# Find the distance from the eye at which a coin of 2 cm

Question:

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:

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,

$\theta=31^{\prime}=\left(\frac{31}{60}\right)^{\circ}=\left(\frac{31}{60} \times \frac{\pi}{180}\right)$ radians

$\theta=\frac{\text { Arc }}{\text { Radius }}$

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

$\Rightarrow$ Radius $=\frac{180 \times 60 \times 2 \times 7}{31 \times 22}$

$=221.7 \mathrm{~cm}$ or $2.217 \mathrm{~m}$