JEE Main 2022 — Heights and Distances Question with Solution

Given a vertical tower $AB$ of height $2h$ stands on a horizontal ground and from a point $P$ on the ground a man can see upto height $h$ of the tower with an angle of elevation $2\alpha$. When from $P$, he moves a distance $d$ in the direction of $\overrightarrow{AP}$, he can see the top $B$ of the tower with an angle of elevation $\alpha$. If $d=\sqrt{7}h$, then $\tan \alpha$ is equal to

Now on plotting the value in diagram we have,

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Now in $△PAC$

$\tan 2\alpha =\frac{h}{x} .....\left(1\right)$

And in $△HAB$ we have

$\tan \alpha =\frac{2h}{x+\sqrt{7h}}$

$\Rightarrow \tan \alpha =\frac{2h}{h\cot 2\alpha +\sqrt{7}h}$ $\left(\text{from equation }\left(1\right)\right)$

$\Rightarrow \tan \alpha =\frac{2}{\frac{\left(1-{\tan }^2\alpha \right)}{2\tan \alpha }+\sqrt{7}}$

Now put $\tan \alpha =t$ and on simplifying we get,

$\Rightarrow t=\frac{2}{\frac{\left(1-t^2\right)}{2t}+\sqrt{7}}$

$\Rightarrow t\left(\frac{1-t^2}{2t}+\sqrt{7}\right)=2$

$\Rightarrow \frac{1-t^2}{2}+\sqrt{7}t=2$

$\Rightarrow t^2-2\sqrt{7}t+3=0$

$\Rightarrow t=\sqrt{7}-2$

$\Rightarrow \tan \alpha =\sqrt{7}-2$