Question

A sphere with centre $O$ sits on the top of a pole as shown in the figure

An observer on the ground is at a distance $50m$ from the foot of the pole. She notes that the angles of elevation from the observer to points $P$ and $Q$ on the sphere are $30^{\circ }$ and $60^{\circ }$, respectively. Then, the radius of the sphere in metres is

• A. $100\left(1-\frac{1}{\sqrt{3}}\right)$
• B. $\frac{50\sqrt{6}}{3}$
• C. $50\left(1-\frac{1}{\sqrt{3}}\right)$
• D. $\frac{100\sqrt{6}}{3}$

Answer 1

Answer: (C)

Given, $OP=OQ=r=AR$

$AB=50$
In $\Delta BRQ,\tan 60^{\circ }$
$=\frac{RQ}{BR}=\frac{h+r}{50-r}$
$\Rightarrow \sqrt{3}\left(50-r\right)=h+r$
$\Rightarrow h=50\sqrt{3}-r\left(\sqrt{3}+1\right)\dots \left(i\right)$
ln $\Delta APB$,
$\tan 30^{\circ }=\frac{AP}{AB}=\frac{h}{50}$
$\Rightarrow h=\frac{50}{\sqrt{3}}\dots \left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$, we get
$\frac{50}{\sqrt{3}}=50\sqrt{3}-r\left(\sqrt{3}+1\right)$
$\Rightarrow r\left(\sqrt{3}+1\right)=\frac{100}{\sqrt{3}}$
$\Rightarrow r=\frac{100}{\sqrt{3}\left(\sqrt{3}+1\right)}$
$=\frac{100\left(3-\sqrt{3}\right)}{9-3}$
$=50\left(1-\frac{1}{\sqrt{3}}\right)$