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'.

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,
$\theta =31\text{'}={\left(\frac{31}{60}\right)}^{°}=\left(\frac{31}{60}\times \frac{\mathrm{\pi }}{180}\right)\mathrm{radians}$
$$\begin{gathered} \theta =\frac{\mathrm{Arc}}{\mathrm{Radius}} \\ \Rightarrow \frac{31}{60}\times \frac{\mathrm{\pi }}{180}=\frac{2}{\mathrm{Radius}} \\ \Rightarrow \mathrm{Radius}=\frac{180\times 60\times 2\times 7}{31\times 22} \\ =221.7\mathrm{cm}\mathrm{or}2.217\mathrm{m} \end{gathered}$$