## Thomas' Calculus 13th Edition

$60^{\circ}$ or $\pi/3$
The result of exercise 31 tells us that ${\bf n}=a{\bf i} + b{\bf j}$ is perpendicular to the lines $ax+by=c$ Rewrite $(y=\sqrt{3}x-1)$ as $(\sqrt{3}x-y=1)$ ${\bf n_{1}}=(\sqrt{3}){\bf i} + (-1){\bf j}=\langle\sqrt{3},-1\rangle$ is perpendicular to $\sqrt{3}x-y=1$ Rewrite $(y= -\sqrt{3}x+2)$ as $(\sqrt{3}x+y=2)$ ${\bf n_{2}}=(\sqrt{3}){\bf i} + (1){\bf j}=\langle\sqrt{3},1\rangle$ is perpendicular to $\sqrt{3}x+y=2.$ $\displaystyle \theta=\cos^{-1}(\frac{{\bf n_{1}}\cdot{\bf n_{2}}}{|{\bf n_{1}}||{\bf n_{2}}|})$ $=\displaystyle \cos^{-1}(\frac{3-1}{\sqrt{3+1}\cdot\sqrt{3+1}})$ $=\displaystyle \cos^{-1}(\frac{2}{4})$ $=\displaystyle \cos^{-1}(\frac{1}{2})=60^{\circ}$ or $\pi/3$