Question

A man standing on a horizontal plane, observes the angle of elevation of the top of a tower to be $\alpha$. After walking a distance equal to double the height of the tower towards the tower, the angle of the elevation becomes $2\alpha$, then $\alpha$ is equal to:

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{12}$
• D. $\frac{\pi }{18}$

Answer 1

Answer: (C)

$\frac{\pi }{12}$

REF Image

Let $h$ be the height of the tower $AB$.

Given that: $CD$ $=2h$

In $\Delta$ $DAB$,

$\tan \alpha =\frac{h}{AD}$

$\Rightarrow \tan \alpha =\frac{h}{AC+2h}$ ………..$\left(1\right)$

In $\Delta$ CAB,

$\tan 2\alpha =\frac{h}{AC}$

We know that , $\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }=\frac{h}{AC}$

$\Rightarrow \frac{2\frac{h}{AC+2h}}{1-{\left(\frac{h}{AC+2h}\right)}^2}=\frac{h}{AC}$ [from eq. $\left(1\right)$]

$\Rightarrow \frac{\frac{2h}{AC+2h}}{\frac{(AC+2h{)}^2-h^2}{(AC+2h{)}^2}}=\frac{h}{AC}$

$\Rightarrow \frac{2h(AC+2h)}{(AC+2h{)}^2-h^2}=\frac{h}{AC}$

$\Rightarrow \frac{2AC+4h}{A{C}^2+4h^2+4hAC-h^2}=\frac{1}{AC}$

$\Rightarrow 2A{C}^2+4hAC=A{C}^2+3h^2+4hAC$

$\Rightarrow A{C}^2=3h^2$

$\Rightarrow AC=\sqrt{3}h$

From $\left(1\right)$,

$\tan \alpha =\frac{h}{\sqrt{3}h+2h}$

$\tan \alpha =\frac{h}{(\sqrt{3}+2)h}$

$\alpha ={\tan }^{-1}\left(\frac{1}{\sqrt{3}+2}\right)$ (By rationalizing now)

$\alpha ={\tan }^{-1}\left(\frac{\sqrt{3}-2}{3-4}\right)$

$\alpha ={\tan }^{-1}\left(\frac{\sqrt{3}-2}{-1}\right)$

$\alpha ={\tan }^{-1}(2-\sqrt{3})$

$\alpha ={\tan }^{-1}\tan \frac{\pi }{12}$ $[\because \tan \frac{\pi }{12}=2-\sqrt{3}]$

$\alpha =\frac{\pi }{12}$.