Answer
$\approx 67.62^{\circ} $ or $1.18$ 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 $ax+by=c$
${\bf n_{1}}=(12){\bf i} + (5){\bf j}=\langle 12,5\rangle$
is perpendicular to $12x+5y=1$
${\bf n_{2}}=(2){\bf i} + (-2){\bf j}=\langle 2,-2\rangle$
is perpendicular to $2x-2y=3.$
$\displaystyle \theta=\cos^{-1}(\frac{{\bf n_{1}}\cdot{\bf n_{2}}}{|{\bf n_{1}}||{\bf n_{2}}|})$
$=\displaystyle \cos^{-1}(\frac{24-10}{\sqrt{144+25}\cdot\sqrt{4+4}})$
$=\displaystyle \cos^{-1}(\frac{14}{13\cdot 2\sqrt{2}})\approx 67.62^{\circ} $ or $1.18$ radians
