**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.