Question

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

Answer 1

`"Let AB = 1.5 m be the observer and CD = 30 m be the tower."`
`"Let the angle of elevation of the top of the tower be α"` 

`CD= CE+ED` 

`⇒ CD=CE+AB` 

`⇒30=CE+1.5` 

`⇒CE=30-1.5=28.5 m`

In ΔCEB, 

`tan ∝ = (CE)/(BE)=28.5/28.5` 

`⇒ tan ∝ =1`

`⇒ tan ∝= tan 45° ` 

`⇒∝=45°`