The angle of elevation of the top P of a vertical tower P Q of height 10 from a point A on the horizontal…

As per the given information, the figure would be 

Here, 

$RA=d\cos 30°=\frac{\sqrt{3} d}{2}$

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

$BR=d\sin 30°=\frac{d}{2}$

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

$\Rightarrow \sqrt{3}=\frac{20-d}{20-\sqrt{3}d}$

$\Rightarrow 20\sqrt{3}-3d=20-d$

$\Rightarrow 2d=20\left(\sqrt{3}-1\right)$

$\Rightarrow d=10\left(\sqrt{3}-1\right)$

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

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

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

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

i.e. $\alpha =25$