A spherical gas balloon of radius 16 meter subtends an angle 60 °...

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 $OAP, \sin 30 ° = \dfrac{16}{OA}$

$⇒ OA = 16 cosec 30 ° = 32$

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

$⇒ OR = OAsin75 °$

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

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

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