Question

An observer finds the angle of elevation of the top of the tower from a certain point on the ground as 30^°. If the observe moves 20 m towards the base of the tower, the angle of elevation of the top increases by 15^°, find the height of the tower.

Answer 1

Let the observer be at point C on the ground.

∴ ∠C = 30^°

He moves 20 m towards the tower and reaches point D.

∴ ∠D = 30^° + 15^° = 45^°

Let the distance BD be x m and the height of the tower be h m.

In ΔABD

`tan 45^@ = (AB)/(BD) = h/x`

`=> 1 = h/x`

`=> h = x`

In ΔABC

`tan 30^@ = (AB)/(BC) = h/(x + 20) = h/(h + 20)`

`=> 1/sqrt3 = h/(h + 20)`

`=> h + 20 = sqrt3h`

`=> h = 20/(sqrt3 - 1) =10(sqrt3 + 1) m`

Hence, the height of the tower is `10(sqrt3+1)` m.