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, r be the distance, at which coin in placed. So that it completely conceals the full moon.

Let, E be the eye of the observer.

Now,

$\theta ={31}^{\prime }={\left(\frac{31}{60}\right)}^{\circ }[\because {60}^{\prime }=1^{\circ }]$

Also,

$\widehat{AB}$ = arc AB = 2 cm = 0.02 m.

Now,

by $\theta =\frac{\text{arc}}{\text{radius}}$

$=\frac{31\pi }{60\times 180}=\frac{0.02}{r}\Rightarrow r=\frac{0.02\times 60\times 180}{31\pi }=2.217m[\because \pi =\frac{22}{7}]$

Thus,

The coin should be placed at a distance of 2.217 m from the eye.