Question

Choose the correct answer of the following question:

From the top of a cliff 20 m high, the angle of elevation of the top of a tower is found to be equal to the angle of depression of the foot of the
tower. The height of the tower is

(a) 20 m (b) 40 m (c) 60 m (d) 80 m [CBSE 2013C]

Answer 1

Let AB be the cliff and CD be the tower.

We have,

$$\begin{gathered} \mathrm{AB}=20\mathrm{m} \\ \mathrm{Also},\mathrm{CE}=\mathrm{AB}=20\mathrm{m} \\ \mathrm{Let}\angle \mathrm{ACB}=\angle \mathrm{CAE}=\angle \mathrm{DAE}=\theta \\ \mathrm{In}∆\mathrm{ABC}, \\ \tan \theta =\frac{\mathrm{AB}}{\mathrm{BC}} \\ \Rightarrow \tan \theta =\frac{20}{\mathrm{BC}} \\ \Rightarrow \tan \theta =\frac{20}{\mathrm{AE}}\left(\mathrm{As},\mathrm{BC}=\mathrm{AE}\right) \\ \Rightarrow \mathrm{AE}=\frac{20}{\tan \theta }.....\left(\mathrm{i}\right) \\ \mathrm{Also},\mathrm{in}∆\mathrm{ADE}, \\ \tan \theta =\frac{\mathrm{DE}}{\mathrm{AE}} \\ \Rightarrow \tan \theta =\frac{\mathrm{DE}}{\left({\displaystyle \frac{20}{\tan \theta }}\right)}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow \tan \theta =\frac{\mathrm{DE}\times \tan \theta }{20} \\ \Rightarrow \mathrm{DE}=\frac{20\times \tan \theta }{\tan \theta } \\ \Rightarrow \mathrm{DE}=20\mathrm{m} \\ \mathrm{Now},\mathrm{CD}=\mathrm{DE}+\mathrm{CE} \\ =20+20 \\ \therefore \mathrm{CD}=40\mathrm{m} \end{gathered}$$

Hence, the correct answer is option (b).

Disclaimer: The answer given in the textbook is incorrect. The same has been rectified above.