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(B) Let $O$ be the center of the sphere and $A$ be the observer's eye. Let $r = 16 \ m$ be the radius. The tangents from $A$ to the sphere touch at $P

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $O$ be the center of the sphere and $A$ be the observer's eye. Let $r = 16 \ m$ be the radius. The tangents from $A$ to the sphere touch at $P$ and $Q$. The angle $\angle PAQ = 60^{\circ}$,so $\angle OAP = 30^{\circ}$.In $	riangle OAP$,$\sin(30^{\circ}) = \frac{OP}{OA} \implies \frac{1}{2} = \frac{16}{OA} \implies OA = 32 \ m$.The angle of elevation of the center $O$ from $A$ is $75^{\circ}$. Let $H$ be the height of the center $O$ from the horizontal level of the observer's eye. Then $H = OA \sin(75^{\circ}) = 32 \sin(45^{\circ} + 30^{\circ}) = 32 \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} ight) = 32 \left( \frac{\sqrt{3}+1}{2\sqrt{2}} ight) = 8\sqrt{2}(\sqrt{3}+1) = 8(\sqrt{6}+\sqrt{2}) \ m$.The height of the top most point $T$ from the level of the observer's eye is $H + r = 8(\sqrt{6}+\sqrt{2}) + 16 = 8(\sqrt{6}+\sqrt{2}+2) \ m$.

$A$ spherical gas balloon of radius $16 \ m$ 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 $m$) of the top most point of the balloon from the level of the observer's eye is:

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