Answer 1

Let PQ be the diameter of the coin and E be the eye of the observer.
Also, let the coin be kept at a distance r from the eye of the observer to hide the moon completely.
Now,

\[\theta = 31' = \left( \frac{31}{60} \right)^\circ = \left( \frac{31}{60} \times \frac{\pi}{180} \right) \text{ radians }\]

\[\theta = \frac{\text{Arc}}{\text{Radius}}\]

\[ \Rightarrow \frac{31}{60} \times \frac{\pi}{180} = \frac{2}{\text{Radius}}\]

\[ \Rightarrow\text{ Radius }= \frac{180 \times 60 \times 2 \times 7}{31 \times 22}\]

\[ = 221 . 7\text{ cm or }2 . 217 m\]