Question

If the angular diameter of the moon is ${30}^o$, how far from the eye a coin of diameter $2.2$ cm can be kept to hide the moon?

Answer 1

Suppose the coin is kept at a distance $r$ from the eye to hide the moon completely.
Let $E$ be the eye of the observer and let $AB$ be the diameter of the coin.
then arc$APB=$ diameter $AB=2.2$cm
Now $\theta ={30}^0$
$\left(\frac{30}{60}\right)=(\frac{1}{2}.\frac{\pi }{180})=\left(\frac{\pi }{360}\right)$
$\therefore \theta =\frac{arc}{radius}$
$\Rightarrow \frac{\pi }{360}=\frac{2.2}{r}\Rightarrow r=\frac{2.2\times 360}{\pi }$
$\Rightarrow r=\frac{2.2\times 252\times 7}{22}=252$