Question

An observer , 1.7 m tall , is 20$\sqrt{3}$ m away from a tower . The angle of elevation from the eye of an observer to the top of tower is 30⁰ . Find the height of the tower.

Answer 1

Let AB be the height of the observer and EC be the height of the tower.

Given:
AB = 1.7 m ⇒ CD = 1.7 m
BC = 20$\sqrt{3}$ m

Let ED be h m.



In ∆ADE,

$$\begin{gathered} \tan 30°=\frac{\mathrm{ED}}{\mathrm{AD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20\sqrt{3}} \\ \Rightarrow h=20\mathrm{m} \end{gathered}$$

∴ EC = ED + DC = (h + 1.7) m = 21.7 m

Hence, the height of the tower is 21.7 m.