Question

The angle of elevation of the top of a vertical tower from a point on the ground is 60° . From another point 10 m vertically above the first, its angle of elevation is 30° .Find the height of the tower.

Answer 1

Let PQ be the tower
We have,
AB =10m,MAP = 30° and ∠PBQ = 60°
Also, MQ = AB =10m
Let BQ = x and PQ = h
So, AM = BQ = x and PM = PQ - MQ =  h -10
In ΔBPQ,

` tan 60° = (PQ)/(BQ)`

`⇒ sqrt(3) = h/x` 

`⇒ x = h/sqrt(3) `                      ..............(1)

Now , in Δ AMP

` tan 30°  = (PM) /( AM)`

`⇒ 1/ sqrt(3) = (h-10) /x`

`⇒h sqrt(3)-10 sqrt(3)=x`

`⇒ h sqrt(3) - 10 sqrt(3) = h/ sqrt(3) `                          [ Using (1)]

⇒ 3h  - 30=h

⇒ 3h - h = 30

⇒ 2h = 30

`⇒ h 30/2 `

∴ h = 15 m

So, the height of the tower is 15 m.