Question

The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building.

Answer 1

In ΔABD,

`tan 60^circ = "AB"/"BD"`

`sqrt(3) = 60/x`

⇒ `x = 60/sqrt(3)`

= `20sqrt(3)  m`

Now, in ΔBCD, ∠D = 90°

`tan 30^circ = "CD"/"BD"`

`1/sqrt(3) = h/(20sqrt(3))`

⇒ `h = (20sqrt(3))/sqrt(3)`

⇒ h = 20 m

Hence, the height of the building = 20 m.