Question

The angle of depression form the top of a tower of a point A on the ground is 30° . On moving a distance of 20 meters from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower to from the point B is 60° . Find the height of the tower and its distance from the point A.

Answer 1

Let PQ be the tower.
We have,
AB = 20m, ∠PAQ = 30°  and ∠PBQ = 60°
Let BQ = x and PQ =  h
In  ΔPBQ,

` tan 60° = (PQ)/(BQ) `

`⇒ sqrt(3) = h/x`

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

Also, in ΔAPQ ,

` tan 30° = (PQ)/(AQ)`

`⇒ 1/ sqrt(3) = h/(AB +BQ)`

`⇒ 1/ sqrt(3) = (x sqrt(3))/(20+x)`                      [Using (1)]

⇒ 20+ x = 3x 

⇒ 3x -x = 20 

⇒ 2x = 20 

`⇒ =20/2`

⇒ x = 10 m

From (i),
`h = 10 sqrt(3)=10xx1.732 =17.32m`
Also, AQ = AB+BQ = 20 + 10 = 30m
So, the height of the tower is 17. 32 m and its distance from the point A is 30 m.