JEE Main 2021 — Heights and Distances Question with Solution

Assume, $O$ is the  centre of sphere

and $P,Q$ are the point of contact of tangents from $A$

Let $T$ be top most point of balloon & $R$ be foot of perpendicular from $O$ to ground.

From triangle $OAP, \sin 30°=\frac{16}{OA}$

$\Rightarrow OA=16cosec30°=32$

From triangle $ARO, \sin 75°=\frac{OR}{OA}$

$\Rightarrow OR=OA\sin 75°$

$\Rightarrow 32\left[\frac{(\sqrt{3}+1)}{2\sqrt{2}}\right]=8\sqrt{6}+8\sqrt{2}$

So the height (in meter) of the top most point of the balloon from the level of the observer's eye is $OR+OT$

$=8(\sqrt{6}+\sqrt{2}+2)$