Find the acute angle between the lines $2x - y + 3 = 0\,\,{\text{and}}\,\,x - 3y + 2 = 0$.

Hint:Here before solving this question we need to know the following formula: -
Angle between two lines

$\operatorname{Tan} \theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|\,\,\,\,\,\,\,\,\,\,...(1)$

Where,
${m_1}\,{\text{and}}\,{m_2}$are the slopes of the given line.


Complete step by step solution:
According to this question we have,

$\begin{gathered}
  2x - y + 3 = 0 \\
  x - 3y + 2 = 0 \\
\end{gathered} $

Converting these lines into slope-intercept form, we get

$\begin{gathered}
  y = 2x + 3\,...(2) \\
  3y = x + 2 \\
  y = \dfrac{x}{3} + \dfrac{2}{3}...(3) \\
\end{gathered} $

Comparing the both equation \[\left( 2 \right)\] and \[\left( 3 \right)\] with$y = mx + c$, where$m$ is slope and $c$ is intercept,
So, ${m_1} = 2\,and\,{m_2} = \dfrac{1}{3}$
Substitute all the values in the equation\[\left( 1 \right)\].
\[\begin{gathered}
  \operatorname{Tan} \theta = \left| {\dfrac{{2 - \dfrac{1}{3}}}{{1 + \left( 2 \right)\left( {\dfrac{1}{3}} \right)}}} \right| \\
  \operatorname{Tan} \theta = \left| {\dfrac{{\dfrac{5}{3}}}{{\dfrac{5}{3}}}} \right| = 1 \\
\end{gathered} \]
$\theta = \dfrac{\pi }{4}$
Thus, the acute angle is $\theta = \dfrac{\pi }{4}$

Note: The coefficient of \[x\]in \[y = MX + c\]represents the slope of the line. Here the major emphasis must be laid on finding the slope of the line. The modulus in formula is used because in question acute angle is asked.