Question

The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of depression from the top of the tower to the foot of the hill is 30°. If the tower is 50 m high, find the height of the hill ?

Answer 1

Let AB be the hill and CD be the tower.

Angle of elevation of the hill at the foot of the tower is 60°, i.e., ∠ADB = 60° and the angle of depression of the foot of hill from the top of the tower is 30°, i.e., ∠CBD = 30°.

Now in right angled ΔCBD:

`tan30^@=(CD)/(BD)`

`rArrBD=(CD)/tan30^@`

`rArrBD=50/((1/sqrt3))`

`rArrBD =50sqrt3`

In right ΔABD:

`tan60^@=(AD)/(BD)`

`rArr AB=BDxxtan60^@`

`=50sqrt3xxsqrt3m`

`=50xx3m`

`=150m`

Hence, the height of the hill is 150 m.