Find the diameter of the sun in km supposing that it subtends an angle of 32' at the eye of an observer.

Let PQ be the diameter of the Sun and E be the eye of the observer.

Because the distance between the Sun and the Earth is quite large, we will take PQ as arc PQ.

Now,

$r=91 \times 10^{6} \mathrm{~km}$

$\theta=32^{\prime}=\left(\frac{32}{60}\right)^{\circ}=\left(\frac{32}{60} \times \frac{\pi}{180}\right)$ radians

$\theta=\frac{\text { Arc }}{\text { Radius }}$

$\Rightarrow \frac{32}{60} \times \frac{\pi}{180}=\frac{d}{91 \times 10^{6}}$

$\Rightarrow d=\frac{32 \times 91 \times 10^{6} \times 22}{60 \times 180 \times 7}=847407.4 \mathrm{~km}$