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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