$(a)$ How many astronomical units $(AU)$ make $1$ parsec?
$(b)$ Consider a sun-like star at a distance of $2$ parsecs. When it is seen through a telescope with $100$ magnification,what should be the angular size of the star? The Sun appears to be $(\frac{1}{2})^{\circ}$ from the Earth. Due to atmospheric fluctuations,the eye cannot resolve objects smaller than $1$ arc minute.
$(c)$ Mars has approximately half of the Earth's diameter. When it is closest to the Earth,it is at about $\frac{1}{2} AU$ from the Earth. Calculate what size it will appear when seen through the same telescope.

(N/A) By definition,$1$ parsec is the distance at which an arc of length $1 AU$ subtends an angle of $1^{\prime \prime}$.
Since $1^{\circ} = 3600^{\prime \prime}$ and $1^{\circ} = \frac{\pi}{180} \text{ rad}$,we have $1^{\prime \prime} = \frac{\pi}{180 \times 3600} \text{ rad}$.
Thus,$1 \text{ parsec} = \frac{1 AU}{1^{\prime \prime}} = \frac{180 \times 3600}{\pi} AU \approx 2.06 \times 10^{5} AU$.
$(b)$ The angular diameter of the Sun is $0.5^{\circ} = 30^{\prime}$ at $1 AU$. At a distance of $2$ parsecs $(2 \times 2.06 \times 10^{5} AU)$,the angular diameter $\alpha$ of the star is $\alpha = \frac{0.5^{\circ}}{2 \times 2.06 \times 10^{5}} \approx 1.21 \times 10^{-6} \text{ degrees} \approx 0.0044^{\prime \prime}$.
With $100$ magnification,the apparent size is $100 \times 0.0044^{\prime \prime} = 0.44^{\prime \prime}$. Since $0.44^{\prime \prime} < 1^{\prime}$,the star will not be resolved and will appear as a point source due to atmospheric turbulence.
$(c)$ Mars diameter is $0.5 \times D_{Earth}$. Angular size $\theta = \frac{\text{Diameter}}{\text{Distance}}$. At $0.5 AU$,$\theta_{Mars} = \frac{0.5 \times D_{Earth}}{0.5 AU} = \frac{D_{Earth}}{1 AU} \approx 0.5^{\circ} = 30^{\prime}$.
With $100$ magnification,the apparent size is $100 \times 30^{\prime} = 3000^{\prime} = 50^{\circ}$.