JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Heights and Distances

Place $B$ at the origin with $BQ$ horizontal and $BA$ vertical, so $A=(0,30)$. The angle of depression of $Q$ is $60^{\circ }$, giving $\tan 60^{\circ }=\frac{30}{BQ}$, so $BQ=\frac{30}{\sqrt{3}}=10\sqrt{3}$ m. Let $PQ=h$; since the tower is vertical, $P$ sits at horizontal distance $10\sqrt{3}$ and height $h$. The depression of $P$ is $15^{\circ }$, so $\tan 15^{\circ }=\frac{30-h}{10\sqrt{3}}$. With $\tan 15^{\circ }=2-\sqrt{3}$, $30-h=10\sqrt{3}(2-\sqrt{3})=20\sqrt{3}-30$, hence $h=60-20\sqrt{3}$. The point $C$ on $AB$ with $CB=PQ=h$ makes $BCPQ$ a rectangle of width $BQ=10\sqrt{3}$ and height $h=60-20\sqrt{3}$, so its area is $10\sqrt{3}(60-20\sqrt{3})=600\sqrt{3}-600=600(\sqrt{3}-1)$ m${}^2$.