Find the distance from the eye at which a coin of \(2\) cm diameter should be held so as to conceal the full moon whose angular diameter is \(31’\).
Solution:
Diameter of the coin:
\[ s=2 \text{ cm} \]
Angular diameter of the moon:
\[ 31′ \]
Since,
\[ 1^\circ=60′ \]
\[ 31’=\frac{31}{60}^\circ \]
Convert into radians:
\[ \theta=\frac{31}{60}\times\frac{\pi}{180} \]
\[ \theta=\frac{31\pi}{10800} \]
Using,
\[ s=r\theta \]
\[ r=\frac{s}{\theta} \]
\[ r=\frac{2}{31\pi/10800} \]
\[ r=\frac{21600}{31\pi} \]
\[ r\approx221.9 \text{ cm} \]
Therefore, the required distance is:
\[ \boxed{221.9 \text{ cm (approximately)}} \]