The angle of elevation of the top P of a vertical tower PQ of...

As per the given information, the figure would be

Here,

$RA = dcos30 ° = \dfrac{\sqrt{3}d}{2}$

$QR = AQA - RA = 10 - \dfrac{\sqrt{3}d}{2}$

$BR = dsin30 ° = \dfrac{d}{2}$

Now $\tan 60 ° = \dfrac{10 - BR}{QR} = \dfrac{10 - \dfrac{d}{2}}{10 - \dfrac{\sqrt{3}d}{2}}$

$⇒ \sqrt{3} = \dfrac{20 - d}{20 - \sqrt{3}d}$

$⇒ 20\sqrt{3} - 3d = 20 - d$

$⇒ 2d = 20\left(\sqrt{3} - 1\right)$

$⇒ d = 10\left(\sqrt{3} - 1\right)$

Now $ar \left(\text{trap}PQRB\right) = \dfrac{1}{2}\left(PQ + BR\right)QR$

$= \dfrac{1}{2}\left(10 + \dfrac{d}{2}\right)\left(10 - \dfrac{\sqrt{3}d}{2}\right)$

$= \dfrac{1}{2}\left(10 + 5\sqrt{3} - 5\right)\left(10 - 15 + 5\sqrt{3}\right)$

$= \dfrac{1}{2}\left(5\sqrt{3} + 5\right)\left(5\sqrt{3} - 5\right) = \dfrac{1}{2}\left(75 - 25\right) = 25$

i.e. $\alpha = 25$