Question

From the top of a building 60 m high the angles of depression of the top and the bottom of a tower are
observed to be 30° and 60°. Find the height of the tower.​

Answer 1

Answer:

Let BC be the building and AD be the tower.

Let the height of tower, AD be h m.

Angles of depression of the top D and the bottom A of the tower CB are 30° and 60° respectively.

∴ ∠CDE = 30°

∠CAB = 60°

Since, BC = 60 m.

∴ CE = (60 – h) m

Let AB = DE = x m

In ∆DEC,

In ∆CBA,

Equating equation (1) and (2),

⇒ 3 (60 – h) = 60

⇒ 180 – 3h) = 60

⇒ 180 – 60 = 3n

⇒ 120 = 3h

⇒ h = 40

Thus, the height of the tower is 40 m.