JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Heights and Distances

Let the cloud be a height $x$ above $P$ at horizontal distance $y$. The elevation gives $\tan 30^{\circ }=\frac{x}{y}$, so $y=\sqrt{3}x$. The image of the cloud lies as far below the lake surface as the cloud is above it, a depth $x+400$ below $P$ (since $P$ is $200\mathrm{m}$ above the water), and the depression gives $\tan 60^{\circ }=\frac{x+400}{y}$, so $\sqrt{3}y=x+400$. Substituting $y=\sqrt{3}x$ gives $3x=x+400$, hence $x=200$, and then $\sin 30^{\circ }=\frac{x}{PC}$ gives $PC=\frac{200}{1/2}=400\mathrm{m}$.