Question

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

Answer 1

BC is the tower 30 m high, and the observer is at A. E denotes his eyes’ position. 

Let the angle of elevation be θ.

FC = BC – AE = 30 – 1.5 = 28.5 m 

EF = AB = 28.5 m 

In ∆EFC, tan θ = \(\frac{FC}{EF}\) = \(\frac{28.5}{28.5}\) = 1.

∴ θ = 45°.