Answer
$45^{\circ}$, or $\pi/4$
Work Step by Step
The result of exercise 31 tells us that ${\bf n}=a{\bf i} + b{\bf j}$
is perpendicular to the lines $ax+by=c$
${\bf n_{1}}=(3){\bf i} + (1){\bf j}=\langle 3,1\rangle$
is perpendicular to $3x+y=5.$
${\bf n_{2}}=(2){\bf i} + (-1){\bf j}=\langle 2,-1\rangle$
is perpendicular to $2x-y=4.$
$\displaystyle \theta=\cos^{-1}(\frac{{\bf n_{1}}\cdot{\bf n_{2}}}{|{\bf n_{1}}||{\bf n_{2}}|})$
$=\displaystyle \cos^{-1}(\frac{6-1}{\sqrt{9+1}\cdot\sqrt{4+1}})$
$=\displaystyle \cos^{-1}(\frac{5}{\sqrt{50}})$
$=\displaystyle \cos^{-1}(\frac{5}{5\sqrt{2}})$
$=\displaystyle \cos^{-1}(\frac{1}{\sqrt{2}})=45^{\circ}$, or $\pi/4$
