Question

A man standing on a level plane observes the elevation of the top of a pole to be $\theta$. If he walks a distance equal to double the height of the pole towards the pole, the angle of elevation becomes $2\theta .$ Then the value of $\theta$ (in degrees) is

Answer 1

Let $AB$ be the pole with height $h.$
In $△ABC$
$x=h\cot 2\theta \cdots \left(i\right)$
In $△ABD$
$h=(2h+x)\tan \theta \cdots \left(ii\right)$

From equation $\left(i\right)$ and $\left(ii\right)$,
$h=(2h+h\cot 2\theta )\tan \theta \Rightarrow 1=(2+\cot 2\theta )\tan \theta (\because h\ne 0)\Rightarrow 1=(2+\frac{1}{\tan 2\theta })\tan \theta \Rightarrow 1=(2+\frac{1-{\tan }^2\theta }{2\tan \theta })\tan \theta \Rightarrow {\tan }^2\theta -4\tan \theta +1=0\Rightarrow \tan \theta =\frac{4\pm \sqrt{12}}{2}\Rightarrow \tan \theta =2+\sqrt{3},2-\sqrt{3}\Rightarrow \theta ={75}^{\circ },{15}^{\circ }$
As $2\theta$ is an acute angle, so
$\theta ={15}^{\circ }$