Question

A man standing on a level plane observes the angle of elevation of top of pole to be 'a'. He walks a distance equal to double the height of pole towards it and finds that the elevation is 2a then a=

• A. $\frac{\pi }{13}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is A $\frac{\pi }{13}$

$\tan \alpha =\frac{h}{2h+x},\tan 2\alpha =\frac{h}{x}\Rightarrow x=\frac{h}{\tan 2\alpha }$

$\tan \alpha =\frac{h}{2h+\frac{h}{\tan 2\alpha }}=\frac{1}{2+\frac{1}{\tan 2\alpha }}$

$\Rightarrow \tan \alpha =\frac{1}{\frac{2\tan 2\alpha +1}{\tan 2\alpha }}=\frac{\tan 2\alpha }{1+2\tan 2\alpha }$

We know that

$\tan (a+b)=\frac{\tan a+\tan b}{1+\left(\tan a\tan b\right)}$

$\Rightarrow \tan \alpha =\frac{\frac{2\tan \alpha }{1-{\tan }^2\alpha }}{1+\frac{4\tan \alpha }{(1-{\tan }^2\alpha )}}$

$\Rightarrow \tan \alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha +4\tan \alpha }$

$\Rightarrow 1-{\tan }^2\alpha +4\tan \alpha =2$

$\Rightarrow {\tan }^2\alpha -4\tan \alpha +1=0$

$\Rightarrow \tan \alpha =\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm 2\sqrt{3}}{2}$

$\Rightarrow \tan \alpha =2\pm \sqrt{3}$

$\Rightarrow \alpha =15=\frac{\pi }{13}$