Question

A sphere with centre $O$ sits atop 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 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

The correct option is C $50(1-\frac{1}{\sqrt{3}})$


Let $PC=h$
$\tan {30}^{\circ }=\frac{PC}{AC}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{50}\Rightarrow h=\frac{50}{\sqrt{3}}$

$\tan {60}^{\circ }=\frac{QB}{AB}\Rightarrow \sqrt{3}=\frac{h+r}{50-r}\Rightarrow 50\sqrt{3}-\sqrt{3}r=\frac{50}{\sqrt{3}}+r\Rightarrow 150-3r=50+\sqrt{3}r\Rightarrow r=\frac{100}{3+\sqrt{3}}=\frac{100}{\sqrt{3}(\sqrt{3}+1)}\Rightarrow r=\frac{50(\sqrt{3}-1)}{\sqrt{3}}=50(1-\frac{1}{\sqrt{3}})$