Question

An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

Answer 1

Let the angle of elevation of the top of the tower from the eye of the observer is θ

Given that,

AB = 22 m,

PQ = 1.5 m = MB

And QB = PM = 20.5 m

⇒ AM = AB – MB

= 22 – 1.5

= 20.5 m

Now, In ∆APM,

tan θ = `"AM"/"PM" = 20.5/20.5` = 1

⇒ tan θ = tan 45°

∴ θ = 45°

Hence, the required angle of elevation of the top of the tower from the eye of the observer is 45°.