Question

Find the distance from the eye at which a coin of 2cm diameter should be held to conceal the full moon whose angular diameter is 31’.

Answer 1

Hint: Use the fact that the angle subtended by the diameter of the coin at the eye of the observer should be greater or equal to the angle subtended by the moon. Since the angular diameter of the moon is small, we can approximate the length of the diameter to be equal to the corresponding arc and hence find the distance at which the coin should be placed from the eye.

Complete step-by-step answer:

Since the angular diameter of the moon is 31’, we have the $\angle BAC=31'$.

Also since $31'$ is small, we have diameter approximately equal to arc length.

We know that if x is the measure of the angle in degrees, l is the length of the arc and r the radius, then $l=\dfrac{x}{360}\times 2\pi r$

Here, we have $x=(\dfrac{31}{60}){}^\circ ,l=2cm(dia\ of\ the\ coin)$ and r= AG = AD.

Hence, we have

$2=\dfrac{31}{60\times 360}\times 2\pi \times AG$

Solving for AG, we get

$AG=\dfrac{2\times 60\times 360}{31\times 2\times \dfrac{22}{7}}=221.7cm$

Hence the distance from the eye of the observer at which the coin should be placed so that the moon is completely concealed is 221.7cm.

Note: Alternative solution: Using Trigonometry

We have

$\angle DAG=\dfrac{31'}{2}=(\dfrac{31}{2\times 60}){}^\circ =(\dfrac{31}{120}){}^\circ =\dfrac{31}{120}\times \dfrac{\pi }{180}rad$

Now, we have

\[\sin \left( \angle GAD \right)=\dfrac{GD}{AD}\]

Since $\angle GAD$ is small, we have $\sin \left( \angle GAD \right)=\angle GAD$

Hence, we have

$\angle GAD=\dfrac{GD}{AD}\Rightarrow \dfrac{31}{120}\times \dfrac{\pi }{180}=\dfrac{1}{AD}$

Solving for AD, we get

$AD=\dfrac{120\times 180}{31\times \dfrac{22}{7}}=221.70cm$, which is the same as obtained above.