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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