ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are $$\cot^{-1}(3\sqrt{2})$$ and $$\text{cosec}^{-1}(2\sqrt{2})$$ respectively, then the height of the tower (in metres) is

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.