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Question
An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Solution
Let the angle of elevation of the top of the tower from the eye of the observer is θ
Given that,
AB = 22 m,
PQ = 1.5 m = MB
And QB = PM = 20.5 m
⇒ AM = AB – MB
= 22 – 1.5
= 20.5 m
Now, In ∆APM,
tan θ = `"AM"/"PM" = 20.5/20.5` = 1
⇒ tan θ = tan 45°
∴ θ = 45°
Hence, the required angle of elevation of the top of the tower from the eye of the observer is 45°.
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