112. Find the angle between the lines $x+5 y-3=0$ and $2 x-3 y+$ $+4=0$.
Solution. Let's find the coordinates of the normal vectors of the given lines: $\vec{n}_{1}=(1 ; 5), \vec{n}_{2}=(2 ;-3)$. According to formula (5), we get

$$
\begin{gathered}
\cos \varphi=\left|\frac{\vec{n}_{1} \vec{n}_{2}}{\left|\vec{n}_{1}\right| \vec{n}_{2} \mid}\right|=\left|\frac{1 \cdot 2+5 \cdot(-3)}{\sqrt{1+25} \cdot \sqrt{4+9}}\right|=\left|\frac{2-15}{\sqrt{13} \cdot \sqrt{26}}\right|= \\
=\left|\frac{-13}{\sqrt{13} \sqrt{13 \cdot 2}}\right|=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \\
\varphi=\arccos \frac{\sqrt{2}}{2}=\frac{\pi}{4}
\end{gathered}
$$