A star is ;$1.45\ parsec$ light years away. How much parallax would this star show when viewed from two locations of the earth six months apart in its orbit around the sun?

One light year $=$ speed of light $\times$ one year

or $1 ly=3\times 10^8\times (24\times 3600)=94608\times 10^{11} m$

So, $4.29 ly=4.29\times (94608\times 10^{11})=4.058\times 10^{16} m$

As $1$ parsec $=3.08\times 10^{16} m$

(Parsec

is a unit of length used to measure large distances to objects outside our Solar System

)

Thus, $4.29 ly=\dfrac{4.058\times 10^{16}}{3.08\times 10^{16}}=1.32$ parsec

Now angular displacement , $\theta=\dfrac{d}{D}$

where $d=$ diameter of earth's orbit $= 3\times 10^{11} m$ and

$D=$ distance of star from the earth $=4.058\times 10^{16} m$

So, $\theta=\dfrac{3\times 10^{11}}{4.058\times 10^{16}}=7.39\times 10^{-6}$ rad

As $1 sec=4.85\times 10^{-6} rad$ so, $7.39\times 10^{-6} rad=\dfrac{7.39\times 10^{-6}}{4.85\times 10^{-6}}=1.52 sec$