Question

The angle of depression from the top of a tower of a point A on the ground is 30°. On moving a distance of 20 metres from the point A towards the foot of the tower to a point B, the angle of elevation of the top of the tower from the point B is 60°. Find the height of the tower and its distance from the point A. [CBSE 2012]

Answer 1

Let PQ be the tower.

We have,

$$\begin{gathered} \mathrm{AB}=20\mathrm{m},\angle \mathrm{PAQ}=30°\mathrm{and}\angle \mathrm{PBQ}=60° \\ \mathrm{Let}\mathrm{BQ}=x\mathrm{and}\mathrm{PQ}=h \\ \mathrm{In}∆\mathrm{PBQ}, \\ \tan 60°=\frac{\mathrm{PQ}}{\mathrm{BQ}} \\ \Rightarrow \sqrt{3}=\frac{h}{x} \\ \Rightarrow h=x\sqrt{3}.....\left(\mathrm{i}\right) \\ \mathrm{Also},\mathrm{in}∆\mathrm{APQ}, \\ \tan 30°=\frac{\mathrm{PQ}}{\mathrm{AQ}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{AB}+\mathrm{BQ}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{x\sqrt{3}}{20+x}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow 20+x=3x \\ \Rightarrow 3x-x=20 \\ \Rightarrow 2x=20 \\ \Rightarrow x=\frac{20}{2} \\ \Rightarrow x=10\mathrm{m} \\ \mathrm{From}\left(\mathrm{i}\right), \\ h=10\sqrt{3}=10\times 1.732=17.32\mathrm{m} \\ \mathrm{Also},\mathrm{AQ}=\mathrm{AB}+\mathrm{BQ}=20+10=30\mathrm{m} \end{gathered}$$

So, the height of the tower is 17.32 m and its distance from the point A is 30 m.