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
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.
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