Question

The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower. [CBSE 2011]

Answer 1

Let PQ be the tower.

We have,

$$\begin{gathered} \mathrm{AB}=10\mathrm{m},\angle \mathrm{MAP}=30°\mathrm{and}\angle \mathrm{PBQ}=60° \\ \mathrm{Also},\mathrm{MQ}=\mathrm{AB}=10\mathrm{m} \\ \mathrm{Let}\mathrm{BQ}=x\mathrm{and}\mathrm{PQ}=h \\ \mathrm{So},\mathrm{AM}=\mathrm{BQ}=x\mathrm{and}\mathrm{PM}=\mathrm{PQ}-\mathrm{MQ}=h-10 \\ \mathrm{In}∆\mathrm{BPQ}, \\ \tan 60°=\frac{\mathrm{PQ}}{\mathrm{BQ}} \\ \Rightarrow \sqrt{3}=\frac{h}{x} \\ \Rightarrow x=\frac{h}{\sqrt{3}}.....\left(\mathrm{i}\right) \\ \mathrm{Now},\mathrm{in}∆\mathrm{AMP}, \\ \tan 30°=\frac{\mathrm{PM}}{\mathrm{AM}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h-10}{x} \\ \Rightarrow h\sqrt{3}-10\sqrt{3}=x \\ \Rightarrow h\sqrt{3}-10\sqrt{3}=\frac{h}{\sqrt{3}}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow 3h-30=h \\ \Rightarrow 3h-h=30 \\ \Rightarrow 2h=30 \\ \Rightarrow h=\frac{30}{2} \\ \therefore h=15\mathrm{m} \end{gathered}$$

So, the height of the tower is 15 m.