Question

The angles of depression of the top and bottom of a tower as seen from the top of a $60\sqrt{3}$ m high cliff are 45° and 60° respectively. Find the height of the tower.

Answer 1

Let AE be the tower and BD be the cliff of $60\sqrt{3}$ height

In right triangle ABC
$$\begin{gathered} \tan 45°=\frac{\mathrm{AC}}{\mathrm{BC}} \\ \Rightarrow 1=\frac{h}{x} \\ \Rightarrow x=h \end{gathered}$$
Now, In right triangle BCE
$$\begin{gathered} \tan 60°=\frac{EC}{\mathrm{BC}} \\ \Rightarrow \sqrt{3}=\frac{60\sqrt{3}}{x} \\ \Rightarrow x=60\mathrm{m} \\ \Rightarrow x=h=60\mathrm{m} \\ \therefore \mathrm{AE}=\mathrm{AC}+\mathrm{CE}=60+60=120\mathrm{m} \end{gathered}$$
Hence, the height of the tower is 120 m.