Question

If the distance of moon from the earth is $360000 \,km$ and its diameter subtends an angle of $31^{\prime}$ at the eye of the observer, then the diameter of the moon is

• A. $1247.62\, km$
• B. $2247.62\, km$
• C. $3247.62\, km$
• D. None of these

Answer 1

Answer: (C)

Let $A B$ be the diameter of the moon and let $O$ be the eye of the observer. Since, the distance between the earth and the moon is quite large, so we take diameter $A B$ as arc $A B$. Let $d$ be the diameter of the moon. Then, $d=\operatorname{arc} A B$.

Now, $ \theta=31^{\prime}=\left(\frac{31}{60}\right)^{\circ}=\left(\frac{31}{60} \times \frac{\pi}{180}\right)^c$
and $ r=360000\, km$
$\therefore \theta=\frac{\text { arc }}{\text { radius }}=\frac{\text { diameter of the moon }}{\text { moon's distance from the earth }}$
$\Rightarrow \frac{31}{60} \times \frac{\pi}{180}=\frac{d}{360000}$
$\Rightarrow d=\left(\frac{31}{60} \times \frac{\pi}{180} \times 360000\right) km $
$\Rightarrow d =3247.62\, km$
Hence, the diameter of the moon is $3247.62\, km$.