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

Let AB be thee building and PQ be the tower.
We have,

PQ = 60m,∠APB = 30°, ∠PAQ = 60°
In  ΔAPQ,

` tan 60° = (PQ)/(AP)`

`⇒ sqrt(3) = 60/(AP)`

`⇒ AP = 60/sqrt(3)`

`⇒ AP = (60 sqrt(3))/3`

`⇒ AP = 20 sqrt(3) m`

Now, in Δ ABP,

`  tan 30° = (AB)/(AP)`

`⇒1/ sqrt(3) = (AB)/(20 sqrt(3))`

`⇒ AB = (20 sqrt(3))/ sqrt(3)`

∴ AB = 20 m

So, the height of the building is 20 m