When the planet Jupiter is at a distance of 824.7 million km from the earth, its angular diameter is measured to be $35.72\text{'}\text{'}$ of arc. The diameter of Jupiter can be calculated as

Distance of Jupiter from earth (d) = 8247 . million km

$=824.7\times {10}^{6}km$

Angular diameter of Juptier $\left(\theta \right)=35.73"$

But $1^0=60"=(60\times 60)"$

$1"={\left(\frac{1}{60\times 60}\right)}^0=\frac{1}{60\times 60}\times \frac{\pi }{180}rad$

$=\frac{3.14}{60\times 60\times 180}rad$

$1"=4.85\times {10}^{-6}rad$

$\therefore$ Angular diameter of Jupiter $\left(\theta \right)=(35.72\times 4.85\times {10}^{-6})rad$

If D is the diameter of the Jupiter then, angular diameter

$\left(\theta \right)=\frac{D}{d}$

$=(35.72\times 4.85\times {10}^{-6})\times 82.47\times {10}^6$

$=142873$

$=1.429\times {10}^{5}km$