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Trignometry - Formulas and Properties

Useful

In a right-angled triangle, we consider an angle 'x'

Sin(x) = $$\dfrac{opposite}{hypotenuse}$$

Cos(x) = $$\dfrac{adjacent}{hypotenuse}$$

Tan(x) = $$\dfrac{opposite}{adjacent}$$

Trigonometric values for important angles: 0, 30, 45, 60, 90.

Sin(0) = 0, Sin(30) = 1/2, Sin(45) = 1/$$\sqrt{\ 2}$$, sin(60) = $$\frac{\sqrt{\ 3}}{2}$$, sin(90) = 1 

Cos(0) = 1, Cos(30) = $$\frac{\sqrt{\ 3}}{2}$$, cos(45) = 1/$$\sqrt{\ 2}$$, cos(60) = 1/2, cos(90) = 0

Tan(0) = 0, Tan(30) = $$\frac{1}{\sqrt{\ 3}}$$, Tan(45) = 1, Tan(60) = $$\sqrt{\ 3}$$, Tan(90) = infinity.

Sin(90+x) = cos(x)

Cos(90+x) = -sin(x)

sin(-x) = -sin(x)

cos(-x) = cos(x)

$$\sin^2\left(x\right)+\cos^2\left(x\right)=1$$

sin(x+y) = sin(x)cos(y) + cos(x)sin(y)

cos(x+y) = cos(x)cos(y) - sin(x)sin(y)

Question 1

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $$\angle BAC = 45^{\circ}$$ and $$\angle ABC=\theta\ $$, then $$\dfrac{AD}{BE}$$ equals

Question 2

Let C be the circle $$x^{2} + y^{2} + 4x - 6y - 3 = 0$$ and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to $$60^{\circ}$$. Then, the point at which L touches the line $$x$$ = 6 is

Question 3

Let $$\triangle ABC$$ be an isosceles triangle such that AB and AC are of equal length. AD is the altitude from A on BC and BE is the altitude from B on AC. If AD and BE intersect at O such that $$\angle AOB = 105^\circ$$, then $$\frac{AD}{BE}$$ equals

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