Choose the correct answer of the following question:

Let AB be the cliff and CD be the tower.

We have,

$\mathrm{AB}=20 \mathrm{~m}$

Also, $\mathrm{CE}=\mathrm{AB}=20 \mathrm{~m}$

Let $\angle \mathrm{ACB}=\angle \mathrm{CAE}=\angle \mathrm{DAE}=\theta$

In $\Delta \mathrm{ABC}$,

$\tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}}$

$\Rightarrow \tan \theta=\frac{20}{\mathrm{BC}}$

$\Rightarrow \tan \theta=\frac{20}{\mathrm{AF}} \quad(\mathrm{As}, \mathrm{BC}=\mathrm{AE})$

$\Rightarrow \mathrm{AE}=\frac{20}{\tan \theta} \quad \ldots .(\mathrm{i})$

Also, in $\Delta \mathrm{ADE}$,

$\tan \theta=\frac{\mathrm{DE}}{\mathrm{AE}}$

$\Rightarrow \tan \theta=\frac{\mathrm{DE}}{\left(\frac{20}{\tan \theta}\right)} \quad[\mathrm{Using}(\mathrm{i})]$

$\Rightarrow \tan \theta=\frac{\mathrm{DE} \times \tan \theta}{20}$

$\Rightarrow \mathrm{DE}=\frac{20 \times \tan \theta}{\tan \theta}$

$\Rightarrow \mathrm{DE}=20 \mathrm{~m}$

Now, $\mathrm{CD}=\mathrm{DE}+\mathrm{CE}$

$=20+20$

$\therefore \mathrm{CD}=40 \mathrm{~m}$

Hence, the correct answer is option (b).

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