Question

The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, then find the height of the building. [CBSE 2013]

Answer 1

Let AB be the building and PQ be the tower.

We have,

$$\begin{gathered} \mathrm{PQ}=60\mathrm{m},\angle \mathrm{APB}=30°,\angle \mathrm{PAQ}=60° \\ \mathrm{In}∆\mathrm{APQ}, \\ \tan 60°=\frac{\mathrm{PQ}}{\mathrm{AP}} \\ \Rightarrow \sqrt{3}=\frac{60}{\mathrm{AP}} \\ \Rightarrow \mathrm{AP}=\frac{60}{\sqrt{3}} \\ \Rightarrow \mathrm{AP}=\frac{60\sqrt{3}}{3} \\ \Rightarrow \mathrm{AP}=20\sqrt{3}\mathrm{m} \\ \mathrm{Now},\mathrm{in}∆\mathrm{ABP}, \\ \tan 30°=\frac{\mathrm{AB}}{\mathrm{AP}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{AB}}{20\sqrt{3}} \\ \Rightarrow \mathrm{AB}=\frac{20\sqrt{3}}{\sqrt{3}} \\ \therefore \mathrm{AB}=20\mathrm{m} \end{gathered}$$

So, the height of the building is 20 m.