Question

The angles of elevation of the top of a tower from two points at distances of 5 metres and 20 metres from the base of the tower and in the same straight line with it, are complementary. Find the height of the tower. [CBSE 2014]

Answer 1

Let the height of the tower be AB.

We have,
$$\begin{gathered} \mathrm{AC}=5\mathrm{m},\mathrm{AD}=20\mathrm{m} \\ \mathrm{Let}\mathrm{the}\mathrm{angle}\mathrm{of}\mathrm{elevation}\mathrm{of}\mathrm{the}\mathrm{top}\mathrm{of}\mathrm{the}\mathrm{tower}\left(\mathrm{i}.\mathrm{e}.\angle \mathrm{ACB}\right)\mathrm{from}\mathrm{point}\mathrm{C}\mathrm{be}\theta \mathit{.}\mathit{}\mathrm{Then}, \\ \mathrm{the}\mathrm{angle}\mathrm{of}\mathrm{elevation}\mathrm{of}\mathrm{the}\mathrm{top}\mathrm{of}\mathrm{the}\mathrm{tower}\left(\mathrm{i}.\mathrm{e}.\angle \mathrm{ADB}\right)\mathrm{from}\mathrm{point}\mathrm{D}=\left(90°-\theta \right) \\ \mathrm{Now},\mathrm{in}∆\mathrm{ABC}, \\ \tan \theta =\frac{\mathrm{AB}}{\mathrm{AC}} \\ \Rightarrow \tan \theta =\frac{\mathrm{AB}}{5}.....\left(\mathrm{i}\right) \\ \mathrm{Also},\mathrm{in}∆\mathrm{ABD}, \\ \cot \left(90°-\theta \right)=\frac{\mathrm{AD}}{\mathrm{AB}} \\ \Rightarrow \tan \theta =\frac{20}{\mathrm{AB}}.....\left(\mathrm{ii}\right) \\ \mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get} \\ \frac{\mathrm{AB}}{5}=\frac{20}{\mathrm{AB}} \\ \Rightarrow {\mathrm{AB}}^2=100 \\ \Rightarrow \mathrm{AB}=\sqrt{100} \\ \therefore \mathrm{AB}=10\mathrm{m} \end{gathered}$$

So, the height of the tower is 10 m.