Question

A spherical gas balloon of radius $16$ meter subtends an angle ${60}^{\circ }$ at the eye of the observer A while the angle of elevation of its center from the eye of A is ${75}^{\circ }$. Then the height (in meter) of the top most point of the balloon from the level of the observer’s eye is

• A. $8(2+2\sqrt{3}+\sqrt{2})$
• B. $8(\sqrt{6}-\sqrt{2}+2)$
• C. $8(\sqrt{2}+2+\sqrt{3})$
• D. $8(\sqrt{6}+\sqrt{2}+2)$

Answer 1

The correct option is D $8(\sqrt{6}+\sqrt{2}+2)$

Let $O$ be the centre of circle.
$P,Q$ points of contact tangents from $A$
$T$ be the topmost point on ellipse and $R$ be the foot of perpandicular.
In $\Delta OAP$
$OA=16\text{cosec}{30}^0=32$
In $\Delta ABO$
$OA=OA\sin {75}^0=32\times \frac{\sqrt{3}+1}{2\sqrt{2}}$
$OA=16\times \frac{\sqrt{3}+1}{2}\times \sqrt{2}=8\sqrt{2}(\sqrt{3}+1)=8(\sqrt{6}+\sqrt{2})$
So, topmost point is $=OR+OT$
$=8(\sqrt{6}+\sqrt{2}+2)m$