Question

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

Answer 1

Let AB = 1.5 m be the observer and CD = 30 m be the tower.
Let the angle of elevation of the top of the tower be $\alpha$.
CD = CE + ED
$$\begin{gathered} \Rightarrow \mathrm{CD}=\mathrm{CE}+\mathrm{AB} \\ \Rightarrow 30=\mathrm{CE}+1.5 \\ \Rightarrow \mathrm{CE}=30-1.5=28.5\mathrm{m} \end{gathered}$$
In $△$CEB,
$$\begin{gathered} \tan \alpha =\frac{CE}{BE}=\frac{28.5}{28.5} \\ \Rightarrow \tan \alpha =1 \\ \Rightarrow \tan \alpha =\tan 45° \\ \Rightarrow \alpha =45° \end{gathered}$$