Question

A sphere with centre $O$ sits atop of 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 meters is

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

Answer 1

Answer: (C)

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

in $△PST$

$\tan {30}^o=\frac{PS}{50}\Rightarrow PS=\frac{50}{\sqrt{3}}$

Now, in $△QUT$

$QU=PS+r$ and $UT=50-r$

$\tan {60}^o=\frac{QU}{US}\Rightarrow \sqrt{3}=\frac{PS+r}{50-r}\Rightarrow r=\frac{50\sqrt{3}-PS}{\sqrt{3}+1}\Rightarrow r=\frac{50\sqrt{3}-\frac{50}{\sqrt{3}}}{\sqrt{3}+1}=50(1-\frac{1}{\sqrt{3}})$

Therefore, Answer is $50(1-\frac{1}{\sqrt{3}})$