Find the diameter of the sun in km supposing that it subtends an angle of \(32’\) at the eye of an observer. Given that the distance of the sun is \(91\times10^6\) km.
Solution:
Distance of the sun:
\[ r=91\times10^6 \text{ km} \]
Angular diameter:
\[ 32′ \]
Since,
\[ 1^\circ=60′ \]
\[ 32’=\frac{32}{60}^\circ \]
Convert into radians:
\[ \theta=\frac{32}{60}\times\frac{\pi}{180} \]
\[ \theta=\frac{4\pi}{1350} \]
Using,
\[ s=r\theta \]
\[ s=91\times10^6\times\frac{4\pi}{1350} \]
\[ s\approx847197 \text{ km} \]
Therefore, the diameter of the sun is:
\[ \boxed{8.47\times10^5 \text{ km (approximately)}} \]