Question

The angular diameter of a planet measured from earth is $90"$. If the diameter of the planet is $\pi \times 10^6\,m$ , then its distance from the earth is

• A. $3.6\times 10^9\,m$
• B. $7.2\times 10^9\,m$
• C. $3.6\times 10^6\,m$
• D. $7.2\times 10^6\,m$
• E. $1.8\times 10^8\,m$

Answer 1

Answer: (B)

Given $\theta=90^{11}=90 \times \frac{\pi}{180 \times 60 \times 60}=\frac{\pi}{7200}$
$D=\pi \times 10^{6} m d=\frac{D}{\theta}=\frac{\pi \times 10^{6} \times 7200}{\pi}$
$=7.2 \times 10^{9} M$