A sphere with centre O sits on the top of a p | vedclass

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(C) Let $h$ be the height of the pole and $r$ be the radius of the sphere. The point $P$ is the bottom of the sphere,so the height of $P$ from the gro

Vedclass · Accepted Answer

$50 \left(1-\frac{1}{\sqrt{3}}\right)$

Vedclass · Answer

(C) Let $h$ be the height of the pole and $r$ be the radius of the sphere. The point $P$ is the bottom of the sphere,so the height of $P$ from the ground is $h$. The point $Q$ is at the same level as the center $O$,so the height of $Q$ from the ground is $h+r$. The horizontal distance from the observer $B$ to the pole is $AB = 50 \ m$.In $	riangle APB$,$	an 30^{\circ} = \frac{AP}{AB} = \frac{h}{50}$.$\Rightarrow h = 50 	an 30^{\circ} = \frac{50}{\sqrt{3}}$.In the triangle formed by the observer,the point $Q$,and the point $R$ (where $R$ is the projection of $Q$ on the ground),the horizontal distance is $BR = AB - r = 50 - r$ and the vertical height is $RQ = h+r$.$	an 60^{\circ} = \frac{RQ}{BR} = \frac{h+r}{50-r}$.$\Rightarrow \sqrt{3}(50-r) = h+r$.Substitute $h = \frac{50}{\sqrt{3}}$:$\sqrt{3}(50-r) = \frac{50}{\sqrt{3}} + r$.$50\sqrt{3} - r\sqrt{3} = \frac{50}{\sqrt{3}} + r$.$50\sqrt{3} - \frac{50}{\sqrt{3}} = r(1+\sqrt{3})$.$50 \left( \frac{3-1}{\sqrt{3}} ight) = r(1+\sqrt{3})$.$r = \frac{50 	imes 2}{\sqrt{3}(1+\sqrt{3})} = \frac{100}{\sqrt{3}+3} = \frac{100}{\sqrt{3}(1+\sqrt{3})} = \frac{100(3-\sqrt{3})}{3(3-1)} = \frac{100(3-\sqrt{3})}{6} = \frac{50(3-\sqrt{3})}{3} = 50 \left( 1 - \frac{1}{\sqrt{3}} ight)$.

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