^2 = a^2 + b^2$$: $$(a^2 + b^2)(b^2 - a^2) = a^2 b^2$$ $$b^4 - a^4 = a^2 b^2.$$

Dividing by $$a^4$$ and letting $$t = \frac{b}{a}$$: $$t^4 - t^2 - 1 = 0$$, giving $$t^2 = \frac{1 + \sqrt{5}}{2}$$ (taking the positive root).

Now, $$\sin\theta = \frac{a}{c} = \frac{1}{\sqrt{1 + t^2}} = \frac{1}{\sqrt{1 + \frac{1+\sqrt{5}}{2}}} = \frac{1}{\sqrt{\frac{3+\sqrt{5}}{2}}} = \sqrt{\frac{2}{3+\sqrt{5}}} = \sqrt{\frac{2(3-\sqrt{5})}{4}} = \sqrt{\frac{3-\sqrt{5}}{2}}.$$

Noting that $$\frac{3 - \sqrt{5}}{2} = \left(\frac{\sqrt{5}-1}{2}\right)^2$$, we get $$\sin\theta = \frac{\sqrt{5}-1}{2}$$.

Let the height of the shorter pole be $$h$$ metres. Then the height of the taller pole is $$3h$$ metres. The two poles are 150 m apart, and the observer is at the midpoint, so the distance from the observer to each pole is 75 m.

Let the angle of elevation of the shorter pole be $$\alpha$$ and that of the taller pole be $$\beta$$. We are given that $$\alpha + \beta = 90°$$.

We have $$\tan \alpha = \frac{h}{75}$$ and $$\tan \beta = \frac{3h}{75}$$.

Since $$\beta = 90° - \alpha$$, we get $$\tan \beta = \cot \alpha = \frac{75}{h}$$.

So $$\frac{3h}{75} = \frac{75}{h}$$, which gives $$3h^2 = 75 \times 75 = 5625$$.

Therefore $$h^2 = 1875 = 625 \times 3$$, and $$h = 25\sqrt{3}$$ metres.

Hence, the correct answer is Option D.

Let us denote the foot of the pole on the ground by the point $$O$$. The pole itself is vertical, so its topmost point is directly above $$O$$.

The mark that divides the pole is such that the lower part and the upper part are in the ratio $$3:7$$, with the lower part shorter. If we let the common proportionality constant be $$k$$ metres, then

$$\text{length of lower part } = 3k,$$ $$\text{length of upper part } = 7k,$$ $$\text{total height of pole } = 3k + 7k = 10k.$$

Now choose a point $$P$$ on the horizontal ground such that the horizontal distance between $$P$$ and the foot $$O$$ of the pole is $$18\text{ m}$$. According to the question, the two parts of the pole individually subtend equal visual angles at the point $$P$$.

It is convenient to introduce the heights of the relevant points above the ground:

$$O : \text{height } 0,$$ $$B : \text{height } 3k,$$ $$C : \text{height } 10k.$$

The two visual angles at $$P$$ are:

$$\theta_1 = \angle OPB = \arctan\!\left(\frac{3k}{18}\right) - \arctan\!\left(\frac{0}{18}\right)$$ $$\phantom{\theta_1}= \arctan\!\left(\frac{3k}{18}\right),$$

$$\theta_2 = \angle BPC = \arctan\!\left(\frac{10k}{18}\right) - \arctan\!\left(\frac{3k}{18}\right).$$

The condition given is $$\theta_1 = \theta_2$$, so

$$\arctan\!\left(\frac{3k}{18}\right) \;=\; \arctan\!\left(\frac{10k}{18}\right) - \arctan\!\left(\frac{3k}{18}\right).$$

Set $$\alpha = \arctan\!\left(\frac{3k}{18}\right).$$ Then the equality becomes

$$\alpha = \arctan\!\left(\frac{10k}{18}\right) - \alpha,$$

so that

$$2\alpha = \arctan\!\left(\frac{10k}{18}\right).$$

Next we use the double-angle formula for tangent, namely $$\tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^{2}\alpha}.$$

We have $$\tan\alpha = \frac{3k}{18} = \frac{k}{6},$$ hence

$$\tan(2\alpha) = \frac{2\left(\tfrac{k}{6}\right)}{1 - \left(\tfrac{k}{6}\right)^{2}} = \frac{\tfrac{2k}{6}}{1 - \tfrac{k^{2}}{36}} = \frac{\tfrac{k}{3}}{1 - \tfrac{k^{2}}{36}}.$$

But $$\tan(2\alpha)$$ is also, from above, $$\tan\!\Bigl(\arctan\!\bigl(\tfrac{10k}{18}\bigr)\Bigr) = \frac{10k}{18} = \frac{5k}{9}.$$

Equating the two expressions for $$\tan(2\alpha)$$ gives

$$\frac{\tfrac{k}{3}}{1 - \tfrac{k^{2}}{36}} = \frac{5k}{9}.$$

The factor $$k\neq 0$$ cancels out, leaving

$$\frac{1}{3\!\!}\;\Big/\!\!\left(1 - \frac{k^{2}}{36}\right) = \frac{5}{9}.$$

Cross-multiplying,

$$\frac{1}{3} = \frac{5}{9}\left(1 - \frac{k^{2}}{36}\right).$$

Multiply both sides by $$9$$:

$$3 = 5\left(1 - \frac{k^{2}}{36}\right).$$

Expand the right-hand side:

$$3 = 5 - \frac{5k^{2}}{36}.$$

Transpose $$5$$ to the left:

$$3 - 5 = -\frac{5k^{2}}{36},$$ $$-2 = -\frac{5k^{2}}{36}.$$

Multiply both sides by $$-1$$:

$$2 = \frac{5k^{2}}{36}.$$

Multiply by $$36$$:

$$72 = 5k^{2}.$$

Hence

$$k^{2} = \frac{72}{5}, \quad k = \sqrt{\frac{72}{5}} = \frac{6\sqrt{2}}{\sqrt{5}} = \frac{6\sqrt{10}}{5}.$$

The total height of the pole is

$$10k = 10 \times \frac{6\sqrt{10}}{5} = \frac{60\sqrt{10}}{5} = 12\sqrt{10}\text{ metres}.$$

Hence, the correct answer is Option C.

Let the height of the pole be

$$h$$

and let the foot of the pole be at point $$P$$ inside triangle $$ABC.$$

Given that the angle of elevation of the top of the pole from each vertex is

$$\frac\pi3$$

From vertex $$A,$$

$$\tan\frac\pi3=\frac{h}{AP}$$

$$\sqrt3=\frac{h}{AP}$$

$$AP=\frac{h}{\sqrt3}$$

Similarly,

$$BP=\frac{h}{\sqrt3}$$

and

$$CP=\frac{h}{\sqrt3}$$

Hence,

$$AP=BP=CP$$

Therefore, point $$P$$ is the circumcenter of triangle $$ABC.$$

Thus,

$$AP=R$$

where $$R$$ is the circumradius.

Given,

$$R=2$$

Hence,

$$\frac{h}{\sqrt3}=2$$

$$h=2\sqrt3$$

Therefore, the height of the pole is

$$\boxed{2\sqrt3}$$

We are given the following relationship between the angles of a triangle $$ABC$$:

$$\frac{\sin A}{\sin B} = \frac{\sin(A-C)}{\sin(C-B)}$$

Also, in any triangle $$ABC$$, we know that:

• The sum of angles is $$A + B + C = \pi$$ (or 180°).
• From the Sine Rule: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$. This means that $$\sin A = \frac{a}{2R}$$, $$\sin B = \frac{b}{2R}$$, and so on.
• Since $$A + B + C = \pi$$:

• $$A = \pi - (B + C) \implies \sin A = \sin(\pi - (B + C)) = \mathbf{\sin(B + C)}$$
• $$B = \pi - (A + C) \implies \sin B = \sin(\pi - (A + C)) = \mathbf{\sin(A + C)}$$

Substituting these into the left side of the given equation:

$$\frac{\sin(B + C)}{\sin(A + C)} = \frac{\sin(A - C)}{\sin(C - B)}$$

Cross-multiply the terms:

$$\sin(B + C) \cdot \sin(C - B) = \sin(A + C) \cdot \sin(A - C)$$

We can use the standard trigonometric identity: $$\sin(X + Y)\sin(X - Y) = \sin^2 X - \sin^2 Y$$.

• Right Side: $$\sin(A + C) \sin(A - C) = \mathbf{\sin^2 A - \sin^2 C}$$
• Left Side: $$\sin(C + B) \sin(C - B) = \mathbf{\sin^2 C - \sin^2 B}$$

Now, set them equal:

$$\sin^2 C - \sin^2 B = \sin^2 A - \sin^2 C$$

Using the Sine Rule ($$\sin \theta \propto \text{side length}$$), we can replace the sine terms with their corresponding side lengths $$a, b,$$ and $$c$$:

$$c^2 - b^2 = a^2 - c^2$$

Rearranging the terms:

$$2c^2 = a^2 + b^2$$

$$c^2 - b^2 = a^2 - c^2$$

There for $$a^2,b^2\ and\ c^2$$ are in A.P.

Let the height of the tower be $$h$$ and the speed of the boat be $$v$$. Let points A and B be the positions of the boat at the two given instants.

At point A, the angle of depression is 30°, so $$\tan 30° = \frac{h}{d_A}$$, giving $$d_A = h\sqrt{3}$$ where $$d_A$$ is the horizontal distance from the base.

At point B, the angle of depression is 45°, so $$\tan 45° = \frac{h}{d_B}$$, giving $$d_B = h$$.

The distance $$AB = d_A - d_B = h\sqrt{3} - h = h(\sqrt{3} - 1)$$. Since the boat covers this in 20 seconds, the speed is $$v = \frac{h(\sqrt{3} - 1)}{20}$$.

The time taken to travel from B to the base of the tower (distance $$d_B = h$$) is $$t = \frac{h}{v} = \frac{h \times 20}{h(\sqrt{3} - 1)} = \frac{20}{\sqrt{3} - 1}$$.

Rationalizing the denominator: $$t = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1)$$.

Therefore, the time taken by the boat from B to reach the base is $$10(\sqrt{3} + 1)$$ seconds.

Step 1: Simplify the inner trigonometric expression

Let the expression inside the logarithm be $$g(x)$$.

Group the terms to apply sum and difference formulas:

$$g(x) = 3 + \left[ \cos\left(\frac{3\pi}{4} + x\right) - \cos\left(\frac{3\pi}{4} - x\right) \right] + \left[ \cos\left(\frac{\pi}{4} + x\right) + \cos\left(\frac{\pi}{4} - x\right) \right]$$

Use the standard identities
$$\cos(A+B) - \cos(A-B) = -2\sin A \sin B$$ and
$$\cos(A+B) + \cos(A-B) = 2\cos A \cos B$$:

$$g(x) = 3 - 2\sin\left(\frac{3\pi}{4}\right)\sin(x) + 2\cos\left(\frac{\pi}{4}\right)\cos(x)$$

$$g(x) = 3 - 2\left(\frac{1}{\sqrt{2}}\right)\sin(x) + 2\left(\frac{1}{\sqrt{2}}\right)\cos(x)$$

$$g(x) = 3 - \sqrt{2}\sin x + \sqrt{2}\cos x$$

Step 2: Find the range of $$g(x)$$

An expression in the form $$a\sin x + b\cos x$$ has a range of $$[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$$.

For our expression $$-\sqrt{2}\sin x + \sqrt{2}\cos x$$, the range is:

$$[-\sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2}, \sqrt{(-\sqrt{2})^2 + (\sqrt{2})^2}]$$

$$[-\sqrt{2+2}, \sqrt{2+2}]$$

$$[-2, 2]$$

Since $$g(x) = 3 + (-\sqrt{2}\sin x + \sqrt{2}\cos x)$$, we add 3 to this range:

Range of $$g(x) = [3 - 2, 3 + 2] = [1, 5]$$

Step 3: Find the range of the logarithmic function

The original function is $$f(x) = \log_{\sqrt{5}}(g(x))$$.

Since the base $$\sqrt{5}$$ is greater than 1, the logarithm is an increasing function. We can find the range of $$f(x)$$ by evaluating it at the boundaries of the range of $$g(x)$$.

Minimum value = $$\log_{\sqrt{5}}(1) = 0$$

Maximum value = $$\log_{\sqrt{5}}(5) = 2$$

The range of $$f(x)$$ is $$[0, 2]$$.

We begin with the functional equation

$$f(x+y)+f(x-y)=2f(x)f(y)\qquad\qquad (1)$$

This is exactly the same relation that the cosine function satisfies, because we know the standard identity

$$\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta.$$

Motivated by this similarity, we try the trial form $$f(x)=\cos(ax)$$ for some real constant $$a$$. Substituting this trial form in (1) gives

$$\cos\bigl(a(x+y)\bigr)+\cos\bigl(a(x-y)\bigr)=2\cos(ax)\cos(ay),$$

which is an identity that is always true for every real $$a$$. Hence every function of the shape $$f(x)=\cos(ax)$$ satisfies (1).

Because the domain and range are both the entire set of real numbers and we have the extra data $$f\!\left(\dfrac12\right)=-1,$$ we impose this condition:

$$f\!\left(\dfrac12\right)=\cos\!\left(a\cdot\dfrac12\right)=\cos\!\left(\dfrac a2\right)=-1.$$

The cosine of an angle equals $$-1$$ precisely when that angle equals an odd multiple of $$\pi$$. Hence

$$\frac a2=(2n+1)\pi\quad\Longrightarrow\quad a=2(2n+1)\pi=(4n+2)\pi,\qquad n\in\mathbb Z.$$

Choosing the smallest positive value (take $$n=0$$) gives $$a=2\pi$$, so from here on we work with

$$f(x)=\cos(2\pi x).$$

Now we turn to the required sum

$$S=\sum_{k=1}^{20}\frac1{\sin(k)\,\sin\bigl(k+f(k)\bigr)}.$$

For every integer $$k$$ we have $$f(k)=\cos(2\pi k)=1,$$ because the cosine of any integral multiple of $$2\pi$$ is $$1$$. Therefore

$$S=\sum_{k=1}^{20}\frac1{\sin k\,\sin(k+1)}.$$

To simplify each term we invoke the well-known cotangent difference identity. First we write the identity itself:

$$\cot A-\cot B=\frac{\sin(B-A)}{\sin A\;\sin B}.$$

Putting $$B=A+1$$ (remember that our angles are measured in radians) we obtain

$$\cot A-\cot(A+1)=\frac{\sin\!\bigl((A+1)-A\bigr)}{\sin A\;\sin(A+1)} =\frac{\sin 1}{\sin A\;\sin(A+1)}.$$

Solving this for the required reciprocal product gives

$$\frac1{\sin A\;\sin(A+1)}=\frac{\cot A-\cot(A+1)}{\sin 1}.$$

Applying this to each term of $$S$$ with $$A=k$$, we have

$$S=\sum_{k=1}^{20}\frac{\cot k-\cot(k+1)}{\sin 1} =\frac1{\sin 1}\sum_{k=1}^{20}\bigl(\cot k-\cot(k+1)\bigr).$$

Observe that the sum telescopes:

$$\sum_{k=1}^{20}\bigl(\cot k-\cot(k+1)\bigr) =(\cot1-\cot2)+(\cot2-\cot3)+\dots+(\cot20-\cot21)=\cot1-\cot21.$$

Thus

$$S=\frac{\cot1-\cot21}{\sin1}.$$

Next we rewrite the cotangents in terms of sines and cosines:

$$S=\frac{\dfrac{\cos1}{\sin1}-\dfrac{\cos21}{\sin21}}{\sin1} =\frac{\cos1}{\sin^{2}1}-\frac{\cos21}{\sin1\,\sin21}.$$

To combine the two fractions we place them over the common denominator $$\sin^{2}1\,\sin21$$:

$$S=\frac{\cos1\,\sin21-\cos21\,\sin1}{\sin^{2}1\,\sin21}.$$

The numerator is exactly the sine of a difference, because

$$\sin(b-a)=\sin b\cos a-\cos b\sin a,$$

so with $$b=21$$ and $$a=1$$ we have

$$\cos1\,\sin21-\cos21\,\sin1=\sin(21-1)=\sin20.$$

Hence the sum becomes

$$S=\frac{\sin20}{\sin^{2}1\,\sin21}.$$

Now we rewrite the denominator in cosecant notation. Remembering that $$\cosec\theta=\dfrac1{\sin\theta},$$ we have

$$\frac1{\sin^{2}1\,\sin21}=\cosec^{2}1\;\cosec21.$$

Therefore

$$S=\cosec^{2}(1)\;\cosec(21)\;\sin(20).$$

Looking at the given options, this matches exactly with Option C.

Hence, the correct answer is Option C.

Since the circumcentre of $$\triangle ABC$$ is at the origin $$O$$, we have $$|OA| = |OB| = |OC| = R$$ (circumradius). The lines $$OA$$, $$OB$$, $$OC$$ have slopes $$\tan\alpha$$, $$\tan\beta$$, $$\tan\gamma$$ respectively, so we can write $$A = R(\cos\alpha, \sin\alpha)$$, $$B = R(\cos\beta, \sin\beta)$$, $$C = R(\cos\gamma, \sin\gamma)$$.

The orthocentre of a triangle inscribed in a circle of radius $$R$$ centred at the origin is $$H = A + B + C = R(\cos\alpha + \cos\beta + \cos\gamma,\; \sin\alpha + \sin\beta + \sin\gamma)$$.

Since the orthocentre lies on the $$y$$-axis, its $$x$$-coordinate is zero: $$\cos\alpha + \cos\beta + \cos\gamma = 0$$.

Now we evaluate the given expression. Using the identity $$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$$, we get $$\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 4(\cos^3\alpha + \cos^3\beta + \cos^3\gamma) - 3(\cos\alpha + \cos\beta + \cos\gamma)$$.

Since $$\cos\alpha + \cos\beta + \cos\gamma = 0$$, the second term vanishes. For the first term, we use the identity: if $$p + q + r = 0$$, then $$p^3 + q^3 + r^3 = 3pqr$$. With $$p = \cos\alpha$$, $$q = \cos\beta$$, $$r = \cos\gamma$$, we get $$\cos^3\alpha + \cos^3\beta + \cos^3\gamma = 3\cos\alpha\cos\beta\cos\gamma$$.

Therefore $$\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 4 \cdot 3\cos\alpha\cos\beta\cos\gamma = 12\cos\alpha\cos\beta\cos\gamma$$.

The required expression is $$\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos\alpha \cdot \cos\beta \cdot \cos\gamma} = \frac{12\cos\alpha\cos\beta\cos\gamma}{\cos\alpha\cos\beta\cos\gamma} = 12$$.

Therefore $$\left(\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos\alpha \cdot \cos\beta \cdot \cos\gamma}\right)^2 = 12^2 = 144$$.

Step 1: Simplify the given equation

Start with the given trigonometric equation:

$$\sin^4 \theta + \cos^4 \theta - \sin \theta \cos \theta = 0$$

Use the standard algebraic identity $$a^2 + b^2 = (a+b)^2 - 2ab$$ to rewrite the first two terms, where $$a = \sin^2 \theta$$ and $$b = \cos^2 \theta$$:

$$\sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta$$

Apply the fundamental trigonometric identity $$\sin^2 \theta + \cos^2 \theta = 1$$:

$$\sin^4 \theta + \cos^4 \theta = (1)^2 - 2\sin^2 \theta \cos^2 \theta$$
$$\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta$$

Substitute this simplified expression back into the original equation:

$$1 - 2\sin^2 \theta \cos^2 \theta - \sin \theta \cos \theta = 0$$

Step 2: Solve the quadratic equation

To make the equation easier to solve, let us introduce a temporary variable $$t = \sin \theta \cos \theta$$.

$$1 - 2t^2 - t = 0$$

Rearrange the terms into a standard quadratic format:

$$2t^2 + t - 1 = 0$$

Factorize the quadratic equation:

$$2t^2 + 2t - t - 1 = 0$$
$$2t(t + 1) - 1(t + 1) = 0$$
$$(2t - 1)(t + 1) = 0$$

This gives two possible cases for $$t$$:

• Case 1: $$t = \frac{1}{2} \implies \sin \theta \cos \theta = \frac{1}{2}$$
• Case 2: $$t = -1 \implies \sin \theta \cos \theta = -1$$

Multiply both sides of each case by 2 to utilize the double angle identity $$\sin 2\theta = 2\sin\theta\cos\theta$$:

• From Case 1: $$2\sin \theta \cos \theta = 1 \implies \sin 2\theta = 1$$
• From Case 2: $$2\sin \theta \cos \theta = -2 \implies \sin 2\theta = -2$$

Since the value of any sine function must lie perfectly between $$-1$$ and $$1$$, Case 2 ($$\sin 2\theta = -2$$) is mathematically impossible and is rejected.

Step 3: Find the solutions for the angle

We are left with the valid trigonometric equation:

$$\sin 2\theta = 1$$

The problem specifies the domain for $$\theta$$ is $$[0, 4\pi]$$.

Therefore, the domain for the double angle $$2\theta$$ will be exactly twice that interval, which is $$[0, 8\pi]$$.

We need to find all angles within $$[0, 8\pi]$$ where the sine value is exactly $$1$$. These occur at $$\frac{\pi}{2}$$ and repeat every full cycle of $$2\pi$$:

• $$2\theta = \frac{\pi}{2}$$
• $$2\theta = 2\pi + \frac{\pi}{2} = \frac{5\pi}{2}$$
• $$2\theta = 4\pi + \frac{\pi}{2} = \frac{9\pi}{2}$$
• $$2\theta = 6\pi + \frac{\pi}{2} = \frac{13\pi}{2}$$

Divide each value by 2 to find the individual roots for $$\theta$$:

• $$\theta = \frac{\pi}{4}$$
• $$\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$$
• $$\theta = 2\pi + \frac{\pi}{4} = \frac{9\pi}{4}$$
• $$\theta = 3\pi + \frac{\pi}{4} = \frac{13\pi}{4}$$

Step 4: Calculate the sum and final expression

Let $$S$$ be the sum of all these valid roots:

$$S = \frac{\pi}{4} + \frac{5\pi}{4} + \frac{9\pi}{4} + \frac{13\pi}{4}$$
$$S = \frac{28\pi}{4}$$
$$S = 7\pi$$

The problem asks for the value of the expression $$\frac{8S}{\pi}$$. Substitute our calculated sum $$S$$ into this expression:

$$\text{Value} = \frac{8(7\pi)}{\pi}$$
$$\text{Value} = 8 \times 7$$
$$\text{Value} = 56$$

The final answer is 56.

In $$\triangle ABC$$, we have $$AC = b = 12$$, $$AB = c = 5$$, and area $$= 30$$ cm$$^2$$.

The area of a triangle is $$\frac{1}{2} \cdot AB \cdot AC \cdot \sin A = \frac{1}{2}(5)(12)\sin A = 30$$, so $$\sin A = 1$$, meaning $$A = 90°$$.

Since $$A = 90°$$, by the Pythagorean theorem, $$BC = a = \sqrt{5^2 + 12^2} = \sqrt{25+144} = 13$$.

For a right triangle, the circumradius is $$R = \frac{a}{2} = \frac{13}{2}$$.

The semi-perimeter is $$s = \frac{a+b+c}{2} = \frac{13+12+5}{2} = 15$$.

The inradius is $$r = \frac{\text{Area}}{s} = \frac{30}{15} = 2$$.

Therefore $$2R + r = 2 \times \frac{13}{2} + 2 = 13 + 2 = 15$$.

The answer is $$15$$.

We need to find the minimum value of $$\alpha$$ for which the equation $$\frac{4}{\sin x} + \frac{1}{1 - \sin x} = \alpha$$ has at least one solution in $$\left(0, \frac{\pi}{2}\right)$$.

Let $$t = \sin x$$. Since $$x \in \left(0, \frac{\pi}{2}\right)$$, we have $$t \in (0, 1)$$.

We define $$f(t) = \frac{4}{t} + \frac{1}{1 - t}$$ for $$t \in (0, 1)$$. The equation has a solution if and only if $$\alpha$$ is in the range of $$f(t)$$.

To find the minimum of $$f(t)$$, we compute the derivative: $$f'(t) = -\frac{4}{t^2} + \frac{1}{(1-t)^2}$$.

Setting $$f'(t) = 0$$, we get $$\frac{1}{(1-t)^2} = \frac{4}{t^2}$$, so $$t^2 = 4(1-t)^2$$.

Taking the positive square root (since $$t > 0$$ and $$1 - t > 0$$), $$t = 2(1 - t)$$, which gives $$t = 2 - 2t$$, so $$3t = 2$$ and $$t = \frac{2}{3}$$.

Now $$f\left(\frac{2}{3}\right) = \frac{4}{2/3} + \frac{1}{1 - 2/3} = 6 + 3 = 9$$.

We verify this is a minimum by checking the second derivative or noting that $$f(t) \to \infty$$ as $$t \to 0^+$$ and $$t \to 1^-$$, so the critical point must be a minimum.

Hence, the answer is $$9$$.

We begin with the given trigonometric equation

$$ (k+1)\tan^2x-\sqrt{2}\,\lambda \tan x = (1-k), \qquad k\neq -1, \; \lambda\in\mathbb R. $$

Let us put $$t=\tan x.$$ Then the equation becomes a quadratic in $$t$$:

$$ (k+1)t^{2}-\sqrt2\,\lambda\,t-(1-k)=0. $$

The two real roots of this quadratic correspond to

$$ t_{1}= \tan\alpha,\qquad t_{2}= \tan\beta. $$

For a quadratic $$at^{2}+bt+c=0,$$ the standard Vieta relations state

$$\text{Sum of roots}= -\dfrac{b}{a},\qquad\text{Product of roots}= \dfrac{c}{a}.$$

Here we have $$a=k+1,\; b=-\sqrt2\,\lambda,\; c=-(1-k),$$ so

$$ S=t_{1}+t_{2}= -\frac{-\sqrt2\,\lambda}{\,k+1\,}= \frac{\sqrt2\,\lambda}{k+1}, $$

$$ P=t_{1}t_{2}= \frac{-(1-k)}{\,k+1\,}= \frac{k-1}{k+1}. $$

We are given the extra information

$$\tan^{2}(\alpha+\beta)=50.$$

Using the tangent-addition formula,

$$ \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\,\tan\beta} =\frac{S}{1-P}. $$

Therefore

$$ \tan^{2}(\alpha+\beta)=\left(\frac{S}{1-P}\right)^{2}=50. $$

Substituting the expressions for $$S$$ and $$P$$ we get

$$ \left(\frac{\dfrac{\sqrt2\,\lambda}{k+1}}{\,1-\dfrac{k-1}{k+1}\,}\right)^{2}=50. $$

First calculate the denominator:

$$ 1-P = 1-\frac{k-1}{k+1} = \frac{k+1}{k+1}-\frac{k-1}{k+1} = \frac{(k+1)-(k-1)}{k+1} = \frac{2}{k+1}. $$

Hence

$$ \tan^{2}(\alpha+\beta)=\left(\frac{\sqrt2\,\lambda}{k+1}\,\frac{k+1}{2}\right)^{2} =\left(\frac{\sqrt2\,\lambda}{2}\right)^{2}=50. $$

Simplifying the square gives

$$ \left(\frac{\sqrt2\,\lambda}{2}\right)^{2} =\frac{2\lambda^{2}}{4} =\frac{\lambda^{2}}{2} =50. $$

Multiplying by 2,

$$ \lambda^{2}=100. $$

So

$$ \lambda=\pm 10. $$

The question asks for “a value of $$\lambda$$,” and among the options the only value that appears is $$10.$$ Therefore we select that.

Hence, the correct answer is Option B.

We have been given two infinite series

$$x \;=\; \sum_{n=0}^{\infty}(-1)^n \tan^{2n}\theta \qquad\text{and}\qquad y \;=\; \sum_{n=0}^{\infty}\cos^{2n}\theta,$$ with the restriction $$0<\theta<\dfrac{\pi}{4}.$$

Because $$0<\theta<\dfrac{\pi}{4}$$ we know $$\tan\theta<1$$ and $$\cos\theta<1,$$ so $$|\,-\tan^{2}\theta\,|<1$$ and $$|\cos^{2}\theta|<1.$$ Therefore both series are convergent geometric series, and we may apply the standard infinite-geometric-series formula.

Formula used. For an infinite geometric series with first term $$a$$ and common ratio $$r$$ satisfying $$|r|<1,$$ the sum is

$$S \;=\;\dfrac{a}{1-r}\,.$$

Finding $$x$$. For the series that defines $$x$$, the first term is obtained by putting $$n=0$$:

$$a_x \;=\;(-1)^0\tan^{0}\theta \;=\;1.$$

The common ratio is the factor that multiplies one term to obtain the next. Setting $$n\mapsto n+1$$ we see

$$r_x \;=\;-\,\tan^{2}\theta.$$

Applying the formula,

$$x \;=\;\dfrac{a_x}{1-r_x} \;=\;\dfrac{1}{1-(-\tan^{2}\theta)} \;=\;\dfrac{1}{1+\tan^{2}\theta}.$$

We now invoke the well-known trigonometric identity $$1+\tan^{2}\theta=\sec^{2}\theta.$$ Substituting,

$$x \;=\;\dfrac{1}{\sec^{2}\theta} \;=\;\cos^{2}\theta.$$

Finding $$y$$. For the series that defines $$y$$, the first term is obtained by putting $$n=0$$:

$$a_y \;=\;\cos^{0}\theta \;=\;1.$$

The common ratio is

$$r_y \;=\;\cos^{2}\theta.$$

Again using the formula,

$$y \;=\;\dfrac{a_y}{1-r_y} \;=\;\dfrac{1}{1-\cos^{2}\theta}.$$

From the Pythagorean identity $$\sin^{2}\theta +\cos^{2}\theta =1$$ we have $$1-\cos^{2}\theta=\sin^{2}\theta.$$ Substituting,

$$y \;=\;\dfrac{1}{\sin^{2}\theta} \;=\;\csc^{2}\theta.$$

Evaluating the required combinations. The option that interests us most is $$y(1-x).$$ Let us compute it step by step:

$$y(1-x) \;=\;\csc^{2}\theta\bigl(1-\cos^{2}\theta\bigr).$$

But we already know $$1-\cos^{2}\theta=\sin^{2}\theta.$$ Substituting,

$$y(1-x) \;=\;\csc^{2}\theta\,\sin^{2}\theta \;=\;\dfrac{1}{\sin^{2}\theta}\,\sin^{2}\theta \;=\;1.$$

Thus we have obtained the identity

$$y(1-x)=1.$$

No other option yields the value $$1$$, so the relationship demanded by the question corresponds precisely to option B.

Hence, the correct answer is Option B.

Step 1: Apply Complementary Angle Transformations

Notice that the angles $$\frac{\pi}{8}$$ and $$\frac{3\pi}{8}$$ are complementary because they add up to $$\frac{\pi}{2}$$.

Therefore, we can establish the following relationships:

• $$\cos\left(\frac{3\pi}{8}\right) = \sin\left(\frac{\pi}{2} - \frac{3\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right)$$
• $$\sin\left(\frac{3\pi}{8}\right) = \cos\left(\frac{\pi}{2} - \frac{3\pi}{8}\right) = \cos\left(\frac{\pi}{8}\right)$$

Step 2: Substitute and Factor the Expression

The original given expression is:

$$E = \cos^3\left(\frac{\pi}{8}\right) \cos\left(\frac{3\pi}{8}\right) + \sin^3\left(\frac{\pi}{8}\right) \sin\left(\frac{3\pi}{8}\right)$$

Now, substitute the transformed values from Step 1 into this expression:

$$E = \cos^3\left(\frac{\pi}{8}\right) \sin\left(\frac{\pi}{8}\right) + \sin^3\left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8}\right)$$

Notice that both terms share a common factor of $$\sin\left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8}\right)$$. Let us factor that out:

$$E = \sin\left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8}\right) \left[ \cos^2\left(\frac{\pi}{8}\right) + \sin^2\left(\frac{\pi}{8}\right) \right]$$

Step 3: Simplify using Trigonometric Identities

Apply the fundamental Pythagorean identity $$\cos^2 \theta + \sin^2 \theta = 1$$ to the terms inside the bracket:

$$E = \sin\left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8}\right) \times 1$$
$$E = \sin\left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8}\right)$$

To proceed, multiply and divide the entire expression by $$2$$ to create the standard double angle formula $$\sin(2\theta) = 2 \sin \theta \cos \theta$$:

$$E = \frac{1}{2} \left[ 2 \sin\left(\frac{\pi}{8}\right) \cos\left(\frac{\pi}{8}\right) \right]$$
$$E = \frac{1}{2} \sin\left(2 \times \frac{\pi}{8}\right)$$
$$E = \frac{1}{2} \sin\left(\frac{\pi}{4}\right)$$

Step 4: Final Calculation

We know the standard value for $$\sin\left(\frac{\pi}{4}\right)$$ is $$\frac{1}{\sqrt{2}}$$.

$$E = \frac{1}{2} \times \frac{1}{\sqrt{2}}$$
$$E = \frac{1}{2\sqrt{2}}$$

The correct answer is Option B.

Step 1: Calculate the Sum (L + M)

First, let us add the two given expressions together.

$$L = \sin^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)$$
$$M = \cos^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)$$

Adding them yields:

$$L + M = \left[\sin^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)\right] + \left[\cos^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)\right]$$

Group the terms with the same angles:

$$L + M = \left[\sin^2\left(\frac{\pi}{16}\right) + \cos^2\left(\frac{\pi}{16}\right)\right] - 2\sin^2\left(\frac{\pi}{8}\right)$$

Apply the fundamental identity $$\sin^2\theta + \cos^2\theta = 1$$ to the first bracket:

$$L + M = 1 - 2\sin^2\left(\frac{\pi}{8}\right)$$

Now, apply the double angle identity $$\cos(2\theta) = 1 - 2\sin^2\theta$$:

$$L + M = \cos\left(2 \times \frac{\pi}{8}\right)$$
$$L + M = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$$

Step 2: Calculate the Difference (L - M)

Next, let us subtract $$M$$ from $$L$$.

$$L - M = \left[\sin^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)\right] - \left[\cos^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)\right]$$

The $$\sin^2\left(\frac{\pi}{8}\right)$$ terms cancel out:

$$L - M = \sin^2\left(\frac{\pi}{16}\right) - \cos^2\left(\frac{\pi}{16}\right)$$

Factor out a negative sign to match the standard double angle formula:

$$L - M = -\left[\cos^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{16}\right)\right]$$

Apply the double angle identity $$\cos(2\theta) = \cos^2\theta - \sin^2\theta$$:

$$L - M = -\cos\left(2 \times \frac{\pi}{16}\right)$$
$$L - M = -\cos\left(\frac{\pi}{8}\right)$$

Step 3: Solve for M

We now have a simple system of two equations:

1. $$L + M = \frac{1}{\sqrt{2}}$$
2. $$L - M = -\cos\left(\frac{\pi}{8}\right)$$

To isolate $$M$$, subtract equation (2) from equation (1):

$$(L + M) - (L - M) = \frac{1}{\sqrt{2}} - \left(-\cos\left(\frac{\pi}{8}\right)\right)$$
$$2M = \frac{1}{\sqrt{2}} + \cos\left(\frac{\pi}{8}\right)$$

Divide the entire equation by 2 to find the final value of $$M$$:

$$M = \frac{1}{2\sqrt{2}} + \frac{1}{2}\cos\left(\frac{\pi}{8}\right)$$

Hence, the answer is option D.

We begin with the given equation

$$\cos^4\theta+\sin^4\theta+\lambda=0.$$

For real values of $$\theta$$ to exist, the trigonometric part $$\cos^4\theta+\sin^4\theta$$ must be equal to $$-\lambda$$ for some $$\theta$$. So we first find the range of the expression

$$f(\theta)=\cos^4\theta+\sin^4\theta.$$

Write $$\cos^4\theta+\sin^4\theta$$ in terms of $$\cos^2\theta$$ and $$\sin^2\theta$$. Let

$$a=\cos^2\theta,\qquad b=\sin^2\theta.$$

We know the Pythagorean identity $$a+b=1$$ and, of course, $$0\le a\le1,\;0\le b\le1.$$ Using the algebraic identity $$a^2+b^2=(a+b)^2-2ab,$$ we have

$$\cos^4\theta+\sin^4\theta =a^2+b^2 =(a+b)^2-2ab =1^2-2ab =1-2\cos^2\theta\sin^2\theta.$$

Next, recall the double-angle formula $$\sin2\theta=2\sin\theta\cos\theta,$$ so that

$$\sin^22\theta=4\sin^2\theta\cos^2\theta \;\Longrightarrow\; \cos^2\theta\sin^2\theta=\dfrac{\sin^22\theta}{4}.$$

Substituting this into the previous expression gives

$$f(\theta)=1-2\left(\dfrac{\sin^22\theta}{4}\right) =1-\dfrac{1}{2}\sin^22\theta.$$

The term $$\sin^22\theta$$ always lies in the interval $$[0,1]$$ because the square of a sine function never exceeds 1 and is never negative. Hence

$$-\dfrac12\le -\dfrac{1}{2}\sin^22\theta\le0,$$ and adding 1 to every part we obtain the range of $$f(\theta):$$

$$\boxed{\dfrac12\le f(\theta)\le1}.$$

This means the expression $$\cos^4\theta+\sin^4\theta$$ can take any value from $$\dfrac12$$ up to $$1$$, inclusive.

Now the original equation can be written as

$$f(\theta)+\lambda=0 \;\Longrightarrow\; f(\theta)=-\lambda.$$

Thus, for some real $$\theta,$$ the number $$-\lambda$$ must lie in the interval $$\left[\dfrac12,1\right]$$ determined above. Multiplying that interval by $$-1$$ reverses the inequality signs, giving the permissible interval for $$\lambda:$$

$$-\left[\,\dfrac12,1\,\right]=\left[-1,-\dfrac12\right].$$

We include the endpoints because both $$\sin^22\theta=0$$ (giving $$f(\theta)=1$$) and $$\sin^22\theta=1$$ (giving $$f(\theta)=\dfrac12$$) are achievable for suitable values of $$\theta$$ (for example, $$\theta=0$$ and $$\theta=\dfrac{\pi}{4}$$ respectively).

Therefore

$$\boxed{\lambda\in\left[-1,-\dfrac12\right]}.$$

Among the given choices, this corresponds exactly to Option B.

Hence, the correct answer is Option B.

Let the still surface of the lake be a horizontal plane. Choose a point $$A$$ on this surface such that the given point $$P$$ lies vertically above $$A$$. We are told that $$P$$ is 200 m above the lake, so $$PA = 200\text{ m}$$.

Let the cloud be $$C$$ and its image in the lake be $$C'$$. Because an image in a perfectly still lake is formed by reflection in the water surface, $$C$$ and $$C'$$ are on the same vertical line through $$A'$$, with $$A'$$ exactly midway between them. If we denote the height of the cloud above the lake by $$h$$ metres, then

$$A'C = h,\qquad A'C' = h,\qquad \text{and}\qquad C'$$ is $$h$$ metres below the water surface.

Put the horizontal distance of both $$C$$ and $$C'$$ from the foot $$A$$ as $$x$$ metres. Thus the coordinates (taking the lake surface as the $$x$$-axis and up as positive $$y$$) can be visualised as

$$A(0,0),\; P(0,200),\; C(x,h),\; C'(x,-h).$$

The angle of elevation of $$C$$ from $$P$$ is 30°. By the definition of tangent in a right triangle,

$$\tan 30^{\circ} \;=\; \frac{\text{opposite side}}{\text{adjacent side}} \;=\; \frac{(h-200)}{x}.$$

Since $$\tan 30^{\circ} = \frac{1}{\sqrt{3}}$$, we get

$$\frac{1}{\sqrt{3}} = \frac{h-200}{x} \;\;\Longrightarrow\;\; x = \sqrt{3}\,(h-200). \quad (1)$$

Next, the angle of depression of the image $$C'$$ from $$P$$ is 60°. The angle of depression equals the angle the line of sight makes below the horizontal. Using the same tangent definition, we have

$$\tan 60^{\circ} = \frac{\text{vertical drop from }P\text{ to }C'}{\text{horizontal distance }x} = \frac{200+h}{x}.$$

Because $$\tan 60^{\circ} = \sqrt{3}$$, this yields

$$\sqrt{3} = \frac{200+h}{x} \;\;\Longrightarrow\;\; x = \frac{200+h}{\sqrt{3}}. \quad (2)$$

Both equations (1) and (2) represent the same value of $$x$$, so we equate them:

$$\sqrt{3}(h-200) = \frac{200+h}{\sqrt{3}}.$$

Multiplying both sides by $$\sqrt{3}$$ to clear the denominator,

$$3(h-200) = 200 + h.$$

Expanding and rearranging step by step,

$$3h - 600 = 200 + h$$
$$3h - h = 200 + 600$$
$$2h = 800$$
$$h = 400.$$ So the cloud is $$400\text{ m}$$ above the lake.

The vertical separation between $$P$$ and $$C$$ is therefore

$$\text{vertical difference} = h - 200 = 400 - 200 = 200\text{ m}.$$

Substituting $$h = 400$$ into equation (1) (or equation (2)) to find $$x$$,

$$x = \sqrt{3}(h-200) = \sqrt{3}\times 200 = 200\sqrt{3}\text{ m}.$$

Now we know both the horizontal and vertical components of the line segment $$PC$$. Using the Pythagoras theorem, which states $$\text{hypotenuse}^2 = \text{base}^2 + \text{perpendicular}^2,$$ we have

$$PC = \sqrt{(200)^2 + (200\sqrt{3})^2} = \sqrt{40000 + 120000} = \sqrt{160000} = 400\text{ m}.$$

Hence, the correct answer is Option C.

Let us denote the foot of the mountain by point $$B$$, the first observation point on the ground by $$A$$ and the summit of the mountain by $$S$$. We are asked to find the height $$BS$$ (in km) of the mountain.

We first translate the given angles into the language of right-angle trigonometry. From point $$A$$ the angle of elevation of the summit is $$45^\circ$$. Using the definition of the tangent function,

$$\tan 45^\circ \;=\;\frac{\text{opposite side}}{\text{adjacent side}} \;=\;\frac{BS}{AB}.$$

Because $$\tan 45^\circ = 1$$, we obtain the simple relation

$$BS = AB.$$

We now move from $$A$$ a distance of exactly $$1\text{ km}$$ towards the summit along a path that is inclined at $$30^\circ$$ to the horizontal. Let the new point reached on this path be $$C$$. Since the traveller is going “towards the summit”, the path $$AC$$ lies in the plane containing $$AB$$ and $$BS$$. From elementary resolution of a line segment inclined at $$30^\circ$$ we get:

vertical rise from $$A$$ to $$C$$: $$AC \sin 30^\circ = 1 \times \frac12 = 0.5\text{ km},$$

horizontal advance (towards the foot $$B$$): $$AC \cos 30^\circ = 1 \times \frac{\sqrt3}{2} = \frac{\sqrt3}{2}\text{ km}.$$

Hence the height of point $$C$$ above the ground is

$$\text{height of }C = 0.5\text{ km},$$

and the remaining vertical distance from $$C$$ to the summit is

$$SC = BS - 0.5\text{ km}.$$

Meanwhile the horizontal distance from point $$C$$ to the foot $$B$$ has decreased by $$\dfrac{\sqrt3}{2}\text{ km}$$, so it is now

$$BC = AB - \frac{\sqrt3}{2}\text{ km}.$$

But we already know $$AB = BS$$, so

$$BC = BS - \frac{\sqrt3}{2}.$$

From point $$C$$ the angle of elevation of the summit is stated to be $$60^\circ$$. Applying the tangent definition once more, this time in right-angled triangle $$B C S$$, we have

$$\tan 60^\circ = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{SC}{BC} = \frac{BS - 0.5}{\,BS - \dfrac{\sqrt3}{2}\,}.$$

Because $$\tan 60^\circ = \sqrt3$$, we write

$$\sqrt3 = \frac{BS - 0.5}{BS - \dfrac{\sqrt3}{2}}.$$

To remove the denominator, we cross-multiply:

$$\sqrt3\Bigl(BS - \frac{\sqrt3}{2}\Bigr) = BS - 0.5.$$

Expanding the left-hand side gives

$$\sqrt3\,BS - \sqrt3\!\left(\frac{\sqrt3}{2}\right) = BS - 0.5,$$

and since $$\sqrt3 \times \sqrt3 = 3$$, this becomes

$$\sqrt3\,BS - \frac{3}{2} = BS - 0.5.$$

Now we gather the $$BS$$ terms on one side and the constants on the other side:

$$\sqrt3\,BS - BS = \frac{3}{2} - 0.5.$$

The right-hand constant simplifies because $$0.5 = \dfrac{1}{2}$$, so

$$\sqrt3\,BS - BS = \frac{3}{2} - \frac{1}{2} = 1.$$

Factoring $$BS$$ out of the left-hand side, we obtain

$$(\sqrt3 - 1)\,BS = 1.$$

Finally we solve for $$BS$$ (the required height):

$$BS = \frac{1}{\sqrt3 - 1}.$$

This expression exactly matches Option C given in the problem statement. Hence, the correct answer is Option C.

Step 1: Apply the AM GM Inequality

Since exponential functions always yield positive values, both $$2^{\sin x}$$ and $$2^{\cos x}$$ are strictly positive for all real values of $$x$$. This allows us to apply the Arithmetic Mean Geometric Mean inequality, which states that for any two positive numbers $$a$$ and $$b$$:

$$\frac{a + b}{2} \ge \sqrt{ab}$$

Substitute $$a = 2^{\sin x}$$ and $$b = 2^{\cos x}$$:

$$\frac{2^{\sin x} + 2^{\cos x}}{2} \ge \sqrt{2^{\sin x} \cdot 2^{\cos x}}$$

Step 2: Simplify the Right Hand Side

Use the exponent addition property $$a^m \cdot a^n = a^{m+n}$$ to combine the terms inside the square root:

$$\frac{2^{\sin x} + 2^{\cos x}}{2} \ge \sqrt{2^{\sin x + \cos x}}$$

Express the square root as a fractional exponent of $$\frac{1}{2}$$:

$$2^{\sin x} + 2^{\cos x} \ge 2 \cdot (2^{\sin x + \cos x})^{\frac{1}{2}}$$
$$2^{\sin x} + 2^{\cos x} \ge 2^1 \cdot 2^{\frac{\sin x + \cos x}{2}}$$

Combine the base 2 terms on the right side by adding their exponents:

$$2^{\sin x} + 2^{\cos x} \ge 2^{1 + \frac{\sin x + \cos x}{2}}$$

Step 3: Minimize the Exponent

For the entire expression to reach its absolute minimum, the exponent $$1 + \frac{\sin x + \cos x}{2}$$ must be minimized.

Recall the standard range for any trigonometric expression of the form $$a\sin x + b\cos x$$ is $$[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$$.

For our specific term $$\sin x + \cos x$$, the coefficients are $$a=1$$ and $$b=1$$.

Therefore, the minimum possible value of $$\sin x + \cos x$$ is $$-\sqrt{1^2 + 1^2} = -\sqrt{2}$$.

Substitute this minimum value back into the exponent's fraction:

$$\text{Minimum exponent} = 1 + \frac{-\sqrt{2}}{2}$$

Simplify the fraction by recognizing that $$2 = \sqrt{2} \cdot \sqrt{2}$$:

$$1 - \frac{\sqrt{2}}{2} = 1 - \frac{1}{\sqrt{2}}$$

Step 4: Final Evaluation

Substitute the fully minimized exponent back into our inequality from Step 2:

$$2^{\sin x} + 2^{\cos x} \ge 2^{1 - \frac{1}{\sqrt{2}}}$$

This proves that the absolute minimum value of the expression is exactly $$2^{1 - \frac{1}{\sqrt{2}}}$$.

Hence, the answer is option D.

We have to solve the trigonometric equation $$2\cot^2\theta-\dfrac{5}{\sin\theta}+4=0$$ for $$\theta\in(0,2\pi)\setminus\{\pi\}$$, obtain the smallest root $$\theta_1$$ and the largest root $$\theta_2$$ in that interval, and then evaluate the definite integral $$\displaystyle\int_{\theta_1}^{\theta_2}\cos^2 3\theta\,d\theta$$.

First, recall that $$\cot\theta=\dfrac{\cos\theta}{\sin\theta}$$, so $$\cot^2\theta=\dfrac{\cos^2\theta}{\sin^2\theta}$$. Substituting this in the given equation we get

$$2\left(\dfrac{\cos^2\theta}{\sin^2\theta}\right)-\dfrac{5}{\sin\theta}+4=0.$$

To clear denominators we multiply every term by $$\sin^2\theta$$:

$$2\cos^2\theta-5\sin\theta+4\sin^2\theta=0.$$

Now we express $$\cos^2\theta$$ in terms of $$\sin\theta$$ by using the Pythagorean identity $$\cos^2\theta=1-\sin^2\theta$$. Let $$x=\sin\theta$$ for simplicity. Substituting $$\cos^2\theta=1-x^2$$ gives

$$2(1-x^2)-5x+4x^2=0.$$

Expanding and collecting like terms:

$$2-2x^2-5x+4x^2=0,$$

$$2+2x^2-5x=0.$$

Re-ordering, we get the quadratic equation

$$2x^2-5x+2=0.$$

To find its roots we use the quadratic formula. For $$ax^2+bx+c=0$$, the roots are $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$. Here $$a=2,\;b=-5,\;c=2$$, so

$$x=\dfrac{5\pm\sqrt{(-5)^2-4\cdot2\cdot2}}{2\cdot2} =\dfrac{5\pm\sqrt{25-16}}{4} =\dfrac{5\pm\sqrt{9}}{4} =\dfrac{5\pm3}{4}.$$

This gives two numerical roots:

$$x_1=\dfrac{5+3}{4}=2,\qquad x_2=\dfrac{5-3}{4}=\dfrac12.$$

Since $$x=\sin\theta$$ must satisfy $$-1\le x\le1$$, the value $$x=2$$ is impossible. Thus the only admissible solution is

$$\sin\theta=\dfrac12.$$

Within the interval $$(0,2\pi)$$ (excluding $$\pi$$) the equation $$\sin\theta=\dfrac12$$ is satisfied at

$$\theta=\dfrac\pi6\quad\text{and}\quad\theta=\dfrac{5\pi}6.$$

There are no other values in $$\bigl(\pi,2\pi\bigr)$$ with $$\sin\theta=\dfrac12$$ because $$\sin\theta$$ becomes negative beyond $$\pi$$. Therefore,

$$\theta_1=\dfrac{\pi}{6},\qquad \theta_2=\dfrac{5\pi}{6}.$$

Now we have to evaluate the definite integral

$$\int_{\theta_1}^{\theta_2}\cos^2 3\theta\,d\theta =\int_{\pi/6}^{5\pi/6}\cos^2 3\theta\,d\theta.$$

We use the power-reduction identity for the square of cosine:

$$\cos^2\alpha=\dfrac{1+\cos2\alpha}{2}.$$

With $$\alpha=3\theta$$, this becomes

$$\cos^2 3\theta=\dfrac{1+\cos6\theta}{2}.$$

Substituting into the integral gives

$$\int_{\pi/6}^{5\pi/6}\cos^2 3\theta\,d\theta =\int_{\pi/6}^{5\pi/6}\dfrac{1+\cos6\theta}{2}\,d\theta.$$

We can split and integrate term by term:

$$\int_{\pi/6}^{5\pi/6}\dfrac{1}{2}\,d\theta+\int_{\pi/6}^{5\pi/6}\dfrac{\cos6\theta}{2}\,d\theta =\dfrac12\int_{\pi/6}^{5\pi/6}d\theta+\dfrac12\int_{\pi/6}^{5\pi/6}\cos6\theta\,d\theta.$$

The first integral is straightforward:

$$\dfrac12\int_{\pi/6}^{5\pi/6}d\theta=\dfrac12\bigl[\theta\bigr]_{\pi/6}^{5\pi/6} =\dfrac12\left(\dfrac{5\pi}{6}-\dfrac{\pi}{6}\right) =\dfrac12\cdot\dfrac{4\pi}{6} =\dfrac{4\pi}{12} =\dfrac{\pi}{3}.$$

For the second integral we use the standard antiderivative $$\int\cos k\theta\,d\theta=\dfrac{\sin k\theta}{k}.$$ Here $$k=6,$$ so

$$\int\cos6\theta\,d\theta=\dfrac{\sin6\theta}{6}.$$

Therefore,

$$\dfrac12\int_{\pi/6}^{5\pi/6}\cos6\theta\,d\theta =\dfrac12\left[\dfrac{\sin6\theta}{6}\right]_{\pi/6}^{5\pi/6} =\dfrac{1}{12}\bigl[\sin6\theta\bigr]_{\pi/6}^{5\pi/6}.$$

Now we evaluate $$\sin6\theta$$ at the limits:

At $$\theta=\dfrac{5\pi}{6}:$$ $$6\theta=5\pi,$$ so $$\sin5\pi=0.$$

At $$\theta=\dfrac{\pi}{6}:$$ $$6\theta=\pi,$$ so $$\sin\pi=0.$$

Hence

$$\dfrac{1}{12}\bigl[\sin6\theta\bigr]_{\pi/6}^{5\pi/6} =\dfrac{1}{12}(0-0)=0.$$

Adding the two parts, the value of the integral is

$$\dfrac{\pi}{3}+0=\dfrac{\pi}{3}.$$

Hence, the correct answer is Option A.

Let the foot of the hill be the point $$B$$ and let the top of the hill be the point $$T$$. We take the horizontal plane through $$B$$ as our reference ground level.

We first stand at a point $$P$$ on this horizontal plane such that the angle of elevation of $$T$$ from $$P$$ is $$45^\circ$$. If the horizontal distance $$BP$$ is denoted by $$x$$ metres and the vertical height of the hill $$BT$$ is denoted by $$h$$ metres, then by the definition of tangent we have

$$\tan 45^\circ=\frac{\text{opposite}}{\text{adjacent}}=\frac{h}{x}.$$

Since $$\tan 45^\circ=1$$, this at once gives

$$h=x.$$

Now we walk from $$P$$ directly towards the top, along a path that is inclined at $$30^\circ$$ to the horizontal, and we cover a distance of $$80$$ metres along this path. Let the new point of observation be $$Q$$. Because the path makes an angle of $$30^\circ$$ with the horizontal, we can resolve the $$80$$ m into horizontal and vertical components using the standard relations

$$\text{horizontal component}=80\cos30^\circ,\qquad\text{vertical component}=80\sin30^\circ.$$

Recalling that $$\cos30^\circ=\dfrac{\sqrt3}{2}$$ and $$\sin30^\circ=\dfrac12$$, we obtain

$$\text{horizontal component}=80\left(\frac{\sqrt3}{2}\right)=40\sqrt3,$$ $$\text{vertical component}=80\left(\frac12\right)=40.$$

Thus, compared with point $$P$$, the point $$Q$$ is $$40\sqrt3$$ m closer to the foot $$B$$ horizontally and $$40$$ m higher vertically. Consequently the horizontal distance from $$Q$$ to $$B$$ equals

$$x-40\sqrt3,$$

and the height of $$Q$$ above the ground plane equals

$$40\text{ m}.$$

From $$Q$$ the angle of elevation of the top $$T$$ is now $$75^\circ$$. The vertical separation between $$Q$$ and $$T$$ is

$$h-40,$$

while the horizontal separation is

$$x-40\sqrt3.$$

Applying the definition of tangent again, we write

$$\tan75^\circ=\frac{h-40}{\,x-40\sqrt3\,}.$$

But we already have $$h=x$$ from the first observation, so we substitute $$x=h$$ to get

$$\tan75^\circ=\frac{h-40}{\,h-40\sqrt3\,}.$$

To solve for $$h$$ we first recall the exact value of $$\tan75^\circ$$. Using the angle-addition formula

$$\tan(45^\circ+30^\circ)=\frac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ},$$

and the facts $$\tan45^\circ=1$$ and $$\tan30^\circ=\dfrac1{\sqrt3},$$ we find

$$\tan75^\circ=\frac{1+\dfrac1{\sqrt3}}{1-1\cdot\dfrac1{\sqrt3}} =\frac{\sqrt3+1}{\sqrt3-1}.$$

Introduce the shorthand $$t=\tan75^\circ=\dfrac{\sqrt3+1}{\sqrt3-1}.$$ Our equation becomes

$$\frac{h-40}{h-40\sqrt3}=t.$$

Cross-multiplying,

$$h-40=t(h-40\sqrt3).$$

Expand the right side:

$$h-40=th-40t\sqrt3.$$

Rearrange to collect the $$h$$ terms on the left and constants on the right:

$$h-th=40-40t\sqrt3.$$

Factor out $$h$$ on the left:

$$h(1-t)=40(1-t\sqrt3).$$

Finally, divide by the coefficient $$1-t$$ (noting that $$t\neq1$$) to obtain $$h$$ explicitly:

$$h=40\frac{1-t\sqrt3}{1-t}.$$

We now substitute the exact value of $$t$$. First compute $$1-t$$ and $$1-t\sqrt3$$ separately:

$$1-t=1-\frac{\sqrt3+1}{\sqrt3-1} =\frac{\sqrt3-1-(\sqrt3+1)}{\sqrt3-1} =\frac{-2}{\sqrt3-1},$$

$$1-t\sqrt3=1-\sqrt3\left(\frac{\sqrt3+1}{\sqrt3-1}\right) =\frac{\sqrt3-1-(3+\sqrt3)}{\sqrt3-1} =\frac{-4}{\sqrt3-1}.$$

Taking the ratio,

$$\frac{1-t\sqrt3}{1-t} =\frac{\dfrac{-4}{\sqrt3-1}}{\dfrac{-2}{\sqrt3-1}} =\frac{-4}{-2}=2.$$

Therefore

$$h=40\times2=80.$$

Thus the height of the hill is $$80\text{ metres}$$.

So, the answer is $$80$$.

We start with the given transcendental equation

$$\log_{\frac12}\!\bigl|\sin x\bigr| \;=\; 2 \;-\; \log_{\frac12}\!\bigl|\cos x\bigr|$$

in the interval $$[0,\,2\pi]$$. Here the base of every logarithm is $$\dfrac12,$$ which lies between 0 and 1, so all logarithms are well-defined only when their arguments are positive; that already guarantees $$\sin x\neq0$$ and $$\cos x\neq0$$ for each admissible solution.

First we transpose the term $$-\log_{\frac12}\!\bigl|\cos x\bigr|$$ to the left side:

$$\log_{\frac12}\!\bigl|\sin x\bigr| \;+\; \log_{\frac12}\!\bigl|\cos x\bigr| \;=\; 2.$$

Now we invoke the well-known product rule for logarithms, stated as

$$\log_b m + \log_b n \;=\; \log_b(mn)\quad\text{for the same base }b.$$

Applying this rule with $$b=\dfrac12,\;m=|\sin x|,\;n=|\cos x|,$$ we obtain

$$\log_{\frac12}\!\bigl(|\sin x|\,|\cos x|\bigr) \;=\; 2.$$

Next we convert this logarithmic statement to its equivalent exponential form. For any base $$b\;(0<b\neq1)$$ we have the definition

$$\log_b A = C \;\Longleftrightarrow\; A = b^{\,C}.$$

Hence

$$|\sin x|\,|\cos x| \;=\; \Bigl(\tfrac12\Bigr)^{2} \;=\; \tfrac14.$$

Because the left-hand side contains the product of absolute values, we rewrite it through the double‐angle identity. We recall the formula

$$\sin 2x \;=\; 2\sin x\cos x,$$

and from it we find

$$|\sin x|\,|\cos x| \;=\; \tfrac12\,|2\sin x\cos x| \;=\; \tfrac12\,|\sin 2x|.$$

Substituting this expression into the equation $$|\sin x|\,|\cos x|=\tfrac14$$ gives

$$\tfrac12\,|\sin 2x| \;=\; \tfrac14,$$

and multiplying both sides by 2 yields the simpler condition

$$|\sin 2x| \;=\; \tfrac12.$$

Thus our task reduces to solving

$$\bigl|\sin 2x\bigr| = \frac12$$

for $$x$$ in the original domain $$[0,\,2\pi].$$ Set $$\theta = 2x;$$ then $$\theta$$ runs through the interval $$[0,\,4\pi]$$ while $$x$$ runs through $$[0,\,2\pi].$$ Therefore we must find all $$\theta\in[0,\,4\pi]$$ such that

$$|\sin\theta| = \frac12.$$

We know that $$\sin\theta = \pm\dfrac12$$ at the angles

$$\theta = \frac{\pi}{6},\;\frac{5\pi}{6},\;\frac{7\pi}{6},\;\frac{11\pi}{6}$$

within the first full cycle $$[0,\,2\pi].$$ Since the sine function has period $$2\pi,$$ the same four angles recur once more when we add $$2\pi,$$ giving in the second cycle

$$\theta = \frac{\pi}{6}+2\pi,\;\frac{5\pi}{6}+2\pi,\;\frac{7\pi}{6}+2\pi,\;\frac{11\pi}{6}+2\pi.$$

Altogether, in the whole interval $$[0,\,4\pi]$$ we therefore obtain eight distinct solutions for $$\theta.$$ Explicitly, they are

$$\theta = \frac{\pi}{6},\;\frac{5\pi}{6},\;\frac{7\pi}{6},\;\frac{11\pi}{6},\; \frac{13\pi}{6},\;\frac{17\pi}{6},\;\frac{19\pi}{6},\;\frac{23\pi}{6}.$$

Since $$\theta = 2x,$$ each such $$\theta$$ corresponds to a unique $$x=\dfrac{\theta}{2}$$ lying in $$[0,\,2\pi].$$ None of these $$x$$-values makes $$\sin x$$ or $$\cos x$$ equal to zero, so all are admissible for the original logarithmic equation. Consequently there are exactly eight distinct $$x$$ satisfying the given equation in the stated interval.

So, the answer is $$8$$.

We have been given two separate relations and both angles lie in the first quadrant, that is $$0<\alpha<\dfrac{\pi}{2}$$ and $$0<\beta<\dfrac{\pi}{2}$$, so all trigonometric ratios of these angles are positive.

First of all, consider the equation

$$\dfrac{\sqrt{2}\,\sin\alpha}{\sqrt{1+\cos 2\alpha}}=\dfrac{1}{7}.$$

We recall the double-angle identity

$$\cos 2\alpha = 2\cos^2\alpha - 1,$$

which can be rearranged to give

$$1 + \cos 2\alpha = 2\cos^2\alpha.$$

Substituting this into the denominator we obtain

$$\sqrt{1+\cos 2\alpha} = \sqrt{2\cos^2\alpha} = \sqrt{2}\,\cos\alpha,$$

because $$\cos\alpha>0$$ in the given interval and therefore the absolute value is not required.

Now we rewrite the left‐hand side of the original equation:

$$\dfrac{\sqrt{2}\,\sin\alpha}{\sqrt{2}\,\cos\alpha} = \dfrac{\sin\alpha}{\cos\alpha} = \tan\alpha.$$

Thus we deduce

$$\tan\alpha = \dfrac{1}{7}.$$

Next, consider the second relation

$$\sqrt{\dfrac{1-\cos 2\beta}{2}} = \dfrac{1}{\sqrt{10}}.$$

We recall the familiar identity

$$\sin^2\beta = \dfrac{1-\cos 2\beta}{2}.$$

Taking the positive square root (again justified by $$0<\beta<\dfrac{\pi}{2}$$) we get

$$\sin\beta = \dfrac{1}{\sqrt{10}}.$$

Using the Pythagorean relation $$\sin^2\beta + \cos^2\beta = 1,$$ we find

$$\cos\beta = \sqrt{1 - \sin^2\beta} = \sqrt{1 - \dfrac{1}{10}} = \sqrt{\dfrac{9}{10}} = \dfrac{3}{\sqrt{10}}.$$

Hence

$$\tan\beta = \dfrac{\sin\beta}{\cos\beta} = \dfrac{\dfrac{1}{\sqrt{10}}}{\dfrac{3}{\sqrt{10}}} = \dfrac{1}{3}.$$

Our goal is to compute $$\tan(\alpha + 2\beta).$$ We already know $$\tan\alpha$$ and $$\tan\beta,$$ so we next find $$\tan 2\beta.$$ The double-angle formula for tangent is

$$\tan 2\beta = \dfrac{2\tan\beta}{1-\tan^2\beta}.$$

Substituting $$\tan\beta = \dfrac{1}{3}$$ gives

$$\tan 2\beta = \dfrac{2\left(\dfrac{1}{3}\right)}{1-\left(\dfrac{1}{3}\right)^2} = \dfrac{\dfrac{2}{3}}{1-\dfrac{1}{9}} = \dfrac{\dfrac{2}{3}}{\dfrac{8}{9}} = \dfrac{2}{3}\times\dfrac{9}{8} = \dfrac{3}{4}.$$

Now we use the angle‐addition formula for tangent:

$$\tan(\alpha + 2\beta) = \dfrac{\tan\alpha + \tan 2\beta}{1 - \tan\alpha\,\tan 2\beta}.$$

We substitute $$\tan\alpha = \dfrac{1}{7}$$ and $$\tan 2\beta = \dfrac{3}{4}$$:

First the numerator:

$$\tan\alpha + \tan 2\beta = \dfrac{1}{7} + \dfrac{3}{4} = \dfrac{4}{28} + \dfrac{21}{28} = \dfrac{25}{28}.$$

Then the denominator:

$$1 - \tan\alpha\,\tan 2\beta = 1 - \left(\dfrac{1}{7}\right)\left(\dfrac{3}{4}\right) = 1 - \dfrac{3}{28} = \dfrac{25}{28}.$$

Hence

$$\tan(\alpha + 2\beta) = \dfrac{\dfrac{25}{28}}{\dfrac{25}{28}} = 1.$$

Hence, the correct answer is Option 1.

We are given the equality

$$\sin^4\alpha + 4\cos^4\beta + 2 = 4\sqrt2\;\sin\alpha\cos\beta\,,\qquad \alpha,\,\beta\in[0,\pi].$$

Because $$\sin\alpha\ge 0$$ for every $$\alpha\in[0,\pi],$$ we set

$$\sin\alpha=\sqrt{s}\quad(s\ge 0),\qquad \cos\beta = C\quad(C\in[-1,1]).$$

Then $$s=\sin^2\alpha\in[0,1]$$ and $$C^2=\cos^2\beta\in[0,1].$$ Substituting these symbols in the given equation we obtain

$$s^2 + 4C^4 + 2 = 4\sqrt2\,\sqrt{s}\,C. \quad -(1)$$

The sign of the right-hand side is the sign of $$C$$. The left-hand side is clearly positive. Hence $$C$$ must be non-negative, i.e. $$C\ge 0\;(0\le\beta\le\pi/2).$$ Therefore we can write $$C=\sqrt{c}\,(c=C^2).$$ It is more convenient to remove the radicals completely, so we introduce two new non-negative variables

$$v=\sqrt{s}=\sin\alpha,\qquad u=\sqrt2\,C=\sqrt2\,\cos\beta.$$

Because $$0\le s\le 1$$ and $$0\le C\le 1,$$ we have $$0\le v\le 1,\;0\le u\le\sqrt2.$$ Now (1) becomes

$$v^4+u^4+2=4uv. \quad -(2)$$

To analyse (2) we separate two non-negative parts.

First write

$$v^4+u^4=(v^2-u^2)^2+2v^2u^2,$$ so that

$$(2)\;\Longrightarrow\;(v^2-u^2)^2+2v^2u^2+2-4uv=0. \quad -(3)$$

We now estimate the second bracket in (3) with the Arithmetic-Geometric Mean inequality. For any non-negative numbers $$x,y$$ we have $$\dfrac{x+1}{2}\ge\sqrt{x}.$$ Putting $$x=v^2u^2$$ yields

$$v^2u^2+1\;\ge\;2vu\quad\Longrightarrow\quad 2v^2u^2+2\;\ge\;4vu.$$

Hence

$$2v^2u^2+2-4vu\;\ge\;0.$$

Both summands in (3) are therefore non-negative, and their sum equals zero. Consequently each must vanish.

1. $$(v^2-u^2)^2=0\;\Longrightarrow\;v^2=u^2\;\Longrightarrow\;v=u.$$

2. $$2v^2u^2+2-4vu=0\;\Longrightarrow\;v^2u^2+1=2vu.$$ Substituting $$u=v$$ from step 1 we get $$v^4+1=2v^2\;\Longrightarrow\;(v^2-1)^2=0\;\Longrightarrow\;v=1.$$

Thus

$$v=1,\qquad u=1.$$ Reverting to the original trigonometric quantities, we have

$$\sin\alpha=v=1\;\Longrightarrow\;\alpha=\dfrac{\pi}{2},$$

$$\sqrt2\,\cos\beta=u=1\;\Longrightarrow\;\cos\beta=\dfrac1{\sqrt2}\; \Longrightarrow\;\beta=\dfrac{\pi}{4}.$$

Now we evaluate the required expression. First state the identity that will be used:

$$\cos(A+B)-\cos(A-B)=-2\sin A\sin B.$$

Using $$A=\alpha,\;B=\beta$$ together with $$\sin\alpha=1,\;\sin\beta=\dfrac1{\sqrt2},$$ we get

$$\cos(\alpha+\beta)-\cos(\alpha-\beta)= -2\sin\alpha\sin\beta =-2\left(1\right)\left(\dfrac1{\sqrt2}\right)=-\sqrt2.$$

Hence, the correct answer is Option B.

First, recall the well-known double-angle identity for sine:

$$\sin(2\theta)=2\sin\theta\cos\theta.$$

We shall apply this identity successively to break the sine of a larger angle into sines and cosines of smaller and smaller angles until the desired product appears.

Take $$\theta=\dfrac{\pi}{2^{2}}.$$ Then $$2\theta=\dfrac{\pi}{2},$$ and the identity gives

$$\sin\!\Bigl(\dfrac{\pi}{2}\Bigr)=2\sin\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr).$$

But $$\sin\!\bigl(\dfrac{\pi}{2}\bigr)=1,$$ so we obtain

$$1=2\sin\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr). \quad -(1)$$

Now apply the same identity once more, this time to the angle $$\dfrac{\pi}{2^{3}}.$$ Since $$2\!\left(\dfrac{\pi}{2^{3}}\right)=\dfrac{\pi}{2^{2}},$$ we get

$$\sin\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr)=2\sin\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr). \quad -(2)$$

Substituting the value of $$\sin\!\bigl(\dfrac{\pi}{2^{2}}\bigr)$$ from (2) into (1), we have

$$1=2\Bigl[2\sin\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\Bigr]\cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr) =2^{2}\sin\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr).$$

We can clearly see a pattern forming. Repeating this process again and again, each time halving the angle, produces additional cosine factors and an additional power of 2 in front. After proceeding down to the angle $$\dfrac{\pi}{2^{10}},$$ the cumulative result is

$$1 =2^{9}\sin\!\Bigl(\dfrac{\pi}{2^{10}}\Bigr)\, \cos\!\Bigl(\dfrac{\pi}{2^{10}}\Bigr) \cos\!\Bigl(\dfrac{\pi}{2^{9}}\Bigr) \cos\!\Bigl(\dfrac{\pi}{2^{8}}\Bigr) \cdots \cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr).$$

Writing the product of cosines in ascending order of the denominators and grouping conveniently, this is

$$1 =2^{9}\Bigl[\sin\!\Bigl(\dfrac{\pi}{2^{10}}\Bigr)\prod_{k=2}^{10}\cos\!\Bigl(\dfrac{\pi}{2^{k}}\Bigr)\Bigr].$$

Observe that the bracketed expression is exactly the quantity asked for in the question, namely

$$\cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\cdots\cos\!\Bigl(\dfrac{\pi}{2^{10}}\Bigr)\sin\!\Bigl(\dfrac{\pi}{2^{10}}\Bigr).$$

Denote this required value by $$P.$$ From the equation just obtained we therefore have

$$1 = 2^{9} P \quad\Longrightarrow\quad P = \dfrac{1}{2^{9}} = \dfrac{1}{512}.$$

Hence, the correct answer is Option B.

We have to find the value of the expression $$\cos^2 10^\circ-\cos 10^\circ\cos 50^\circ+\cos^2 50^\circ.$$

First, we convert each squared cosine term by using the identity

$$\cos^2\theta=\frac{1+\cos 2\theta}{2}.$$

Applying this to the two squared terms gives

$$\cos^2 10^\circ=\frac{1+\cos 20^\circ}{2},\qquad \cos^2 50^\circ=\frac{1+\cos 100^\circ}{2}.$$

Substituting these into the original expression, we obtain

$$\cos^2 10^\circ-\cos 10^\circ\cos 50^\circ+\cos^2 50^\circ =\frac{1+\cos 20^\circ}{2}-\cos 10^\circ\cos 50^\circ +\frac{1+\cos 100^\circ}{2}.$$

Now we combine the two half-fractions:

$$\frac{1+\cos 20^\circ}{2}+\frac{1+\cos 100^\circ}{2} =\frac{2+\cos 20^\circ+\cos 100^\circ}{2} =1+\frac{\cos 20^\circ+\cos 100^\circ}{2}.$$

So the whole expression turns into

$$1+\frac{\cos 20^\circ+\cos 100^\circ}{2}-\cos 10^\circ\cos 50^\circ.$$

Next we simplify $$\cos 100^\circ$$. Since $$100^\circ=180^\circ-80^\circ,$$ we use $$\cos(180^\circ-\alpha)=-\cos\alpha$$ to get

$$\cos 100^\circ=-\cos 80^\circ.$$

Hence $$\cos 20^\circ+\cos 100^\circ=\cos 20^\circ-\cos 80^\circ,$$ and the expression becomes

$$1+\frac{\cos 20^\circ-\cos 80^\circ}{2}-\cos 10^\circ\cos 50^\circ.$$

To handle the product $$\cos 10^\circ\cos 50^\circ,$$ we employ the product-to-sum identity

$$\cos A\cos B=\frac{\cos(A+B)+\cos(A-B)}{2}.$$

Taking $$A=10^\circ, B=50^\circ$$ gives

$$\cos 10^\circ\cos 50^\circ =\frac{\cos(10^\circ+50^\circ)+\cos(10^\circ-50^\circ)}{2} =\frac{\cos 60^\circ+\cos(-40^\circ)}{2}.$$

Because $$\cos(-\alpha)=\cos\alpha,$$ we have $$\cos(-40^\circ)=\cos 40^\circ,$$ and since $$\cos 60^\circ=\frac12,$$ we get

$$\cos 10^\circ\cos 50^\circ =\frac{\frac12+\cos 40^\circ}{2} =\frac14+\frac{\cos 40^\circ}{2}.$$

Substituting this back, we arrive at

$$1+\frac{\cos 20^\circ-\cos 80^\circ}{2}-\left(\frac14+\frac{\cos 40^\circ}{2}\right) =1-\frac14+\frac{\cos 20^\circ-\cos 80^\circ-\cos 40^\circ}{2}.$$

The constant terms combine to give $$\frac34,$$ so we now focus on the bracket:

$$\cos 20^\circ-\cos 80^\circ-\cos 40^\circ.$$

We first simplify $$\cos 20^\circ-\cos 40^\circ$$ using the difference identity

$$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}.$$

Here $$A=20^\circ, B=40^\circ,$$ so

$$\cos 20^\circ-\cos 40^\circ =-2\sin\frac{20^\circ+40^\circ}{2}\sin\frac{20^\circ-40^\circ}{2} =-2\sin 30^\circ\sin(-10^\circ).$$

Because $$\sin 30^\circ=\frac12$$ and $$\sin(-10^\circ)=-\sin 10^\circ,$$ we get

$$\cos 20^\circ-\cos 40^\circ =-2\left(\frac12\right)(-\sin 10^\circ) =\sin 10^\circ.$$

So the whole bracket becomes

$$\sin 10^\circ-\cos 80^\circ.$$

But $$\cos 80^\circ=\sin 10^\circ$$ (since $$\cos\theta=\sin(90^\circ-\theta)$$), therefore

$$\sin 10^\circ-\cos 80^\circ=\sin 10^\circ-\sin 10^\circ=0.$$

Thus the entire numerator of the fraction is zero, and the fractional part vanishes. We are left with

$$\frac34+0=\frac34.$$

Hence, the correct answer is Option A.

We begin with the inequality

$$2^{\sqrt{\sin^2 x - 2\sin x + 5}}\;\cdot\;\frac{1}{4^{\sin^2 y}}\;\le\;1.$$

The denominator involves a power of $$4$$, and it is useful to rewrite it with the same base $$2$$. Since $$4 = 2^{2}$$, we have the identity

$$4^{\sin^2 y}\;=\;\bigl(2^2\bigr)^{\sin^2 y}\;=\;2^{2\sin^2 y}.$$

Substituting this into the given expression gives

$$2^{\sqrt{\sin^2 x - 2\sin x + 5}}\;\cdot\;2^{-2\sin^2 y}\;\le\;1,$$

which, by the basic rule $$a^m\cdot a^n = a^{m+n}$$, simplifies to

$$2^{\;\sqrt{\sin^2 x - 2\sin x + 5}\;-\;2\sin^2 y}\;\le\;1.$$

Now we recall the fact that for any base $$a \gt 1$$, the inequality

$$a^{k}\;\le\;1\quad\Longleftrightarrow\quad k\;\le\;0.$$

Using this fact with the base $$a=2$$, the exponent must satisfy

$$\sqrt{\sin^2 x - 2\sin x + 5}\;-\;2\sin^2 y\;\le\;0.$$

Thus

$$\sqrt{\sin^2 x - 2\sin x + 5}\;\le\;2\sin^2 y. \quad -(1)$$

We next simplify the quantity under the square-root. Completing the square gives

$$$ \sin^2 x \;-\;2\sin x \;+\;5 \;=\; \bigl(\sin x -1\bigr)^2 \;+\;4. $$$

Hence

$$\sqrt{\sin^2 x - 2\sin x + 5} \;=\; \sqrt{\bigl(\sin x -1\bigr)^2 + 4}.$$

The term $$(\sin x-1)^2$$ is always non-negative, so

$$$ (\sin x -1)^2 + 4 \;\ge\; 4, \quad\text{and therefore}\quad \sqrt{(\sin x -1)^2 + 4}\;\ge\;2. $$$

Returning to inequality (1), we know the left-hand side is at least $$2$$, so the right-hand side must also be at least $$2$$. Thus

$$2\sin^2 y\;\ge\;2 \;\;\Longrightarrow\;\; \sin^2 y\;\ge\;1.$$

But for every real number $$y$$ we have $$\sin^2 y \le 1$$. Combining $$\sin^2 y \ge 1$$ with $$\sin^2 y \le 1$$ forces

$$\sin^2 y = 1,$$ so $$\sin y = \pm 1.$$

With $$\sin^2 y = 1$$ we obtain $$2\sin^2 y = 2$$, and inequality (1) becomes

$$\sqrt{(\sin x -1)^2 + 4}\;\le\;2.$$

Since the square-root is never less than $$2$$, equality must occur:

$$\sqrt{(\sin x -1)^2 + 4} = 2.$$

Squaring both sides yields

$$(\sin x -1)^2 + 4 = 4 \quad\Longrightarrow\quad (\sin x -1)^2 = 0,$$

which simplifies to

$$\sin x - 1 = 0 \quad\Longrightarrow\quad \sin x = 1.$$

Hence every permissible pair $$(x,y)$$ satisfies

$$\sin x = 1,\qquad \sin y = \pm 1,$$ which in turn gives

$$|\sin x| = 1 = |\sin y|.$$

Therefore the common relation that all such pairs obey is

$$|\sin x| = |\sin y|,$$ which is exactly the statement in Option C.

Hence, the correct answer is Option C.

We have the family of functions $$f_k(x)=\dfrac{1}{k}\bigl(\sin^{\,k}x+\cos^{\,k}x\bigr)$$ for positive integers $$k$$. We need the expression $$f_4(x)-f_6(x)$$ for every real $$x$$.

First we write the two required members of the family explicitly:

$$f_4(x)=\dfrac{1}{4}\bigl(\sin^{4}x+\cos^{4}x\bigr),\qquad f_6(x)=\dfrac{1}{6}\bigl(\sin^{6}x+\cos^{6}x\bigr).$$

To simplify $$\sin^{4}x+\cos^{4}x$$, we start from the square of the Pythagorean identity. Since $$\sin^{2}x+\cos^{2}x=1$$, squaring both sides gives

$$\bigl(\sin^{2}x+\cos^{2}x\bigr)^{2}=1^{2}=1.$$

Expanding the left side,

$$\sin^{4}x+2\sin^{2}x\cos^{2}x+\cos^{4}x=1.$$

Rearranging,

$$\sin^{4}x+\cos^{4}x=1-2\sin^{2}x\cos^{2}x.$$

Next we simplify $$\sin^{6}x+\cos^{6}x$$. We use the algebraic identity for the sum of cubes, namely $$a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$$. Let

$$a=\sin^{2}x,\quad b=\cos^{2}x.$$

Then

$$\sin^{6}x+\cos^{6}x=(\sin^{2}x)^{3}+(\cos^{2}x)^{3} =(a+b)\bigl(a^{2}-ab+b^{2}\bigr).$$

We already know $$a+b=\sin^{2}x+\cos^{2}x=1$$, so only $$a^{2}-ab+b^{2}$$ remains. Observe

$$a^{2}-ab+b^{2}=\sin^{4}x-\sin^{2}x\cos^{2}x+\cos^{4}x.$$

Substituting the earlier result $$\sin^{4}x+\cos^{4}x=1-2\sin^{2}x\cos^{2}x$$ gives

$$a^{2}-ab+b^{2}=(1-2\sin^{2}x\cos^{2}x)-\sin^{2}x\cos^{2}x =1-3\sin^{2}x\cos^{2}x.$$

Hence

$$\sin^{6}x+\cos^{6}x=(a+b)\bigl(a^{2}-ab+b^{2}\bigr) =1\bigl(1-3\sin^{2}x\cos^{2}x\bigr)=1-3\sin^{2}x\cos^{2}x.$$

For brevity, let us denote $$\sin^{2}x\cos^{2}x$$ by $$t$$. Therefore

$$\sin^{4}x+\cos^{4}x=1-2t,\qquad \sin^{6}x+\cos^{6}x=1-3t.$$

We now substitute these into $$f_4(x)$$ and $$f_6(x)$$:

$$f_4(x)=\dfrac{1}{4}(1-2t),\qquad f_6(x)=\dfrac{1}{6}(1-3t).$$

The desired difference becomes

$$f_4(x)-f_6(x)=\dfrac{1}{4}(1-2t)-\dfrac{1}{6}(1-3t).$$

To combine the fractions, we take the common denominator $$12$$:

$$f_4(x)-f_6(x)=\dfrac{3}{12}(1-2t)-\dfrac{2}{12}(1-3t).$$

Expanding the numerators,

$$=\dfrac{3(1-2t)-2(1-3t)}{12} =\dfrac{3-6t-2+6t}{12}.$$

The $$-6t$$ and $$+6t$$ cancel out, leaving

$$=\dfrac{1}{12}.$$

This final expression contains no $$x$$, confirming that the difference is the same for all real $$x$$. Hence, the correct answer is Option A.

We begin with the given equation

$$\cos 2x + \alpha \sin x \;=\; 2\alpha - 7.$$

First, we rewrite $$\cos 2x$$ in terms of $$\sin x$$ by using the standard identity

$$\cos 2x \;=\; 1 - 2\sin^{2}x.$$

Substituting this identity, we obtain

$$1 - 2\sin^{2}x + \alpha \sin x \;=\; 2\alpha - 7.$$

For simplicity, let us set

$$y \;=\; \sin x,$$

remembering that $$y$$ must always satisfy $$-1 \le y \le 1.$$ With this substitution the equation becomes

$$1 - 2y^{2} + \alpha y \;=\; 2\alpha - 7.$$

We bring every term to the left-hand side so that the expression equals zero:

$$1 - 2y^{2} + \alpha y - 2\alpha + 7 \;=\; 0.$$

Combining the constant terms $$1 + 7 = 8,$$ we have

$$-2y^{2} + \alpha y + 8 - 2\alpha = 0.$$

Multiplying by $$-1$$ to make the coefficient of $$y^{2}$$ positive, we get

$$2y^{2} - \alpha y - 8 + 2\alpha = 0,$$

or, more neatly,

$$2y^{2} - \alpha y + 2\alpha - 8 = 0.$$

This is a quadratic equation in $$y$$. We want some $$y$$ in the interval $$[-1,\,1]$$ to satisfy it. To find all such $$\alpha,$$ we solve the quadratic for $$\alpha$$ in terms of $$y$$ (treating $$y$$ as the variable and $$\alpha$$ as the unknown constant). Collecting the $$\alpha$$ terms, we write

$$- \alpha y + 2\alpha = \alpha(2 - y).$$

Shifting the remaining terms to the other side gives

$$\alpha(2 - y) = 8 - 2y^{2}.$$

Provided $$y \ne 2$$ (and indeed $$y$$ is never $$2$$ because $$-1 \le y \le 1$$), we can divide to obtain

$$\displaystyle \alpha = \frac{8 - 2y^{2}}{\,2 - y\,}.$$

Factoring a $$2$$ from the numerator yields the simpler expression

$$\alpha = 2\,\frac{4 - y^{2}}{2 - y}.$$\p>

Define the function

$$g(y) = 2\,\frac{4 - y^{2}}{2 - y}, \qquad -1 \le y \le 1.$$

The set $$S$$ that we seek is precisely the set of all values taken by $$g(y)$$ as $$y$$ varies over $$[-1,\,1].$$ To determine this set we examine the monotonicity of $$g(y).$$ Write

$$g(y) = 2\,\frac{N(y)}{D(y)},$$

where $$N(y) = 4 - y^{2}$$ and $$D(y) = 2 - y.$$ Using the quotient rule, the derivative of the inner fraction is

$$\frac{d}{dy}\!\left(\frac{N(y)}{D(y)}\right) = \frac{N'(y)D(y) - N(y)D'(y)}{D(y)^{2}}.$$

We compute each part:

$$N'(y) = -2y, \qquad D'(y) = -1.$$

Substituting these, we obtain

$$\frac{d}{dy}\!\left(\frac{N}{D}\right) = \frac{(-2y)(2 - y) - (4 - y^{2})(-1)}{(2 - y)^{2}} = \frac{-4y + 2y^{2} + 4 - y^{2}}{(2 - y)^{2}} = \frac{y^{2} - 4y + 4}{(2 - y)^{2}} = \frac{(y - 2)^{2}}{(2 - y)^{2}}.$$

Since both numerator and denominator are perfect squares, their ratio is always non-negative and equals zero only when $$y = 2,$$ which is outside the interval $$[-1,\,1].$$ Hence the derivative is strictly positive on $$[-1,\,1]$$ and $$g(y)$$ is strictly increasing on that interval.

Because $$g$$ is increasing, its minimum and maximum values on $$[-1,\,1]$$ occur at the endpoints:

For $$y = -1:$$

$$g(-1) = 2\,\frac{4 - (-1)^{2}}{2 - (-1)} = 2\,\frac{4 - 1}{3} = 2 \times \frac{3}{3} = 2.$$

For $$y = 1:$$

$$g(1) = 2\,\frac{4 - 1^{2}}{2 - 1} = 2\,\frac{3}{1} = 6.$$

Because every intermediate value is attained by continuity and monotonicity, the range of $$g$$ is the closed interval

$$[2,\,6].$$

Therefore the set $$S$$ of all real numbers $$\alpha$$ for which the original trigonometric equation admits a solution is

$$S = [2,\,6].$$

Hence, the correct answer is Option B.

We are asked to find every $$\theta$$ lying in the closed interval $$[-2\pi,\,2\pi]$$ that satisfies the trigonometric equation

$$2\cos^2\theta + 3\sin\theta = 0.$$

First, we convert the entire equation into a single trigonometric function. We recall the Pythagorean identity

$$\sin^2\theta + \cos^2\theta = 1 \quad\Longrightarrow\quad \cos^2\theta = 1 - \sin^2\theta.$$

Substituting $$\cos^2\theta = 1 - \sin^2\theta$$ in the given equation, we obtain

$$2(1 - \sin^2\theta) + 3\sin\theta = 0.$$

Simplifying term by term, we have

$$2 - 2\sin^2\theta + 3\sin\theta = 0.$$

Let us set $$x = \sin\theta$$ for convenience. Rewriting the equation in terms of $$x$$ gives

$$2 - 2x^2 + 3x = 0.$$

Next, we collect like terms and arrange them in descending powers of $$x$$:

$${-2x^2} + 3x + 2 = 0.$$

Multiplying every term by $$-1$$ (which does not change the solution set) yields the more familiar quadratic form

$$2x^2 - 3x - 2 = 0.$$

We solve this quadratic equation with the quadratic formula. For $$ax^2 + bx + c = 0,$$ the roots are

$$x = \frac{-b \pm \sqrt{\,b^2 - 4ac\,}}{2a}.$$

Here $$a = 2,\; b = -3,\; c = -2.$$ Substituting these values, we find

$$x = \frac{-(-3) \pm \sqrt{\,(-3)^2 - 4(2)(-2)\,}}{2(2)} = \frac{3 \pm \sqrt{\,9 + 16\,}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4}.$$

This gives two possible roots:

$$x_1 = \frac{3 + 5}{4} = \frac{8}{4} = 2, \qquad x_2 = \frac{3 - 5}{4} = \frac{-2}{4} = -\frac12.$$

Since $$x = \sin\theta$$ must lie in the interval $$[-1,\,1],$$ the value $$x = 2$$ is impossible and must be discarded. Thus the only admissible value is

$$\sin\theta = -\frac12.$$

We now solve $$\sin\theta = -\dfrac12$$ for all $$\theta$$ in the range $$[-2\pi,\,2\pi].$$ On the unit circle, the sine function equals $$-\dfrac12$$ at the following standard positions:

$$\theta = -\frac\pi6 + 2k\pi, \qquad\theta = -\frac{5\pi}6 + 2k\pi,$$

where $$k \in \mathbb Z.$$ These are the fourth- and third-quadrant angles corresponding to sine $$-\dfrac12.$$ We list integral values of $$k$$ that keep $$\theta$$ within $$[-2\pi,\,2\pi]$$ and record each distinct angle.

Taking $$k = 0$$

$$\theta_1 = -\frac\pi6, \qquad \theta_2 = -\frac{5\pi}6.$$

Taking $$k = 1$$

$$\theta_3 = -\frac\pi6 + 2\pi = -\frac\pi6 + \frac{12\pi}6 = \frac{11\pi}6,$$

$$\theta_4 = -\frac{5\pi}6 + 2\pi = -\frac{5\pi}6 + \frac{12\pi}6 = \frac{7\pi}6.$$

Taking $$k = -1$$ would produce

$$\theta = -\frac\pi6 - 2\pi = -\frac{13\pi}6 \lt -2\pi,$$

$$\theta = -\frac{5\pi}6 - 2\pi = -\frac{17\pi}6 \lt -2\pi,$$

both of which lie outside the prescribed interval, so they are excluded. Similarly, $$k = 2$$ gives angles greater than $$2\pi.$$ Therefore the complete solution set is

$$S = \Bigl\{-\frac{5\pi}6,\; -\frac\pi6,\; \frac{7\pi}6,\; \frac{11\pi}6\Bigr\}.$$

We now compute the sum of these four angles:

$$\Bigl(-\frac{5\pi}6\Bigr) + \Bigl(-\frac\pi6\Bigr) + \frac{7\pi}6 + \frac{11\pi}6 = \left(-\frac{6\pi}6\right) + \frac{18\pi}6 = -\pi + 3\pi = 2\pi.$$

Thus the sum of all elements of $$S$$ equals $$2\pi.$$ Among the given options, this value corresponds to Option D.

Hence, the correct answer is Option D.

We need all angles $$\theta$$ lying in the open interval $$(0,\dfrac{\pi}{2})$$ that satisfy the trigonometric equation

$$\sin^2 2\theta+\cos^4 2\theta=\dfrac34.$$

First, to simplify the expression, we put

$$x=2\theta.$$

Because $$\theta\in(0,\dfrac{\pi}{2})$$, doubling every number in this interval gives

$$x\in(0,\pi).$$

Under this substitution the equation becomes

$$\sin^2 x+\cos^4 x=\dfrac34.$$

Next, we use the Pythagorean identity $$\sin^2 x+\cos^2 x=1.$$ Solving this for $$\sin^2 x$$ gives

$$\sin^2 x=1-\cos^2 x.$$

Substituting this in the transformed equation produces

$$\bigl(1-\cos^2 x\bigr)+\cos^4 x=\dfrac34.$$

Now let us introduce another temporary variable to avoid writing $$\cos$$ repeatedly. Put

$$y=\cos^2 x.$$

Because $$x\in(0,\pi),$$ the value of $$\cos x$$ ranges from ±1 down to 0, so $$y=\cos^2 x$$ must satisfy $$0<y<1.$$

Under this change of variable, the equation simplifies to a quadratic:

$$1-y+y^2=\dfrac34.$$

We move every term to the left so that the right-hand side becomes zero:

$$1-y+y^2-\dfrac34=0.$$

Combining the constant terms on the left, we have

$$y^2-y+\bigl(1-\dfrac34\bigr)=0,$$

so

$$y^2-y+\dfrac14=0.$$

Observe that the left side is a perfect square. Indeed,

$$(y-\dfrac12)^2=y^2-y+\dfrac14.$$

Therefore we can rewrite the quadratic as

$$(y-\dfrac12)^2=0.$$

Solving this gives a single real root

$$y=\dfrac12.$$

But $$y=\cos^2 x,$$ so

$$\cos^2 x=\dfrac12.$$

Taking square roots on both sides, we obtain

$$\cos x=\pm\dfrac1{\sqrt2}.$$

Now we must find every $$x$$ in the interval $$(0,\pi)$$ that satisfies this relation.

We know from basic trigonometry that

$$\cos x=\dfrac1{\sqrt2}\quad\text{at}\quad x=\dfrac{\pi}{4},$$

and

$$\cos x=-\dfrac1{\sqrt2}\quad\text{at}\quad x=\dfrac{3\pi}{4}.$$

Both of these angles indeed lie strictly between $$0$$ and $$\pi,$$ so the corresponding values of $$x$$ are

$$x_1=\dfrac{\pi}{4},\qquad x_2=\dfrac{3\pi}{4}.$$

We now reverse the substitution $$x=2\theta$$ to back-substitute for $$\theta$$:

For $$x_1$$:

$$2\theta_1=\dfrac{\pi}{4}\;\Longrightarrow\;\theta_1=\dfrac{\pi}{8}.$$

For $$x_2$$:

$$2\theta_2=\dfrac{3\pi}{4}\;\Longrightarrow\;\theta_2=\dfrac{3\pi}{8}.$$

Both values $$\dfrac{\pi}{8}$$ and $$\dfrac{3\pi}{8}$$ lie inside the required interval $$(0,\dfrac{\pi}{2}),$$ so they are the only solutions for $$\theta.$$

The question asks for the sum of all such $$\theta.$$ Adding the two values we get

$$\theta_1+\theta_2=\dfrac{\pi}{8}+\dfrac{3\pi}{8}=\dfrac{4\pi}{8}=\dfrac{\pi}{2}.$$

Hence, the correct answer is Option A.

We have to find the numerical value of the product $$\sin 10^\circ\,\sin 30^\circ\,\sin 50^\circ\,\sin 70^\circ.$$

First, notice that the three angles $$10^\circ,\;50^\circ,\;70^\circ$$ can be written in the symmetric form $$x,\;60^\circ-x,\;60^\circ+x$$ if we choose $$x = 10^\circ.$$ This observation suggests using a standard identity for the product $$\sin x\;\sin(60^\circ-x)\;\sin(60^\circ+x).$$

To derive that identity, we begin with the product-to-sum formula

$$\sin A\;\sin B = \tfrac12\bigl[\cos(A-B)-\cos(A+B)\bigr].$$

Put $$A = 60^\circ+x,\;B = 60^\circ-x.$$ Then

$$\sin(60^\circ+x)\,\sin(60^\circ-x) = \tfrac12\Bigl[\cos\bigl((60^\circ+x)-(60^\circ-x)\bigr) -\cos\bigl((60^\circ+x)+(60^\circ-x)\bigr)\Bigr].$$

Simplifying the arguments of the cosines, we get

$$\cos\bigl((60^\circ+x)-(60^\circ-x)\bigr)=\cos(2x),$$ $$\cos\bigl((60^\circ+x)+(60^\circ-x)\bigr)=\cos(120^\circ).$$

Because $$\cos120^\circ=-\tfrac12,$$ the product becomes

$$\sin(60^\circ+x)\,\sin(60^\circ-x) = \tfrac12\bigl[\cos 2x - (-\tfrac12)\bigr] = \tfrac12\cos 2x + \tfrac14.$$

Multiplying this result by $$\sin x,$$ we obtain

$$\sin x\,\sin(60^\circ-x)\,\sin(60^\circ+x) = \sin x\bigl(\tfrac12\cos 2x + \tfrac14\bigr).$$

Next we rewrite $$\cos 2x$$ via the double-angle identity $$\cos 2x = 1 - 2\sin^2 x.$$ Substituting,

$$\tfrac12\cos 2x + \tfrac14 = \tfrac12(1 - 2\sin^2 x) + \tfrac14 = \tfrac12 - \sin^2 x + \tfrac14 = \tfrac34 - \sin^2 x.$$

Hence

$$\sin x\,\sin(60^\circ-x)\,\sin(60^\circ+x) = \sin x\bigl(\tfrac34 - \sin^2 x\bigr) = \tfrac34\sin x - \sin^3 x.$$

Now we recall the triple-angle formula

$$\sin 3x = 3\sin x - 4\sin^3 x.$$

Dividing both sides of this formula by 4 gives

$$\tfrac14\sin 3x = \tfrac34\sin x - \sin^3 x.$$

Comparing with the expression just obtained, we see that

$$\boxed{\;\sin x\,\sin(60^\circ-x)\,\sin(60^\circ+x) = \tfrac14\sin 3x\;}.$$

Now set $$x = 10^\circ.$$ Then $$60^\circ-x = 50^\circ$$ and $$60^\circ+x = 70^\circ.$$ Therefore

$$\sin10^\circ\,\sin50^\circ\,\sin70^\circ = \tfrac14\sin 3(10^\circ) = \tfrac14\sin30^\circ.$$

Because $$\sin30^\circ = \tfrac12,$$ the triple product equals

$$\sin10^\circ\,\sin50^\circ\,\sin70^\circ = \tfrac14 \times \tfrac12 = \tfrac18.$$

The given expression contains one more factor $$\sin30^\circ=\tfrac12.$$ Multiplying this with the above result, we find

$$\sin10^\circ\,\sin30^\circ\,\sin50^\circ\,\sin70^\circ = \bigl(\tfrac18\bigr)\times\bigl(\tfrac12\bigr) = \tfrac1{16}.$$

Hence, the correct answer is Option B.

We need to find the number of values of $$x$$ in $$[0, \frac{\pi}{2})$$ satisfying $$\sin x - \sin 2x + \sin 3x = 0$$.

We use the sum-to-product identity. Group the first and third terms:

$$\sin x + \sin 3x = 2 \sin\left(\frac{x + 3x}{2}\right) \cos\left(\frac{3x - x}{2}\right) = 2 \sin 2x \cos x$$

Substituting back into the equation:

$$2 \sin 2x \cos x - \sin 2x = 0$$

Factoring out $$\sin 2x$$:

$$\sin 2x (2\cos x - 1) = 0$$

This gives us two cases:

Case 1: $$\sin 2x = 0$$

$$2x = n\pi$$ where $$n$$ is an integer, so $$x = \frac{n\pi}{2}$$.

In the interval $$[0, \frac{\pi}{2})$$: $$x = 0$$ (when $$n = 0$$) is valid. Note that $$x = \frac{\pi}{2}$$ (when $$n = 1$$) is NOT included since the domain is strictly less than $$\frac{\pi}{2}$$.

Case 2: $$2\cos x - 1 = 0$$

$$\cos x = \frac{1}{2}$$

$$x = \frac{\pi}{3}$$

Since $$\frac{\pi}{3} \approx 1.047$$ and $$\frac{\pi}{2} \approx 1.571$$, we have $$0 < \frac{\pi}{3} < \frac{\pi}{2}$$, so $$x = \frac{\pi}{3}$$ is valid.

Therefore, the valid solutions in $$[0, \frac{\pi}{2})$$ are $$x = 0$$ and $$x = \frac{\pi}{3}$$.

The number of values of $$x$$ is $$2$$.

The correct answer is Option C: 2.

Let the three sides of the triangle be denoted by $$a$$, $$b$$ and $$c$$, where $$c$$ is the side whose length is explicitly named in the question. We are told that the sum and product of the other two sides satisfy

$$a+b = x, \qquad ab = y,$$

and that these same numbers obey the relation

$$x^{2}-c^{2}=y.$$

To reach the circumradius we shall first determine the angle opposite side $$c$$. For any triangle the Law of Cosines states

$$c^{2}=a^{2}+b^{2}-2ab\cos C,$$

where $$C$$ is the angle between sides $$a$$ and $$b$$ (i.e. the angle opposite side $$c$$).

We wish to rewrite $$a^{2}+b^{2}$$ in terms of the given quantities $$x$$ and $$y$$. Using the identity

$$(a+b)^{2}=a^{2}+b^{2}+2ab,$$

we isolate $$a^{2}+b^{2}$$:

$$a^{2}+b^{2}=(a+b)^{2}-2ab=x^{2}-2y.$$

Substituting this back into the Law of Cosines gives

$$c^{2}=\,\bigl(x^{2}-2y\bigr)\,-2y\cos C.$$

We now move every term except the cosine term to the other side:

$$2y\cos C \;=\;x^{2}-2y-c^{2}.$$

But the question itself supplies $$x^{2}-c^{2}=y$$, so we replace that pair on the right:

$$2y\cos C \;=\;y-2y.$$

Simplifying,

$$2y\cos C = -\,y.$$

Provided the triangle is non-degenerate, $$y\neq 0$$, hence we can divide by $$y$$:

$$2\cos C = -1 \;\;\;\Longrightarrow\;\;\; \cos C = -\tfrac12.$$

The only angle of a triangle whose cosine equals $$-\,\frac12$$ is

$$C = 120^{\circ}.$$

With the angle known, we invoke the standard formula that links a side to the circumradius $$R$$:

$$\text{(Circumradius formula)}\quad c = 2R\sin C.$$

Using $$C=120^{\circ}$$ and recalling $$\sin120^{\circ}=\sin60^{\circ}=\frac{\sqrt3}{2},$$ we have

$$c = 2R\left(\frac{\sqrt3}{2}\right) = R\sqrt3.$$

Solving for $$R$$ yields

$$R = \frac{c}{\sqrt3}.$$

Among the listed choices this matches Option 2.

Hence, the correct answer is Option 2.

We have to find the greatest value that the expression

$$E(\theta)=3\cos\theta+5\sin\left(\theta-\frac{\pi}{6}\right)$$

can take for any real $$\theta$$. To compare unlike trigonometric functions, we first express everything in terms of the basic $$\sin\theta$$ and $$\cos\theta$$ functions.

We recall the compound-angle identity for sine:

$$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.$$

Using this with $$\alpha=\theta$$ and $$\beta=\dfrac{\pi}{6}$$, we write

$$\sin\left(\theta-\frac{\pi}{6}\right)=\sin\theta\cos\frac{\pi}{6}-\cos\theta\sin\frac{\pi}{6}.$$

Now we substitute the exact values $$\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}$$ and $$\sin\dfrac{\pi}{6}=\dfrac12$$:

$$\sin\left(\theta-\frac{\pi}{6}\right)=\sin\theta\left(\frac{\sqrt3}{2}\right)-\cos\theta\left(\frac12\right).$$

Next, we insert this expansion into the original expression:

$$E(\theta)=3\cos\theta+5\Bigg[\sin\theta\left(\frac{\sqrt3}{2}\right)-\cos\theta\left(\frac12\right)\Bigg].$$

Multiplying the 5 inside the brackets term-by-term gives

$$E(\theta)=3\cos\theta+\frac{5\sqrt3}{2}\sin\theta-\frac52\cos\theta.$$

Now we collect the like $$\cos\theta$$ terms together:

$$3\cos\theta-\frac52\cos\theta=\left(3-\frac52\right)\cos\theta=\frac12\cos\theta.$$

Hence the entire expression simplifies to

$$E(\theta)=\frac12\cos\theta+\frac{5\sqrt3}{2}\sin\theta.$$

For clarity, we rename the coefficients:

Let $$A=\frac{5\sqrt3}{2} \quad\text{and}\quad B=\frac12.$$

Thus we have

$$E(\theta)=A\sin\theta+B\cos\theta.$$

There is a standard result for an expression of the form

$$A\sin\theta+B\cos\theta,$$

namely:

It can always be rewritten as

$$\sqrt{A^{2}+B^{2}}\;\sin(\theta+\phi)$$

for some phase angle $$\phi$$ (because $$\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi$$). Since the sine function itself is bounded between $$-1$$ and $$1$$, the maximum possible value is precisely the amplitude, i.e.

$$\max\bigl(A\sin\theta+B\cos\theta\bigr)=\sqrt{A^{2}+B^{2}}.$$

Therefore we compute

$$A^{2}+B^{2}=\left(\frac{5\sqrt3}{2}\right)^{2}+\left(\frac12\right)^{2} =\frac{25\cdot3}{4}+\frac{1}{4} =\frac{75+1}{4} =\frac{76}{4} =19.$$

Taking the square root gives

$$\sqrt{A^{2}+B^{2}}=\sqrt{19}.$$

This is the largest value that $$E(\theta)$$ can ever reach for any real $$\theta$$.

Hence, the correct answer is Option A.

We have to solve the trigonometric equation

$$1+\sin^{4}x=\cos^{2}3x$$

for all real $$x$$ lying in the closed interval $$\left[-\dfrac{5\pi}{2},\dfrac{5\pi}{2}\right]$$.

First, note the possible ranges of the two sides.

For any real number $$y$$, the basic facts are

$$0\le\sin^{2}y\le1\quad\text{and hence}\quad0\le\sin^{4}y\le1.$$

Therefore

$$1+\sin^{4}x\in[1,2].$$

Similarly, for any real number $$z$$, we know

$$-1\le\cos z\le1\quad\Longrightarrow\quad0\le\cos^{2}z\le1.$$

So the right‐hand side satisfies

$$\cos^{2}3x\in[0,1].$$

Now we compare the two ranges. The left‐hand side is always at least $$1$$, while the right‐hand side is at most $$1$$. Hence equality is possible only when both quantities are exactly $$1$$. Therefore we must have

$$1+\sin^{4}x=1\quad\text{and}\quad\cos^{2}3x=1.$$

The first condition gives

$$\sin^{4}x=0\quad\Longrightarrow\quad\sin x=0.$$

The standard solutions of $$\sin x=0$$ are

$$x=n\pi,\qquad n\in\mathbb Z.$$

Next, we verify the second condition at these points. Substituting $$x=n\pi$$ in $$\cos^{2}3x$$, we get

$$\cos^{2}3x=\cos^{2}(3n\pi)=\bigl(\cos(3n\pi)\bigr)^{2}.$$

But $$\cos(3n\pi)=(-1)^{3n}=(-1)^{n},$$ and squaring eliminates the sign, giving

$$\cos^{2}(3n\pi)=\bigl((-1)^{n}\bigr)^{2}=1.$$

Thus every $$x=n\pi$$ automatically satisfies $$\cos^{2}3x=1$$, so all such points are indeed solutions.

We now count how many integers $$n$$ give $$x=n\pi$$ in the required interval. Divide the inequality

$$-\dfrac{5\pi}{2}\le n\pi\le\dfrac{5\pi}{2}$$

by $$\pi$$ (which is positive, so the inequality sign does not change):

$$-\dfrac{5}{2}\le n\le\dfrac{5}{2}.$$

The integers in the interval $$[-2.5,\,2.5]$$ are

$$n=-2,\,-1,\,0,\,1,\,2.$$

Counting them gives exactly $$5$$ values of $$n$$, hence $$5$$ solutions $$x=n\pi$$ in the given domain.

Hence, the correct answer is Option A.

1. Find the sides in terms of a constant

Let us set the given equations to a constant $$k$$.

$$\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$$

This gives us a system of three simple equations:

• $$b + c = 11k$$
• $$c + a = 12k$$
• $$a + b = 13k$$

Add all three equations together to find the perimeter in terms of $$k$$:

$$2(a + b + c) = 36k$$
$$a + b + c = 18k$$

Now, subtract each original equation from this sum to find the individual sides $$a$$, $$b$$, and $$c$$:

• $$a = 18k - (b + c) = 18k - 11k = 7k$$
• $$b = 18k - (c + a) = 18k - 12k = 6k$$
• $$c = 18k - (a + b) = 18k - 13k = 5k$$

2. Apply the Cosine Rule

Since we know the sides are proportional to 7, 6, and 5, we can drop the $$k$$ for the ratio calculations or keep it. It will cancel out anyway. Let us use the standard cosine formulas.

For angle A:

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$

$$\cos A = \frac{36k^2 + 25k^2 - 49k^2}{2(6k)(5k)}$$

$$\cos A = \frac{12k^2}{60k^2} = \frac{1}{5}$$

For angle B:

$$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$$

$$\cos B = \frac{49k^2 + 25k^2 - 36k^2}{2(7k)(5k)}$$

$$\cos B = \frac{38k^2}{70k^2} = \frac{19}{35}$$

For angle C:

$$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$

$$\cos C = \frac{49k^2 + 36k^2 - 25k^2}{2(7k)(6k)}$$

$$\cos C = \frac{60k^2}{84k^2} = \frac{5}{7}$$

3. Equate to the given condition

We are given that $$\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$$.

This implies the ratio $$\alpha : \beta : \gamma$$ is exactly equal to the ratio $$\cos A : \cos B : \cos C$$.

Let us write out that ratio:

$$\alpha : \beta : \gamma = \frac{1}{5} : \frac{19}{35} : \frac{5}{7}$$

To convert these fractions into integers, multiply the entire ratio by 35 (the least common multiple of the denominators 5, 35, and 7).

$$\alpha : \beta : \gamma = \left(\frac{1}{5} \times 35\right) : \left(\frac{19}{35} \times 35\right) : \left(\frac{5}{7} \times 35\right)$$
$$\alpha : \beta : \gamma = 7 : 19 : 25$$

The ordered triad $$(\alpha, \beta, \gamma)$$ is $$(7, 19, 25)$$.

Let us use Napier's tangent analogy for triangle $$ABC$$:
$$\tan\left(\frac{A - B}{2}\right) = \frac{a - b}{a + b} \cot\left(\frac{C}{2}\right)$$

First, calculate the difference and the sum of the sides $$a$$ and $$b$$:
$$a - b = (\sqrt{3} + 1) - (\sqrt{3} - 1) = 2$$
$$a + b = (\sqrt{3} + 1) + (\sqrt{3} - 1) = 2\sqrt{3}$$

Thus, the ratio of the difference to the sum of the sides is:
$$\frac{a - b}{a + b} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$$

Next, find angle $$C$$ using the sum of angles in a triangle. We are given:
$$A + B = 120^\circ$$

Since the sum of all angles in triangle $$ABC$$ is $$180^\circ$$:
$$C = 180^\circ - (A + B) = 180^\circ - 120^\circ = 60^\circ$$

Therefore, half of angle $$C$$ is:
$$\frac{C}{2} = 30^\circ$$

Substitute these values into Napier's analogy:
$$\tan\left(\frac{A - B}{2}\right) = \frac{1}{\sqrt{3}} \cot(30^\circ)$$

Since $$\cot(30^\circ) = \sqrt{3}$$, we get:
$$\tan\left(\frac{A - B}{2}\right) = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1$$

This gives:
$$\frac{A - B}{2} = 45^\circ$$
$$A - B = 90^\circ$$

We now have a system of linear equations for angles $$A$$ and $$B$$:
$$A + B = 120^\circ$$
$$A - B = 90^\circ$$

Adding the two equations:
$$2A = 210^\circ \implies A = 105^\circ$$

Subtracting the second equation from the first:
$$2B = 30^\circ \implies B = 15^\circ$$

Finally, calculate the ratio of angle $$A$$ to angle $$B$$:
$$\frac{A}{B} = \frac{105^\circ}{15^\circ} = 7$$

Therefore, the ratio of angle $$A$$ to angle $$B$$ is $$7 : 1$$.

ALTERNATE :

We have a triangle $$\triangle ABC$$ with the usual notation: side $$a$$ is opposite angle $$A$$ and side $$b$$ is opposite angle $$B$$. The given data are

$$\angle A+\angle B = 120^{\circ}, \qquad a = \sqrt{3}+1, \qquad b = \sqrt{3}-1.$$

From the Law of Sines, which states $$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C},$$ we obtain the proportionality

$$\dfrac{a}{b}=\dfrac{\sin A}{\sin B}. \quad -(1)$$

Substituting the given side lengths into (1) gives

$$\dfrac{\sin A}{\sin B}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}.$$

To simplify the right-hand side we rationalise the denominator:

$$\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\times\dfrac{\sqrt{3}+1}{\sqrt{3}+1} =\dfrac{(\sqrt{3}+1)^{2}}{(\sqrt{3})^{2}-1^{2}} =\dfrac{3+1+2\sqrt{3}}{3-1} =\dfrac{4+2\sqrt{3}}{2} =2+\sqrt{3}.$$

Hence

$$\dfrac{\sin A}{\sin B}=2+\sqrt{3}. \quad -(2)$$

Now let us set $$\angle A = x^{\circ}.$$ Because $$\angle A+\angle B=120^{\circ},$$ we can write $$\angle B = 120^{\circ}-x^{\circ}.$$ Substituting these into equation (2) yields

$$\dfrac{\sin x}{\sin(120^{\circ}-x)} = 2+\sqrt{3}. \quad -(3)$$

We next expand the denominator using the sine subtraction formula $$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.$$ Taking $$\alpha=120^{\circ}$$ and $$\beta=x$$ gives

$$\sin(120^{\circ}-x)=\sin120^{\circ}\cos x-\cos120^{\circ}\sin x.$$

Because $$\sin120^{\circ}=\dfrac{\sqrt{3}}{2}$$ and $$\cos120^{\circ}=-\dfrac12,$$ we have

$$\sin(120^{\circ}-x)=\dfrac{\sqrt{3}}{2}\cos x-\left(-\dfrac12\right)\sin x =\dfrac{\sqrt{3}}{2}\cos x+\dfrac12\sin x =\dfrac{\sqrt{3}\cos x+\sin x}{2}.$$

Substituting this expression back into (3) we get

$$\dfrac{\sin x}{\dfrac{\sqrt{3}\cos x+\sin x}{2}}=2+\sqrt{3}.$$ Multiplying numerator and denominator by $$2$$ yields

$$\dfrac{2\sin x}{\sqrt{3}\cos x+\sin x}=2+\sqrt{3}.$$

Cross-multiplying, we obtain

$$2\sin x=(2+\sqrt{3})(\sqrt{3}\cos x+\sin x).$$

Expanding the right-hand side gives

$$2\sin x=(2+\sqrt{3})\sqrt{3}\cos x+(2+\sqrt{3})\sin x =(2\sqrt{3}+3)\cos x+(2+\sqrt{3})\sin x.$$

Bringing all terms to the left and combining like terms:

$$2\sin x-(2+\sqrt{3})\sin x-(2\sqrt{3}+3)\cos x=0,$$ $$\bigl[2-(2+\sqrt{3})\bigr]\sin x-(2\sqrt{3}+3)\cos x=0,$$ $$-\sqrt{3}\sin x-(2\sqrt{3}+3)\cos x=0.$$

Multiplying by $$-1$$ simplifies this to

$$\sqrt{3}\sin x+(2\sqrt{3}+3)\cos x=0.$$

Dividing throughout by $$\cos x$$ (which is non-zero for the angle we will obtain) gives an equation in $$\tan x$$:

$$\sqrt{3}\tan x+(2\sqrt{3}+3)=0,$$ $$\tan x=-\dfrac{2\sqrt{3}+3}{\sqrt{3}}.$$

Splitting the numerator and simplifying:

$$-\dfrac{2\sqrt{3}}{\sqrt{3}}-\dfrac{3}{\sqrt{3}} =-(2+\sqrt{3}).$$

Thus

$$\tan x=-(2+\sqrt{3}). \quad -(4)$$

We recognise that $$\tan75^{\circ}=2+\sqrt{3},$$ so equation (4) becomes

$$\tan x=-\tan75^{\circ}.$$

The principal value that satisfies this while lying between $$0^{\circ}$$ and $$120^{\circ}$$ is

$$x=105^{\circ},$$ because $$\tan105^{\circ}=\tan(180^{\circ}-75^{\circ})=-\tan75^{\circ}.$$

Therefore

$$\angle A = 105^{\circ}, \qquad \angle B = 120^{\circ}-105^{\circ}=15^{\circ}.$$

The required ratio is then

$$\angle A:\angle B = 105^{\circ}:15^{\circ}=7:1.$$

Hence, the correct answer is Option A.

We are told that the park is the isosceles triangle $$\triangle ABC$$ with $$AB = AC = 100\text{ m}$$. Let $$D$$ be the mid-point of the base $$BC$$ and let the vertical tower stand on $$D$$. Denote the top of the tower by $$T$$ and its height by $$h$$, so $$TD = h$$.

Because $$AB = AC$$, the median $$AD$$ is also the altitude and perpendicular bisector of $$BC$$. Hence $$AD \perp BC$$ and $$BD = DC$$. Write

$$BD = DC = x,\qquad AD = y.$$

The right triangle $$\triangle ABD$$ therefore has

$$AB = 100,\; BD = x,\; AD = y.$$

From the statement of the problem the angle of elevation of $$T$$ as seen from $$A$$ is $$\alpha = \cot^{-1}(3\sqrt2)$$. First convert that to a tangent value:

$$\cot\alpha = 3\sqrt2 \;\Longrightarrow\; \tan\alpha = \frac1{3\sqrt2}.$$

By the definition of the tangent in the right triangle $$\triangle ADT$$,

$$\tan\alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{TD}{AD} = \frac{h}{y}.$$

So

$$\frac{h}{y} = \frac1{3\sqrt2}\;\Longrightarrow\; h = \frac{y}{3\sqrt2}. \quad -(1)$$

Next, the angle of elevation of $$T$$ as seen from $$B$$ is $$\beta = \csc^{-1}(2\sqrt2)$$. That gives

$$\csc\beta = 2\sqrt2 \;\Longrightarrow\; \sin\beta = \frac1{2\sqrt2}.$$

Compute $$\cos\beta$$ using $$\sin^2\beta + \cos^2\beta = 1$$:

$$\sin^2\beta = \frac1{(2\sqrt2)^2} = \frac1{8},\qquad \cos^2\beta = 1 - \frac18 = \frac78.$$

Hence

$$\cos\beta = \sqrt{\frac78} = \frac{\sqrt7}{2\sqrt2}.$$

Now obtain $$\tan\beta$$:

$$\tan\beta = \frac{\sin\beta}{\cos\beta} = \frac{1/(2\sqrt2)}{\sqrt7/(2\sqrt2)} = \frac1{\sqrt7}.$$

In the right triangle $$\triangle BDT$$ we again apply the tangent definition:

$$\tan\beta = \frac{\text{opposite}}{\text{adjacent}} = \frac{TD}{BD} = \frac{h}{x}.$$

Therefore

$$\frac{h}{x} = \frac1{\sqrt7}\;\Longrightarrow\; h = \frac{x}{\sqrt7}. \quad -(2)$$

Both equations (1) and (2) give the same quantity $$h$$, so equate them:

$$\frac{y}{3\sqrt2} = \frac{x}{\sqrt7} \;\Longrightarrow\; y = \frac{3\sqrt2}{\sqrt7}\,x. \quad -(3)$$

Now use the Pythagorean theorem in right triangle $$\triangle ABD$$:

$$AD^2 + BD^2 = AB^2.$$ That is, $$y^2 + x^2 = 100^2 = 10000. \quad -(4)$$

Substitute the value of $$y$$ from (3) into (4):

$$\left(\frac{3\sqrt2}{\sqrt7}x\right)^2 + x^2 = 10000$$ $$\frac{9\cdot2}{7}x^2 + x^2 = 10000$$ $$\frac{18}{7}x^2 + x^2 = 10000$$ $$\left(\frac{18}{7} + 1\right)x^2 = 10000$$ $$\frac{25}{7}x^2 = 10000$$ $$x^2 = 10000 \times \frac{7}{25} = 2800.$$

Thus

$$x = \sqrt{2800} = \sqrt{28 \times 100} = 10\sqrt{28} = 10 \times 2\sqrt7 = 20\sqrt7\text{ m}. \quad -(5)$$

Finally compute $$h$$ from (2):

$$h = \frac{x}{\sqrt7} = \frac{20\sqrt7}{\sqrt7} = 20\text{ m}.$$

Hence, the correct answer is Option B.

Let us imagine the calm surface of the lake as the horizontal reference line. The cloud is somewhere vertically above this surface, and its mirror image (the reflection) is the same vertical distance below the surface. The point $$P$$ is on a tower or pole that rises $$25\text{ m}$$ above the lake.

Call the unknown height of the cloud above the lake $$H\text{ m}$$. Draw a vertical line through the cloud down to the lake and further down to its reflection; call the foot of that vertical line on the lake’s surface $$A$$. Join $$P$$ to the actual cloud and also join $$P$$ to the reflection. Let the horizontal distance from $$P$$ to $$A$$ be $$x\text{ m}$$.

We now have two right-angled triangles that share the same base $$x$$:

1. Triangle $$P A$$-cloud, lying above the lake.
2. Triangle $$P A$$-reflection, lying below the lake.

The problem gives two angles:

• The angle of elevation of the cloud from $$P$$ is $$30^{\circ}$$.
• The angle of depression of the reflection from $$P$$ is $$60^{\circ}$$.

The basic trigonometric fact we will use is the definition of the tangent of an angle in a right-angled triangle:

$$\tan \theta \;=\;\frac{\text{opposite side}}{\text{adjacent side}}.$$

First apply this to the cloud itself. In triangle $$P A$$-cloud, the side opposite the $$30^{\circ}$$ angle is the vertical segment from $$P$$ down to the lake’s surface (that length is $$H-25$$, because $$H$$ is the cloud’s height above the lake and $$25\text{ m}$$ is the height of $$P$$ above the lake). The adjacent side is the horizontal segment $$P A$$ of length $$x$$. Hence

$$\tan 30^{\circ} \;=\;\frac{H-25}{x}.$$

We know $$\tan 30^{\circ}=\dfrac{1}{\sqrt3}$$, so

$$\frac{1}{\sqrt3} \;=\;\frac{H-25}{x},$$

which gives

$$x \;=\;\sqrt3\,(H-25).$$

Next apply the same idea to the reflection. In triangle $$P A$$-reflection, the angle at $$P$$ is a $$60^{\circ}$$ depression; its opposite side is the full vertical drop from $$P$$ down to the reflection. That drop equals $$25+H$$ (down $$25\text{ m}$$ to the lake and then a further $$H\text{ m}$$ to the reflection). The adjacent side is again the same horizontal length $$x$$. Therefore

$$\tan 60^{\circ} \;=\;\frac{25+H}{x}.$$

Because $$\tan 60^{\circ}=\sqrt3$$, we have

$$\sqrt3 \;=\;\frac{25+H}{x},$$

so that

$$x \;=\;\frac{25+H}{\sqrt3}.$$

Both expressions represent the same $$x$$, so we equate them:

$$\sqrt3\,(H-25) \;=\;\frac{25+H}{\sqrt3}.$$

Multiply both sides by $$\sqrt3$$ to clear the denominator:

$$3\,(H-25) \;=\;25+H.$$

Expand the left side:

$$3H - 75 \;=\;25 + H.$$

Bring the $$H$$ terms together and the constants together:

$$3H - H \;=\;25 + 75,$$

$$2H \;=\;100.$$

Divide by $$2$$:

$$H \;=\;50.$$

Thus the cloud is $$50\text{ m}$$ above the surface of the lake.

Hence, the correct answer is Option A.

We are given a triangle $$ABC$$ whose sides are $$AB = 7\text{ m}$$, $$BC = 5\text{ m}$$ and $$CA = 6\text{ m}$$. The point $$D$$ is the mid-point of $$AC$$, so $$AD = DC = \dfrac{6}{2} = 3\text{ m}$$. A vertical lamp-post is erected at $$D$$; let its height be $$h\text{ m}$$. From point $$B$$ the top of the lamp-post is seen under an angle of elevation of $$30^{\circ}$$.

Because the lamp-post is vertical, the right-angled triangle that matters is $$\triangle BDT$$, where $$T$$ is the top of the lamp-post. In this right-angled triangle, the horizontal leg is $$BD$$ and the vertical leg is $$DT = h$$. The angle at $$B$$ is $$30^{\circ}$$. By the definition of the tangent of an angle, we have

$$\tan 30^{\circ} \;=\; \dfrac{\text{opposite side}}{\text{adjacent side}} =\dfrac{h}{BD}.$$

So, if we can determine $$BD$$, we can immediately obtain $$h$$. The point $$D$$ is the mid-point of $$AC$$, hence $$BD$$ is a median drawn from vertex $$B$$ to the side $$AC$$. There is a standard formula for the length of a median in a triangle:

For a triangle with sides $$a,\,b,\,c$$ opposite to vertices $$A,\,B,\,C$$ respectively, the length of the median $$m_b$$ from vertex $$B$$ to side $$AC$$ (whose length is $$b$$) is

$$m_b \;=\;\dfrac{1}{2}\sqrt{2a^{2}+2c^{2}-b^{2}}.$$

In our triangle we identify $$a = BC = 5,\quad b = AC = 6,\quad c = AB = 7.$$

Substituting these values into the median formula, we get

$$ \begin{aligned} BD &= m_b \\ &= \dfrac{1}{2}\sqrt{2a^{2}+2c^{2}-b^{2}} \\ &= \dfrac{1}{2}\sqrt{2(5)^{2}+2(7)^{2}-(6)^{2}} \\ &= \dfrac{1}{2}\sqrt{2\cdot25 + 2\cdot49 - 36} \\ &= \dfrac{1}{2}\sqrt{50 + 98 - 36} \\ &= \dfrac{1}{2}\sqrt{112} \\ &= \dfrac{1}{2}\sqrt{16\cdot7} \\ &= \dfrac{1}{2}\cdot4\sqrt{7} \\ &= 2\sqrt{7}\text{ m}. \end{aligned} $$

Now we return to the right-angled triangle $$\triangle BDT$$. We have already stated

$$\tan 30^{\circ} = \dfrac{h}{BD}.$$

We also know the trigonometric value $$\tan 30^{\circ} = \dfrac{1}{\sqrt{3}}$$. Therefore

$$ \begin{aligned} \dfrac{1}{\sqrt{3}} &= \dfrac{h}{2\sqrt{7}} \\[4pt] \Rightarrow h &= \dfrac{2\sqrt{7}}{\sqrt{3}}. \end{aligned} $$

To give the height in a more conventional radical form, we rationalise the denominator:

$$ \begin{aligned} h &= \dfrac{2\sqrt{7}}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}} \\[4pt] &= \dfrac{2\sqrt{7}\,\sqrt{3}}{3} \\[4pt] &= \dfrac{2\sqrt{21}}{3}\text{ m}. \end{aligned} $$

So the lamp-post is $$\dfrac{2}{3}\sqrt{21}\text{ m}$$ tall.

Hence, the correct answer is Option B.

Let the three sides, taken in ascending order, be $$(b-d),\;b,\;b+d$$ where $$b>d>0$$ so that they form an arithmetic progression of common difference $$d$$. Call the angles opposite these sides $$\alpha,\;\beta,\;\gamma$$ respectively, so $$\gamma$$ is the largest and $$\alpha$$ is the smallest.

The statement “the greatest angle is double the smallest” translates to $$\gamma = 2\alpha.$$ Because a triangle’s angles sum to $$180^\circ,$$ we have

$$\alpha+\beta+\gamma=\alpha+\beta+2\alpha=180^\circ\;\Longrightarrow\;\beta = 180^\circ-3\alpha.$$

Invoke the Law of Sines:

$$\frac{b-d}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{b+d}{\sin2\alpha}.$$

For convenience set the smallest side $$b-d$$ equal to one unit, i.e. $$b-d=1.$$ Then, writing the equalities in pairwise form,

$$\frac{b}{1}= \frac{\sin\beta}{\sin\alpha}$$ $$\text{and}$$ $$\frac{b+d}{1}= \frac{\sin2\alpha}{\sin\alpha}.$$

First simplify the right-hand ratios. Using $$\beta=180^\circ-3\alpha$$ and the identity $$\sin(180^\circ-\theta)=\sin\theta,$$

$$\frac{\sin\beta}{\sin\alpha}=\frac{\sin3\alpha}{\sin\alpha}.$$ Write $$\sin3\alpha=3\sin\alpha-4\sin^3\alpha,$$ so

$$\frac{\sin3\alpha}{\sin\alpha}=3-4\sin^2\alpha.$$

Similarly, $$\frac{\sin2\alpha}{\sin\alpha}= \frac{2\sin\alpha\cos\alpha}{\sin\alpha}=2\cos\alpha.$$

Hence

$$\boxed{b = 3-4\sin^2\alpha}$$ and $$\boxed{\,b+d = 2\cos\alpha\,}.$$

Because $$\sin^2\alpha + \cos^2\alpha =1,$$ convert the first boxed equation entirely into $$\cos\alpha$$ :

$$b = 3-4(1-\cos^2\alpha)=4\cos^2\alpha-1.$$

Now relate $$b+d$$ to $$b$$ and the unit length choice. Since $$b-d=1,$$ we have $$d=b-1,$$ so

$$b+d = b+(b-1)=2b-1.$$

But $$b+d$$ also equals $$2\cos\alpha,$$ therefore

$$2b-1 = 2\cos\alpha.$$

Insert $$b = 4\cos^2\alpha-1$$ from above:

$$2\left(4\cos^2\alpha-1\right)-1 = 2\cos\alpha.$$

Simplify step by step:

$$8\cos^2\alpha-2-1 = 2\cos\alpha$$

$$8\cos^2\alpha - 3 = 2\cos\alpha$$

$$8\cos^2\alpha - 2\cos\alpha - 3 = 0.$$

Let $$x=\cos\alpha.$$ The quadratic becomes

$$8x^2 - 2x - 3 = 0.$$

Apply the quadratic formula:

$$x = \frac{2 \pm \sqrt{2^2 - 4(8)(-3)}}{16}= \frac{2 \pm \sqrt{4+96}}{16}= \frac{2 \pm 10}{16}.$$

This gives two roots, $$x=\dfrac{12}{16}=\dfrac34$$ or $$x=-\dfrac12.$$ Because the smallest angle $$\alpha$$ must be acute, $$\cos\alpha>0,$$ so retain $$\cos\alpha=\dfrac34.$$

Consequently,

$$\sin\alpha = \sqrt{1-\left(\dfrac34\right)^2}= \sqrt{1-\dfrac9{16}}=\sqrt{\dfrac7{16}}=\dfrac{\sqrt7}{4}.$$

Now compute the side lengths relative to the unit smallest side.

Middle side:

$$b = 4\cos^2\alpha - 1 = 4\left(\dfrac34\right)^2 - 1 = 4\cdot\dfrac9{16}-1 = \dfrac9{4}-1 = \dfrac54.$$

Largest side:

$$b+d = 2b-1 = 2\left(\dfrac54\right)-1 = \dfrac{10}{4}-\dfrac4{4}= \dfrac6{4}=\dfrac32.$$

Hence, with $$b-d=1,\;b=\dfrac54,\;b+d=\dfrac32,$$ the proportional sides are

$$1 : \dfrac54 : \dfrac32.$$

Multiply every length by $$4$$ to clear denominators:

$$4 : 5 : 6.$$

Thus the three sides are in the ratio $$4:5:6,$$ which corresponds to Option D.

Hence, the correct answer is Option D.

Let us denote the foot of the tower by $$O$$ and the top of the tower by $$T$$. The given point on the horizontal plane is $$A$$, and a point $$B$$ is located vertically above $$A$$ at a height of $$30\text{ m}$$. We denote the horizontal distance $$OA$$ by $$x\text{ m}$$ and the height of the tower $$OT$$ by $$h\text{ m}$$.

We first use the information from point $$A$$. The angle of elevation of $$T$$ from $$A$$ is $$45^\circ$$. For a right-angled triangle, the basic trigonometric definition is $$\tan(\theta)=\dfrac{\text{opposite side}}{\text{adjacent side}}.$$ Applying this to $$\triangle OAT$$, we have

$$\tan 45^\circ=\dfrac{OT}{OA}\;.$$

Since $$\tan 45^\circ=1$$, this yields

$$1=\dfrac{h}{x}\quad\Longrightarrow\quad h=x.$$ So the height of the tower equals the horizontal distance $$OA$$.

Next we use the information from point $$B$$, which is $$30\text{ m}$$ directly above $$A$$. Hence $$AB=30\text{ m}$$, and the horizontal distance from $$B$$ to the foot of the tower is still $$x\text{ m}$$ (because $$B$$ is vertically above $$A$$).

The angle of elevation of $$T$$ from $$B$$ is $$30^\circ$$, so for right-angled triangle $$\triangle OBT$$ we again apply $$\tan(\theta)=\dfrac{\text{opposite side}}{\text{adjacent side}}.$$ Here the vertical side opposite the angle at $$B$$ is $$TB$$, whose length is the difference between the tower’s height and the height of $$B$$ above the plane, namely $$h-30$$. Thus

$$\tan 30^\circ=\dfrac{TB}{OB}=\dfrac{h-30}{x}\;.$$

Since $$\tan 30^\circ=\dfrac{1}{\sqrt3}$$, we write

$$\dfrac{1}{\sqrt3}=\dfrac{h-30}{x}\;.$$

But from the first part we already have $$h=x$$. Substituting $$h=x$$ into the above gives

$$\dfrac{1}{\sqrt3}=\dfrac{x-30}{x}\;.$$

Now we solve this equation for $$x$$ step by step. First multiply both sides by $$x$$:

$$\dfrac{x}{\sqrt3}=x-30\;.$$

Next, bring the term $$x$$ on the right to the left side:

$$\dfrac{x}{\sqrt3}-x=-30\;.$$

Factor out $$x$$ on the left:

$$x\left(\dfrac{1}{\sqrt3}-1\right)=-30\;.$$

Combine the terms inside the parentheses by taking a common denominator $$\sqrt3$$:

$$x\left(\dfrac{1-\sqrt3}{\sqrt3}\right)=-30\;.$$

Now multiply both sides by $$\dfrac{\sqrt3}{1-\sqrt3}$$ to isolate $$x$$:

$$x=-30\cdot\dfrac{\sqrt3}{1-\sqrt3}\;.$$

Notice that the numerator is negative because $$1-\sqrt3<0$$, so the negatives cancel. We rationalise the denominator by multiplying top and bottom by $$1+\sqrt3$$:

$$x=30\cdot\dfrac{\sqrt3}{\sqrt3-1} =30\cdot\dfrac{\sqrt3(1+\sqrt3)}{(\sqrt3-1)(1+\sqrt3)} =30\cdot\dfrac{\sqrt3(1+\sqrt3)}{3-1} =30\cdot\dfrac{\sqrt3(1+\sqrt3)}{2}\;.$$

Simplify the factor $$\sqrt3(1+\sqrt3)=\sqrt3+\sqrt3\cdot\sqrt3=\sqrt3+3$$, giving

$$x=\dfrac{30}{2}\,\bigl(3+\sqrt3\bigr) =15\bigl(3+\sqrt3\bigr)\;.$$

Thus the horizontal distance of point $$A$$ from the foot of the tower is

$$OA=x=15(3+\sqrt3)\text{ m}.$$

Hence, the correct answer is Option A.

We have a triangle $$\triangle ABC$$ in which the three interior angles $$A,\,B,\,C$$ are in arithmetic progression (A.P.).

If three numbers are in A.P., the middle term equals the average of the other two. Hence we may write

$$A,\;A+d,\;A+2d$$

for the three angles, where $$d$$ is their common difference. Their sum must be $$180^{\circ}$$, so

$$A+(A+d)+(A+2d)=180^{\circ}$$

$$3A+3d=180^{\circ}$$

Dividing every term by $$3$$ gives

$$A+d=60^{\circ} \quad\Longrightarrow\quad B=60^{\circ}.$$

Thus

$$A=60^{\circ}-d,\qquad C=60^{\circ}+d.$$

Next, the ratio of the sides opposite these angles is given: $$a:b=1:\sqrt{3}$$. We invoke the Sine Rule, which states

$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R,$$

where $$R$$ is the circumradius. From the first two fractions we get

$$\frac{a}{b}=\frac{\sin A}{\sin B}.$$

Substituting $$a:b=1:\sqrt{3}$$ and $$B=60^{\circ}$$ yields

$$\frac{1}{\sqrt{3}}=\frac{\sin A}{\sin 60^{\circ}}.$$

Because $$\sin 60^{\circ}=\dfrac{\sqrt{3}}{2},$$ we have

$$\frac{1}{\sqrt{3}}=\frac{\sin A}{\sqrt{3}/2}.$$

Multiplying both sides by $$\dfrac{\sqrt{3}}{2}$$ gives

$$\sin A=\frac{1}{2}.$$ Hence $$A=30^{\circ}$$ (the possibility $$150^{\circ}$$ is impossible, since it would make the third angle negative).

Since $$B=60^{\circ},$$ the remaining angle is

$$C=180^{\circ}-A-B=180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ}.$$

Thus $$\triangle ABC$$ is a right-angled triangle with $$C=90^{\circ}.$$ The side opposite the right angle, namely $$c=AB,$$ is the hypotenuse.

For a $$30^{\circ}\!-\!60^{\circ}\!-\!90^{\circ}$$ triangle the sides satisfy the well-known ratio

$$a:b:c = 1:\sqrt{3}:2.$$

Given $$c=4\text{ cm},$$ equating the “2” part to $$4$$ shows

$$2k = 4 \;\Longrightarrow\; k = 2.$$

Therefore

$$a = k = 2\text{ cm},\qquad b = k\sqrt{3} = 2\sqrt{3}\text{ cm}.$$

Because the legs $$a$$ and $$b$$ are perpendicular (the right angle is at $$C$$), the area $$\Delta$$ of the triangle is

$$\Delta = \frac{1}{2}\,a\,b.$$

Substituting the lengths just found,

$$\Delta = \frac{1}{2}\times 2 \times 2\sqrt{3}=2\sqrt{3}\;\text{sq.\;cm}.$$

Hence, the correct answer is Option A.

Let us denote the bases of the two vertical poles by the points $$A$$ and $$B$$ on the common horizontal ground.

The pole at $$A$$ has height $$5\ \text{m}$$, so its top is the point $$P$$ such that $$AP = 5$$.

The pole at $$B$$ has height $$10\ \text{m}$$, so its top is the point $$Q$$ with $$BQ = 10$$.

Let the horizontal distance between the two bases be $$AB = x$$ metres. Our aim is to find this unknown $$x$$.

We are told that the straight line joining the tops, that is the segment $$PQ$$, makes an angle of $$15^{\circ}$$ with the horizontal ground. This means that, if we travel from the shorter top $$P$$ (height $$5$$) to the taller top $$Q$$ (height $$10$$) along the segment $$PQ$$, we are ascending at an inclination of $$15^{\circ}$$ above the horizontal.

Now observe the right-angled triangle whose horizontal leg is $$AB$$, vertical leg is the difference of heights $$BQ - AP$$, and hypotenuse is $$PQ$$. The vertical rise from $$P$$ to $$Q$$ is

$$\text{rise} \;=\; 10 - 5 \;=\; 5\ \text{m}.$$

The horizontal run is exactly the distance $$x = AB$$ that we want.

By definition, for any right-angled triangle, the tangent of the angle made by the hypotenuse with the horizontal base equals

$$\tan(\text{angle}) \;=\; \dfrac{\text{opposite side (rise)}}{\text{adjacent side (run)}}.$$

Applying this to our triangle with angle $$15^{\circ}$$, we have

$$\tan 15^{\circ} \;=\; \dfrac{5}{x}.$$

Solving this equation for $$x$$ gives

$$x \;=\; \dfrac{5}{\tan 15^{\circ}}.$$

Next, we recall the exact trigonometric value

$$\tan 15^{\circ} = 2 - \sqrt{3}.$$

Substituting this into the expression for $$x$$, we get

$$x \;=\; \dfrac{5}{2 - \sqrt{3}}.$$

To simplify, we rationalise the denominator. Multiply numerator and denominator by the conjugate $$(2 + \sqrt{3})$$:

$$x \;=\; 5 \times \dfrac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})}.$$

The denominator is a difference of squares:

$$(2 - \sqrt{3})(2 + \sqrt{3}) = 2^{2} - (\sqrt{3})^{2} = 4 - 3 = 1.$$

Therefore, the fraction simplifies beautifully to

$$x \;=\; 5(2 + \sqrt{3}).$$

This value represents the distance between the two poles, in metres.

Hence, the correct answer is Option C.

We begin with the three real-valued functions defined on the indicated interval:

$$f(x)=\sqrt{x}, \qquad g(x)=\tan x, \qquad h(x)=\frac{1-x^{2}}{1+x^{2}}, \qquad x\in\left(0,\frac{3}{2}\right).$$

The required function is the composition

$$\phi(x)=((h\circ f)\circ g)(x).$$

By the definition of composition, we first apply $$g$$ to $$x,$$ then apply $$f$$ to the result, and finally apply $$h$$. Symbolically,

$$\phi(x)=h\!\left(f\!\left(g(x)\right)\right).$$

Now, step by step:

• First map $$x$$ through $$g$$:

$$g(x)=\tan x.$$

• Next feed this into $$f$$. Because $$f(u)=\sqrt{u},$$ we have

$$f\!\left(g(x)\right)=\sqrt{\tan x}.$$

• Finally, apply $$h$$, whose formula is $$h(y)=\dfrac{1-y^{2}}{1+y^{2}}.$$ Substituting $$y=\sqrt{\tan x}$$ gives

$$\phi(x)=\frac{1-\bigl(\sqrt{\tan x}\bigr)^{2}}{1+\bigl(\sqrt{\tan x}\bigr)^{2}}.$$

Since $$\bigl(\sqrt{\tan x}\bigr)^{2}=\tan x,$$ this simplifies directly to

$$\phi(x)=\frac{1-\tan x}{1+\tan x}.$$

We must now evaluate this expression at $$x=\dfrac{\pi}{3}.$$ Recalling that

$$\tan\left(\dfrac{\pi}{3}\right)=\sqrt{3},$$

we substitute:

$$\phi\!\left(\frac{\pi}{3}\right)=\frac{1-\sqrt{3}}{1+\sqrt{3}}.$$

To recognise this value, we rationalise the denominator. Multiply numerator and denominator by the conjugate $$1-\sqrt{3}$$:

$$ \phi\!\left(\frac{\pi}{3}\right)= \frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})} =\frac{(1-\sqrt{3})^{2}}{1-3}. $$

Expanding the square in the numerator using $$(a-b)^{2}=a^{2}-2ab+b^{2},$$ we obtain

$$ (1-\sqrt{3})^{2}=1-2\sqrt{3}+3=4-2\sqrt{3}. $$

The denominator simplifies to $$1-3=-2.$$ Therefore,

$$ \phi\!\left(\frac{\pi}{3}\right)=\frac{4-2\sqrt{3}}{-2}. $$

Dividing numerator and denominator by $$-2$$ yields

$$ \phi\!\left(\frac{\pi}{3}\right)=-(2-\sqrt{3})=\sqrt{3}-2. $$

Now we compare this value with the tangents listed in the options. We recall the well-known exact values:

$$ \tan\frac{\pi}{12}=2-\sqrt{3}, \qquad \tan\frac{11\pi}{12}=\tan\!\left(\pi-\frac{\pi}{12}\right)=-\tan\frac{\pi}{12}=-(2-\sqrt{3})=\sqrt{3}-2. $$

Thus

$$ \phi\!\left(\frac{\pi}{3}\right)=\sqrt{3}-2=\tan\frac{11\pi}{12}. $$

Hence, the correct answer is Option D.

We are told that in $$\triangle ABC$$ the angle $$A$$ satisfies the relation $$5\cos A + 3 = 0$$.

First, we isolate $$\cos A$$:

$$5\cos A + 3 = 0 \;\Longrightarrow\; 5\cos A = -3 \;\Longrightarrow\; \cos A = -\dfrac{3}{5}.$$

Since $$A$$ is an angle of a triangle, we know $$0 < A < \pi$$. The value $$\cos A = -\dfrac{3}{5}$$ is negative, so $$A$$ lies in the second quadrant, where sine is positive and cosine is negative.

Now we compute $$\sin A$$ using the fundamental Pythagorean identity $$\sin^2\!A + \cos^2\!A = 1$$.

$$\sin^2\!A = 1 - \cos^2\!A = 1 - \left(-\dfrac{3}{5}\right)^2 = 1 - \dfrac{9}{25} = \dfrac{25}{25} - \dfrac{9}{25} = \dfrac{16}{25}.$$

Because $$A$$ is in the second quadrant, $$\sin A$$ is positive, hence

$$\sin A = \dfrac{4}{5}.$$

Next we find the other required trigonometric ratios.

The secant is the reciprocal of cosine, so

$$\sec A = \dfrac{1}{\cos A} = \dfrac{1}{-\dfrac{3}{5}} = -\dfrac{5}{3}.$$

The tangent is the quotient of sine by cosine, so

$$\tan A = \dfrac{\sin A}{\cos A} = \dfrac{\dfrac{4}{5}}{-\dfrac{3}{5}} = -\dfrac{4}{3}.$$

Now we examine the quadratic equation

$$9x^2 + 27x + 20 = 0$$

and test whether $$x = \sec A$$ or $$x = \tan A$$ satisfy it.

Substituting $$x = \sec A = -\dfrac{5}{3}$$:

$$9\left(-\dfrac{5}{3}\right)^2 + 27\left(-\dfrac{5}{3}\right) + 20 = 9\left(\dfrac{25}{9}\right) + 27\left(-\dfrac{5}{3}\right) + 20 = 25 - 45 + 20 = 0.$$

So $$x = \sec A$$ is indeed a root.

Substituting $$x = \tan A = -\dfrac{4}{3}$$:

$$9\left(-\dfrac{4}{3}\right)^2 + 27\left(-\dfrac{4}{3}\right) + 20 = 9\left(\dfrac{16}{9}\right) + 27\left(-\dfrac{4}{3}\right) + 20 = 16 - 36 + 20 = 0.$$

Thus $$x = \tan A$$ is also a root.

Since the two roots of the quadratic are $$\sec A$$ and $$\tan A$$, they correspond exactly to Option B.

Hence, the correct answer is Option B.

We need to find the number of solutions of $$\sin 3x = \cos 2x$$ in the interval $$\left(\frac{\pi}{2}, \pi\right)$$.

We rewrite $$\cos 2x$$ as $$\sin\left(\frac{\pi}{2} - 2x\right)$$, so the equation becomes:

$$\sin 3x = \sin\left(\frac{\pi}{2} - 2x\right)$$

The general solution of $$\sin A = \sin B$$ is $$A = n\pi + (-1)^n B$$, where $$n$$ is an integer.

Case 1: $$3x = 2k\pi + \left(\frac{\pi}{2} - 2x\right)$$ (when $$n$$ is even):

$$5x = 2k\pi + \frac{\pi}{2}$$

$$x = \frac{(4k + 1)\pi}{10}$$

For $$x \in \left(\frac{\pi}{2}, \pi\right)$$: we need $$\frac{\pi}{2} < \frac{(4k+1)\pi}{10} < \pi$$, which gives $$5 < 4k + 1 < 10$$, so $$4 < 4k < 9$$, meaning $$1 < k < 2.25$$.

The only integer value is $$k = 2$$, giving $$x = \frac{9\pi}{10}$$. We verify: $$\frac{\pi}{2} < \frac{9\pi}{10} < \pi$$. This is valid.

Case 2: $$3x = (2k+1)\pi - \left(\frac{\pi}{2} - 2x\right)$$ (when $$n$$ is odd):

$$3x = (2k+1)\pi - \frac{\pi}{2} + 2x$$

$$x = \frac{(4k+1)\pi}{2}$$

For $$x \in \left(\frac{\pi}{2}, \pi\right)$$: we need $$\frac{\pi}{2} < \frac{(4k+1)\pi}{2} < \pi$$, which gives $$1 < 4k + 1 < 2$$, so $$0 < 4k < 1$$, meaning $$0 < k < 0.25$$.

No integer value of $$k$$ satisfies this. So there are no solutions from this case.

The total number of solutions is 1.

The answer is Option D: 1.

First we examine Statement p. The equation given is

$$2\sin\frac{\theta}{2}= \sqrt{\,1+\sin\theta\,}-\sqrt{\,1-\sin\theta\,}.$$

We are told to put $$\theta =240^\circ.$$ Let us evaluate every term one by one.

We have $$\sin 240^\circ=\sin\left(180^\circ+60^\circ\right)=-\sin60^\circ=-\frac{\sqrt3}{2}.$$

Now compute the left-hand side (L.H.S.) of the equation:

$$2\sin\frac{\theta}{2}=2\sin\left(\frac{240^\circ}{2}\right)=2\sin120^\circ.$$

Using $$\sin120^\circ=\sin(60^\circ)=\frac{\sqrt3}{2},$$ we get

$$2\sin120^\circ=2\cdot\frac{\sqrt3}{2}=\sqrt3.$$

So the L.H.S. equals $$\sqrt3\;(\approx1.732).$$

Next we evaluate the right-hand side (R.H.S.):

$$\sqrt{\,1+\sin\theta\,}-\sqrt{\,1-\sin\theta\,}.$$

Substituting $$\sin\theta=-\frac{\sqrt3}{2},$$ we obtain

$$\sqrt{\,1-\frac{\sqrt3}{2}\,}-\sqrt{\,1+\frac{\sqrt3}{2}\,}.$$

Convert the 1 to a fraction with denominator 2:

$$1=\frac{2}{2},$$ so

$$\sqrt{\frac{2-\sqrt3}{2}}-\sqrt{\frac{2+\sqrt3}{2}}.$$

The first square root is approximately $$\sqrt{0.134}\approx0.366,$$ and the second is approximately $$\sqrt{1.866}\approx1.366.$$ Therefore the R.H.S. is roughly

$$0.366-1.366\approx-1.000.$$

We now compare:

$$\text{L.H.S.}=\sqrt3\;(\approx1.732),\qquad\text{R.H.S.}\approx-1.000.$$

The two sides are not equal, so the equation does not hold for $$\theta=240^\circ.$$ Hence Statement p is false.

Now we investigate Statement q. In any quadrilateral ABCD the interior angles satisfy

$$A+B+C+D=360^\circ.$$

We are to test the identity

$$\cos\!\left(\tfrac12(A+C)\right)+\cos\!\left(\tfrac12(B+D)\right)=0.$$

Let us denote

$$X=\tfrac12(A+C).$$

Because the sum of all four angles is $$360^\circ,$$ we have

$$B+D=360^\circ-(A+C).$$

Dividing by 2 gives

$$\tfrac12(B+D)=\tfrac12\!\left[\,360^\circ-(A+C)\right]=180^\circ-\tfrac12(A+C)=180^\circ-X.$$

Now write the left side of the proposed identity in terms of $$X:$$

$$\cos X+\cos(180^\circ-X).$$

We recall the trigonometric fact: $$\cos(180^\circ-\theta)=-\cos\theta.$$ Stating the formula plainly, “cosine changes sign in the second quadrant.” Using this formula with $$\theta=X$$ gives

$$\cos(180^\circ-X)=-\cos X.$$

Substituting, we get

$$\cos X+\bigl(-\cos X\bigr)=0.$$

The expression indeed equals zero for every choice of angles A, B, C and D that add to $$360^\circ.$$ Thus Statement q is true.

We have found:

Statement p: False,   Statement q: True.

The ordered pair of truth values is (F, T), which corresponds to Option A.

Hence, the correct answer is Option A.

We start with the given equation

$$8\cos x\;\Bigl(\cos\!\left(\tfrac{\pi}{6}+x\right)\cos\!\left(\tfrac{\pi}{6}-x\right)-\tfrac12\Bigr)=1,\qquad x\in[0,\pi].$$

First we simplify the expression inside the big parentheses. We recall the product-to-sum identity

$$\cos A\cos B=\tfrac12\bigl[\cos(A+B)+\cos(A-B)\bigr].$$

Here we have $$A=\tfrac{\pi}{6}+x,\;B=\tfrac{\pi}{6}-x.$$ Substituting, we obtain

$$\cos\!\left(\tfrac{\pi}{6}+x\right)\cos\!\left(\tfrac{\pi}{6}-x\right) =\tfrac12\Bigl[\cos\!\bigl((\tfrac{\pi}{6}+x)+(\tfrac{\pi}{6}-x)\bigr)+\cos\!\bigl((\tfrac{\pi}{6}+x)-(\tfrac{\pi}{6}-x)\bigr)\Bigr].$$

Simplifying the angles, $$(\tfrac{\pi}{6}+x)+(\tfrac{\pi}{6}-x)=\tfrac{\pi}{3}$$ and $$(\tfrac{\pi}{6}+x)-(\tfrac{\pi}{6}-x)=2x$$, so

$$\cos\!\left(\tfrac{\pi}{6}+x\right)\cos\!\left(\tfrac{\pi}{6}-x\right) =\tfrac12\,[\cos(\tfrac{\pi}{3})+\cos(2x)].$$

We know $$\cos(\tfrac{\pi}{3})=\tfrac12,$$ hence

$$\cos\!\left(\tfrac{\pi}{6}+x\right)\cos\!\left(\tfrac{\pi}{6}-x\right)=\tfrac12\!\left[\tfrac12+\cos(2x)\right]=\tfrac14+\tfrac12\cos(2x).$$

Now subtract the $$\tfrac12$$ that appears in the original bracket:

$$\cos\!\left(\tfrac{\pi}{6}+x\right)\cos\!\left(\tfrac{\pi}{6}-x\right)-\tfrac12 =\Bigl(\tfrac14+\tfrac12\cos(2x)\Bigr)-\tfrac12 =-\tfrac14+\tfrac12\cos(2x).$$

Substituting this result back into the main equation, we get

$$8\cos x\;\Bigl(-\tfrac14+\tfrac12\cos(2x)\Bigr)=1.$$

Multiplying by $$8\cos x$$ term by term gives

$$8\cos x\left(-\tfrac14\right)+8\cos x\left(\tfrac12\cos(2x)\right)=1,$$

or

$$-2\cos x+4\cos x\cos(2x)=1.$$

We next express $$\cos(2x)$$ in terms of $$\cos x$$ using the double-angle formula

$$\cos(2x)=2\cos^2x-1.$$

Substituting this, we find

$$4\cos x\cos(2x)=4\cos x\bigl(2\cos^2x-1\bigr)=8\cos^3x-4\cos x.$$

Therefore

$$-2\cos x+\bigl(8\cos^3x-4\cos x\bigr)=1,$$ or $$8\cos^3x-6\cos x=1.$$

We notice the expression $$4\cos^3x-3\cos x$$, which is famous from the triple-angle identity

$$\cos(3x)=4\cos^3x-3\cos x.$$

Indeed,

$$8\cos^3x-6\cos x=2(4\cos^3x-3\cos x)=2\cos(3x).$$

So the equation reduces neatly to

$$2\cos(3x)=1\;\;\Longrightarrow\;\;\cos(3x)=\tfrac12.$$

Let us set $$y=3x.$$ Because $$x\in[0,\pi],$$ we have $$y\in[0,3\pi].$$ We now solve

$$\cos y=\tfrac12,\qquad y\in[0,3\pi].$$

The general solutions of $$\cos y=\tfrac12$$ are

$$y=2n\pi\pm\tfrac{\pi}{3},\qquad n\in\mathbb Z.$$

We list the values lying in $$[0,3\pi]$$:

• For $$n=0$$: $$y=\tfrac{\pi}{3}$$ (while $$-\tfrac{\pi}{3}$$ is negative and hence outside the interval).
• For $$n=1$$: $$y=2\pi-\tfrac{\pi}{3}=\tfrac{5\pi}{3}$$ and $$y=2\pi+\tfrac{\pi}{3}=\tfrac{7\pi}{3}$$; both are within $$[0,3\pi]$$.
• $$n=2$$ would give values $$\gt \,3\pi$$, so we stop here.

Thus $$y\in\Bigl\{\tfrac{\pi}{3},\;\tfrac{5\pi}{3},\;\tfrac{7\pi}{3}\Bigr\}.$$

We now convert back to $$x$$ using $$x=\tfrac{y}{3}$$:

$$x_1=\tfrac{\pi}{9},\qquad x_2=\tfrac{5\pi}{9},\qquad x_3=\tfrac{7\pi}{9}.$$

All three lie in the required interval $$[0,\pi]$$, so they are the complete set of solutions.

Their sum is

$$x_1+x_2+x_3=\tfrac{\pi}{9}+\tfrac{5\pi}{9}+\tfrac{7\pi}{9} =\tfrac{(1+5+7)\pi}{9}=\tfrac{13\pi}{9}.$$

Comparing with the form $$k\pi$$, we see $$k=\tfrac{13}{9}.$$

Hence, the correct answer is Option C.

We are told that $$\tan A$$ and $$\tan B$$ are the roots of the quadratic equation $$3x^2-10x-25=0$$. For any quadratic $$ax^2+bx+c=0$$ whose roots are $$\alpha$$ and $$\beta$$, we have the relations

$$\alpha+\beta=-\dfrac{b}{a},\qquad \alpha\beta=\dfrac{c}{a}.$$

Applying this to the given quadratic, we obtain

$$\tan A+\tan B=\dfrac{10}{3},\qquad \tan A\cdot\tan B=-\dfrac{25}{3}.$$

Next we need $$\tan(A+B)$$. Using the tangent-addition formula

$$\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\,\tan B},$$

we substitute the values just found:

$$\tan(A+B)=\dfrac{\dfrac{10}{3}}{1-\!\left(-\dfrac{25}{3}\right)}=\dfrac{\dfrac{10}{3}}{1+\dfrac{25}{3}} =\dfrac{\dfrac{10}{3}}{\dfrac{28}{3}}=\dfrac{10}{28}=\dfrac{5}{14}.$$

Let us denote $$T=\tan(A+B)=\dfrac{5}{14}.$$

The required expression is

$$E=3\sin^2(A+B)-10\sin(A+B)\cos(A+B)-25\cos^2(A+B).$$

To rewrite everything in terms of $$T$$, we recall the identities

$$\sin^2\theta=\dfrac{\tan^2\theta}{1+\tan^2\theta},\qquad \sin\theta\cos\theta=\dfrac{\tan\theta}{1+\tan^2\theta},\qquad \cos^2\theta=\dfrac{1}{1+\tan^2\theta}.$$

Using $$\theta=A+B$$, every term acquires the common denominator $$1+T^2$$:

$$E=\dfrac{3T^2-10T-25}{1+T^2}.$$

Since $$T=\dfrac{5}{14},$$ we first compute

$$T^2=\left(\dfrac{5}{14}\right)^2=\dfrac{25}{196},\qquad 1+T^2=1+\dfrac{25}{196}=\dfrac{196+25}{196}=\dfrac{221}{196}.$$

Now the numerator becomes

$$3T^2-10T-25 =3\!\left(\dfrac{25}{196}\right)-10\!\left(\dfrac{5}{14}\right)-25 =\dfrac{75}{196}-\dfrac{50}{14}-25.$$

To combine, we convert each term to the common denominator 196:

$$\dfrac{50}{14}=\dfrac{50\times14}{14\times14}=\dfrac{700}{196},\qquad 25=\dfrac{25\times196}{196}=\dfrac{4900}{196}.$$

Hence

$$3T^2-10T-25 =\dfrac{75-700-4900}{196} =\dfrac{-5525}{196}.$$

Putting numerator and denominator together, we obtain

$$E=\dfrac{-5525/196}{221/196} =-\dfrac{5525}{221}.$$

Recognising that $$221\times25=5525,$$ we simplify:

$$E=-25.$$

Hence, the correct answer is Option B.

Let us assume that the tower is perfectly vertical and its height is $$h$$ metres. The car is moving on a horizontal road which we shall regard as level ground. From the top of the tower a man looks down at the car; therefore the angle of depression seen by the man is numerically equal to the angle of elevation that the car would see if it looked up at the top of the tower (alternate angles made by a transversal with two parallel horizontal lines).

At an initial instant the angle of depression is $$30^\circ$$. Let the horizontal distance of the car from the foot of the tower at that moment be $$x_1$$ metres. By simple right-triangle trigonometry in the vertical plane that contains the tower and the car we have the tangent relation:

$$\tan 30^\circ \;=\;\frac{\text{opposite side}}{\text{adjacent side}}\;=\;\frac{h}{x_1}.$$

The numerical value of $$\tan 30^\circ$$ is $$\frac{1}{\sqrt3}$$, so

$$\frac{1}{\sqrt3} \;=\;\frac{h}{x_1}.$$

Cross-multiplying gives

$$x_1 \;=\;\sqrt3\,h.$$

After the car has moved nearer, the angle of depression becomes $$45^\circ$$. Let the new horizontal distance be $$x_2$$ metres. Using the same tangent formula, now with $$45^\circ$$, we write

$$\tan 45^\circ \;=\;\frac{h}{x_2}.$$

Since $$\tan 45^\circ = 1$$, we have immediately

$$1 \;=\;\frac{h}{x_2}\quad\Longrightarrow\quad x_2 \;=\;h.$$

Thus the car travels the horizontal distance

$$x_1 - x_2 \;=\;\sqrt3\,h - h \;=\;h(\sqrt3 - 1)$$

while the angle of depression changes from $$30^\circ$$ to $$45^\circ$$. According to the statement of the problem this travel takes $$18$$ minutes. If the uniform speed of the car be $$v$$ metres per minute, then

$$v \;=\;\frac{\text{distance}}{\text{time}}\;=\;\frac{h(\sqrt3 - 1)}{18}.$$

After this instant the car still has to cover the remaining horizontal distance $$x_2$$ (which equals $$h$$) to reach the foot of the tower. The time $$t$$ required for that final leg is found from

$$t \;=\;\frac{\text{distance still to go}}{\text{same uniform speed}} \;=\;\frac{h}{v}.$$

Substituting the expression for $$v$$ just obtained, we get

$$t \;=\;\frac{h}{\dfrac{h(\sqrt3 - 1)}{18}}.$$

We may cancel the common factor $$h$$ in numerator and denominator, giving

$$t \;=\;18\,\frac{1}{\sqrt3 - 1}.$$

To rationalise the denominator we multiply the numerator and the denominator by the conjugate $$(\sqrt3 + 1):$$

$$t \;=\;18\,\frac{\sqrt3 + 1}{(\sqrt3 - 1)(\sqrt3 + 1)}.$$

The product in the denominator is a difference of squares:

$$(\sqrt3 - 1)(\sqrt3 + 1) \;=\;(\sqrt3)^2 - 1^2 \;=\;3 - 1 \;=\;2.$$

Therefore

$$t \;=\;18\,\frac{\sqrt3 + 1}{2} \;=\;9(\sqrt3 + 1).$$

Since all quantities were in minutes, the required time for the car to reach the foot of the tower is $$9(1 + \sqrt3)$$ minutes.

Hence, the correct answer is Option D.

Let us denote the foot of tower T$$_1$$ by $$D$$, its top by $$A$$, the foot of tower T$$_2$$ by $$B$$ and its top by $$C$$. The heights are $$AD = 60 \text{ m}$$ and $$BC = 80 \text{ m}$$. The two feet $$D$$ and $$B$$ lie on the straight road, so $$DB$$ is the required width of the road. We shall denote this width by $$x$$ metres.

Because both towers are perpendicular to the road, the horizontal distance from the top of T$$_1$$ to tower T$$_2$$ is also $$x$$. Thus:

From $$A$$ to $$B$$ we have a right-angled triangle with horizontal side $$AB = x$$ and vertical side $$AD = 60$$. From $$A$$ to $$C$$ we have another right-angled triangle with the same horizontal side $$AC_{\text{horizontal}} = x$$ and vertical side $$(BC - AD) = 80 - 60 = 20$$.

Let $$\theta$$ be the angle of elevation from $$A$$ to the top $$C$$. Then, according to the statement, the angle of depression from $$A$$ to the foot $$B$$ equals $$2\theta$$. Using the definition of tangent (opposite divided by adjacent) we write:

$$\tan\theta = \frac{20}{x} \qquad\text{and}\qquad \tan(2\theta) = \frac{60}{x}.$$

We now employ the double-angle formula for tangent, stated here for clarity: $$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}.$$

Substituting $$\tan\theta = \dfrac{20}{x}$$ into the formula gives

$$\frac{60}{x} = \frac{2\left(\dfrac{20}{x}\right)}{1 - \left(\dfrac{20}{x}\right)^2}.$$

To remove complex fractions, set $$t = \frac{20}{x},$$ so that $$\tan\theta = t.$$ Then the two equations become

$$\tan(2\theta) = \frac{60}{x} = 3t$$ and by the double-angle formula $$\tan(2\theta) = \frac{2t}{1 - t^2}.$$

Equating the two expressions for $$\tan(2\theta)$$ we have

$$3t = \frac{2t}{1 - t^2}.$$

Because $$t \neq 0$$ (the towers are not coincident), we can divide both sides by $$t$$ to get

$$3 = \frac{2}{1 - t^2}.$$

Cross-multiplying yields

$$3(1 - t^2) = 2 \quad\Longrightarrow\quad 3 - 3t^2 = 2.$$

Rearranging, we obtain

$$3t^2 = 1 \quad\Longrightarrow\quad t^2 = \frac{1}{3}.$$

Taking the positive square root (since $$t > 0$$ for positive distances) gives

$$t = \frac{1}{\sqrt3}.$$

Remembering that $$t = \dfrac{20}{x}$$, we substitute back:

$$\frac{20}{x} = \frac{1}{\sqrt3} \quad\Longrightarrow\quad x = 20\sqrt3.$$

Therefore the width of the road between the feet of the two towers is $$20\sqrt3 \text{ metres}$$.

Hence, the correct answer is Option D.

Let us denote the fixed horizontal ground point of observation by $$P$$ and the aeroplane by $$A$$. The aeroplane is flying in a straight line parallel to the ground at a constant height of $$\sqrt{3}\text{ km}$$.

At the first instant, the angle of elevation of $$A$$ from $$P$$ is $$60^\circ$$. Using the right-angled triangle formed by the vertical height, the horizontal distance and the line of sight, we have

$$\tan 60^\circ \;=\;\frac{\text{vertical height}}{\text{horizontal distance }(x_1)}.$$

We know the height is $$\sqrt{3}\text{ km}$$, and $$\tan 60^\circ = \sqrt{3}$$. Substituting,

$$\sqrt{3} \;=\; \frac{\sqrt{3}}{x_1} \;\;\Longrightarrow\;\; x_1 \;=\; \frac{\sqrt{3}}{\sqrt{3}} = 1\text{ km}.$$

Five seconds later, the angle of elevation falls to $$30^\circ$$. Again applying the tangent ratio,

$$\tan 30^\circ \;=\;\frac{\text{vertical height}}{\text{horizontal distance }(x_2)},$$

with $$\tan 30^\circ = \dfrac{1}{\sqrt{3}}$$. Hence,

$$\frac{1}{\sqrt{3}} \;=\; \frac{\sqrt{3}}{x_2} \;\;\Longrightarrow\;\; x_2 \;=\; \sqrt{3}\times\sqrt{3} = 3\text{ km}.$$

The aeroplane therefore travels horizontally from $$x_1 = 1\text{ km}$$ to $$x_2 = 3\text{ km}$$ in the given five-second interval. The horizontal distance covered is

$$\Delta x \;=\; x_2 - x_1 = 3 - 1 = 2\text{ km}.$$

Speed is distance divided by time. First we find the speed in km per second:

$$v \;=\;\frac{\Delta x}{\Delta t} = \frac{2\text{ km}}{5\text{ s}} = 0.4\text{ km/s}.$$

To convert kilometres per second to kilometres per hour, we multiply by the number of seconds in an hour, $$3600$$:

$$v_{\text{(km/hr)}} \;=\; 0.4 \times 3600 = 1440\text{ km/hr}.$$

Hence, the correct answer is Option D.

Let the points Q and R lie on the horizontal ground along the X-axis, and put their coordinates as $$Q(-a,0)$$ and $$R(a,0)$$. Because the tower is erected at the mid-point of QR, the foot of the tower is $$M(0,0)$$ and the top of the tower is $$T(0,h)$$, where $$h$$ is the required height.

Since the park is isosceles with $$PQ = PR = 200\ \text{m}$$, place the vertex P somewhere above the X-axis at $$P(0,y)$$. (We shall soon see that this choice is consistent.)

First, from the given equalities $$PQ = PR$$ we write the two distance formulas:

$$PQ^2 = (0 + a)^2 + (y - 0)^2 = a^2 + y^2,$$

$$PR^2 = (0 - a)^2 + (y - 0)^2 = a^2 + y^2.$$

Both must equal $$200^2$$, so we have the single condition

$$a^2 + y^2 = 200^2. \qquad (1)$$

Now use the data about the angles of elevation.

At Q the horizontal distance to the tower’s foot M is $$QM = a$$. The angle of elevation there is $$30^\circ$$, so, stating the tangent formula $$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}},$$ we have

$$\tan 30^\circ = \frac{h}{a} \;\;\Longrightarrow\;\; \frac{1}{\sqrt3} = \frac{h}{a} \;\;\Longrightarrow\;\; h = \frac{a}{\sqrt3}. \qquad (2)$$

The same angle $$30^\circ$$ is seen from R, and because $$RM = a$$, relation (2) is confirmed.

From P the horizontal ground distance to M is $$PM = y$$, and the angle of elevation is $$45^\circ$$. Using $$\tan 45^\circ = 1$$ gives

$$\tan 45^\circ = \frac{h}{y} \;\;\Longrightarrow\;\; 1 = \frac{h}{y} \;\;\Longrightarrow\;\; h = y. \qquad (3)$$

Equate the two expressions (2) and (3) for the same height h:

$$y = \frac{a}{\sqrt3}. \qquad (4)$$

Substitute (4) into the distance condition (1):

$$a^2 + \left(\frac{a}{\sqrt3}\right)^2 = 200^2 \;\;\Longrightarrow\;\; a^2 + \frac{a^2}{3} = 200^2.$$

Combine like terms:

$$\frac{4a^2}{3} = 200^2 \;\;\Longrightarrow\;\; a^2 = \frac{3}{4}\, 200^2.$$

Compute $$a$$:

$$a^2 = \frac{3}{4}\times 40000 = 30000 \;\;\Longrightarrow\;\; a = \sqrt{30000} = 100\sqrt3.$$

Finally, use (2) or (3) to get $$h$$. Using (2):

$$h = \frac{a}{\sqrt3} = \frac{100\sqrt3}{\sqrt3} = 100.$$

Thus the tower is $$100 \text{ m}$$ high.

Hence, the correct answer is Option B.