Answer
$8.13^{\circ} $ or $0.14$ radians
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 $ax+by=c$
${\bf n_{1}}=(3){\bf i} + (-4){\bf j}=\langle 3,-4\rangle$
is perpendicular to $3x-4y=3$
${\bf n_{2}}=(1){\bf i} + (-1){\bf j}=\langle 1,-1\rangle$
is perpendicular to $x-y=7.$
$\displaystyle \theta=\cos^{-1}(\frac{{\bf n_{1}}\cdot{\bf n_{2}}}{|{\bf n_{1}}||{\bf n_{2}}|})$
$=\displaystyle \cos^{-1}(\frac{3+4}{\sqrt{9+16}\cdot\sqrt{1+1}})$
$=\displaystyle \cos^{-1}(\frac{7}{5\cdot\sqrt{2}})\approx 8.13^{\circ} $ or $0.14$ radians
