# Angle of Elevation

## Angle of Elevation

If ‘O’ be the observer’s eye and OX be the horizontal line through O. if object P is at a higher level than eye, then angle POX is called the angle of elevation.

Example 1: The top of a hill observed from the top and bottom of a building h is at angle pf elevation p and q respectively. The height of hill is

Solution: Let AD be the building of height h and BP be the hill. Then,

$$\tan q=\frac{h+x}{y}$$ … (1)

$$\tan p=\frac{x}{y}$$,

y = x/ tan p

y = x cot p … (2)

Equation (1) and (2)

$$\tan q=\frac{h+x}{x\cot p}$$,

x cot p = (h + x)/ tan q

x cot p = (h + x) cot q

x (cot p – cot q) = h cot q

$$x=\frac{h\cot q}{\cot p-\cot q}$$,

$$h+x=\frac{h\cot q}{\cot p=\cot q}+h$$,

Height of hill = $$\frac{h\cot p}{\cot p-\cot q}$$,

Example 2: At a point on the ground the angle of elevation of a tower is such that its cotangent is 3/5. On walking 32 m towards the town the cotangent of the angle of elevation is 2/5. The height of the tower is

Solution: Given that cotα = 3/5

And cotβ = 2/5

In triangle BCD, tanβ = h/BC

BC = h tanβ

BC = 2h/5

and in triangle ACD, tanα = h/ (32 + BC)

$$h=(32+\frac{2h}{5})\frac{5}{3}$$,

3h = 160 + 2h

3h – 2h = 160

h = 160m.