Question

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 $91\times {10}^{\circ }km$

Answer 1

Let, E be the eye of the observer and s be the sum.

Now,

$\angle AOB=\theta ={32}^{\prime }={\left(\frac{32}{60}\right)}^{\circ }={(\frac{32}{60}\times \frac{\pi }{180})}^c\because \theta =\frac{\text{arc}}{\text{radius}}\Rightarrow \frac{32}{60}\times \frac{\pi }{180}=\frac{AB}{91\times {10}^6}km\Rightarrow AB=\frac{91\times {10}^6\times 32\times \pi }{60\times 180}=8.474074\times {10}^{5}km$
= 847407.4 km

$\therefore$ Distance of sun is 847407.4 km