Let a vertical tower AB of height 2h stands on a horizontal ground....

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 2α. When from $P$, he moves a distance $d$ in the direction of $\overset{\to}{AP}$, he can see the top $B$ of the tower with an angle of elevation α. If $d = \sqrt{7}h$, then tanα is equal to

Now on plotting the value in diagram we have,

Now in $△ PAC$

\tan 2α = \dfrac{h}{x} . . . . . \left(1\right)

And in $△ HAB$ we have

tanα = \dfrac{2h}{x + \sqrt{7h}}

\to tan&alpha; = \dfrac{2h}{hcot2&alpha; + \sqrt{7}h} \left(\text{from equation}\left(1\right)\right)$</p><p>$\to tanα = \dfrac{2}{\dfrac{\left(1 - \tan^{2}α\right)}{2tanα} + \sqrt{7}}$</p><p>Now put$tanα = t$and on simplifying we get,</p><p>$\to t = \dfrac{2}{\dfrac{\left(1 - {t}^{2}\right)}{2t} + \sqrt{7}}$</p><p>$\to t\left(\dfrac{1 - {t}^{2}}{2t} + \sqrt{7}\right) = 2$</p><p>$\to \dfrac{1 - {t}^{2}}{2} + \sqrt{7}t = 2$</p><p>$\to {t}^{2} - 2\sqrt{7}t + 3 = 0$</p><p>$\to t = \sqrt{7} - 2$</p><p>$\to tanα = \sqrt{7} - 2