Question

the angles of the depression from the top of a 60m high building to the base and top of tower are 45°and 30° respectively .find the height of the tower

Answer 1

Let the height be x
ATQ
60:45::x:30
x*45=60*30
$x = \frac{60 \times 30}{45}$
x=40m

Answer 2

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.