Question

The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.

Answer 1

Let distance between the two towers = AB = x m

And height of the other tower = PA = h m

Given, height of the tower = QB = 30 m, ∠QAB = 60° and ∠PBA = 30°

Now, in ∆QAB,

tan 60° = `"QB"/"AB" = 30/x`

⇒ `sqrt(3) = 30/x`

⇒ `x = 30/sqrt(3)`

⇒ `x = 30/sqrt(3) * sqrt(3)/sqrt(3)`

= `(30sqrt(3))/3`

= `10sqrt(3)`

And in ∆PBA,

tan 30° = `"PA"/"AB" = "h"/x`

⇒ `1/sqrt(3) = "h"/(10sqrt(3))`  ...`[∵ x = 10sqrt(3)  "m"]`

⇒ h = 10

Hence, the required distance and height are `10sqrt(3)  "m"` and 10 m, respectively.