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.

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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