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.
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°