Question

An observer, 1.5 m tall, is 28.5 m away from a tower 30 m high. Determine the angle of elevation of the top of the tower from his eye.

Answer 1

Let BE be the observer of 1.5 m tall. And AD be the tower of height 30. Here we have to find the angle of elevation of the top of the tower.

Let ∠ABC = θ

The corresponding figure is as follows

In ΔABC

`=> tan theta = (AC)/(BC)`

`=> tan theta = 28.5/28.5`

`=> tan theta  = 1`

`=> theta = 45^@`

Hence the required angle is 45°