Q. 144.9( 7 Votes )

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

Answer :

Let PQ = 1.5 m, is the height of the observer.

QB = 20.5 m, is the distance of the observer from the tower

AB = 22 m, is the height of the tower

To find θ, the angle of elevation we need to find AM first.

AM = AB – MB

⇒ AM = AB – PQ

⇒ AM = 22 m – 1.5 m = 20.5 m [∵, MB = PQ]

We have the values of PM and AM i.e. 20.5 m and 20.5 m respectively.

In ∆APM,

⇒ θ = 45°

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

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