Question

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

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

Answer 1

Answer: (C)

Let $AB$ 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 $AB$ as arc $AB$. Let $d$ be the diameter of the moon. Then, $d=\mathrm{arc}AB$.

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