A sphere with centre $\mathrm{O}$ sits atop a pole as shown in the figure. An observer on the ground is at a distance $50 \mathrm{~m}$ from the foot of the pole. She notes that the angles of elevation from the observer to points $\mathrm{P}$ and $\mathrm{Q}$ on the sphere are $30^{\circ}$ and $60^{\circ}$, respectively. Then, the radius of the sphere in meters is -

• A

  $100\left(1-\frac{1}{\sqrt{3}}\right)$
• B

  $\frac{50 \sqrt{6}}{3}$
• Correct answer
  C

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

  $\frac{100 \sqrt{6}}{3}$

Solution

$\mathrm{DE}=\mathrm{BC}=\mathrm{r}$ $\tan 30^{\circ}=\frac{\mathrm{h}}{50}$ $\mathrm{~h}=\frac{50}{\sqrt{3}}$ $\tan 60^{\circ}=\frac{\mathrm{h}+\mathrm{r}}{50-\mathrm{r}}$ $\sqrt{3}(50-\mathrm{r})=\mathrm{h}+\mathrm{r}$ $\sqrt{3}(50-\mathrm{r})=\frac{50}{\sqrt{3}}+\mathrm{r}$ $3(50-\mathrm{r})=50+\sqrt{3} \mathrm{r}$ $100=(3+\sqrt{3}) \mathrm{r}$ $r=\frac{100}{3+\sqrt{3}}$ $r=\frac{100(3-\sqrt{3})}{6}=50\left(1-\frac{1}{\sqrt{3}}\right)$

Hence, the correct answer is option: 'C'.