Question

From the top of a 15 m high building, the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower and the distance between tower and the building.

Answer 1

Let AB = tower

CD = building

Such that ∠ACE = 30°

∠ADB = 60°

AE = h m   (say)

EB = CD = 15 m

and BD = x m   (say)

= CE

Now, In ΔAEC, ∠E = 90° we have

`tan 30"°" = h/x`

⇒ `1/sqrt3 = h/x`

⇒ x = `sqrt3 * h`   ...(i)

Again in ΔABD, ∠B = 90° we have

`tan 60"°" = (AB)/(BD)`

⇒ `sqrt3 = (h + 15)/x`

⇒ `sqrt3 * (sqrt3 * h) = h + 15`     ...[∵ from (i) x = `sqrt3 .` h]

⇒ 3h − h = 15

⇒ h = 7.5 m

⇒ x = `sqrt3h = sqrt3 xx 7.5 ≈ 1.732 xx 7.5` m

x ≈ 12.99 m

Hence height of the tower = 7.5 + 15 = 22.5 m

and distance between tower and building = 12.99 m