Question

Find the diameter of the sun in Km supposing that it subtends an angle of 32’ at the eye of the observer. Given that the distance of the sun from the observer $=91\times {{10}^{6}}km$

Answer 1

Hint: Use the fact that since the angular diameter is small, we can approximate the length of the arc which subtends the angle at the eye of the observer is equal to the diameter of the sun. Use the fact that if x is the measure of the angle in degrees, l is the length of the corresponding arc and r the radius of the circle, then $l=\dfrac{x}{360{}^\circ }\times 2\pi r$. Hence find the diameter of the sun.

Complete step-by-step answer:

We have the angle subtended by the sun at the eye of the observer $=\angle BAC=32'=(\dfrac{32}{60}){}^\circ $
Also since the angular diameter is small, the length of the arc which subtends angle equal to BAC can be approximated by the diameter of the sun.
Suppose the diameter of the sun be d metres.
Now we know that if x is the measure of the angle in degrees, l is the length of the corresponding arc and r the radius of the circle, then $l=\dfrac{x}{360{}^\circ }\times 2\pi r$.
Here we have $l=d,x=\dfrac{32}{60}$ and $r=91\times {{10}^{6}}km=91\times {{10}^{9}}m$
Now, we have
$d=\dfrac{32}{60\times 360}\times 2\pi \times 91\times {{10}^{9}}=\dfrac{32}{60\times 360}\times 2\times \dfrac{22}{7}\times 91\times {{10}^{9}}=0.85\times {{10}^{9}}$
Hence the diameter of the sun $=0.85\times {{10}^{9}}m=0.85\times {{10}^{6}}km=8.5\times {{10}^{5}}km$

Note: Alternative solution:
We have $AE=91\times {{10}^{9}}m$ and $\angle FAE=\dfrac{32'}{2}=16'=\dfrac{16}{60}\times \dfrac{\pi }{180}rad$
Now, we have
$\sin \left( \angle FAE \right)=\dfrac{FE}{AE}$
Since the measure of angle FAE is small, we have $\sin \left( \angle FAE \right)=\angle FAE$
Hence, we have
$\angle FAE=\dfrac{FE}{AE}\Rightarrow \dfrac{16}{60}\times \dfrac{\pi }{180}=\dfrac{FE}{91\times {{10}^{9}}}$
Solving for FE, we get
$FE=0.424\times {{10}^{9}}m$
Hence the radius of the sun $=0.424\times {{10}^{9}}m$
Hence, the diameter of the sun $=2\times 0.424\times {{10}^{9}}=0.848\times {{10}^{9}}\approx 0.85\times {{10}^{9}}m=8.5\times {{10}^{5}}km$
Hence the diameter of the sun $=8.5\times {{10}^{5}}km$