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