Q. 565.0( 1 Vote )

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

Answer :

In the fig let DC is the observer of the height 1.5m.



In ∆AED


tan θ =


tan θ =


tan θ = 1


θ = tan-1 1


θ = 45°


Hence the angle of the observation of


the tower from observer’s eye is 45°


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