Answer
$30^{\circ}$ or $\pi/6$
Work Step by Step
The result of exercise 31 tells us that
${\bf v}=\langle a,\ b \rangle$ is perpendicular to lines $ax+by=c$
${\bf n_{1}}=\langle\sqrt{3},-1\rangle$ is perpendicular to $\sqrt{3}x-y=-2$
${\bf n_{2}}=\langle 1,-\sqrt{3}\rangle$ is perpendicular to $x-\sqrt{3}y=1.$
$\displaystyle \theta=\cos^{-1}(\frac{{\bf n_{1}}\cdot{\bf n_{2}}}{|{\bf n_{1}}||{\bf n_{2}}|})$
$=\displaystyle \cos^{-1}(\frac{\sqrt{3}+\sqrt{3}}{\sqrt{3+1}\cdot\sqrt{1+3}})$
$=\displaystyle \cos^{-1}(\frac{2\sqrt{3}}{4})$
$=\displaystyle \cos^{-1}(\frac{\sqrt{3}}{2})=30^{\circ}$ or $\pi/6$
