Find the acute angle between the pair of straight lines $\;$ $x^2 - 7xy + 12y^2 = 0$
Equation of given pair of lines: $\;$ $x^2 - 7xy + 12y^2 = 0$
Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ gives
$a = 1, \;\; h = \dfrac{-7}{2}, \;\; b = 12$
Let $\theta$ be the angle between the given pair of lines.
Then, $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{h^2 - ab}}{a + b}\right|$
i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\left(\dfrac{-7}{2}\right)^2 - 1 \times 12}}{1 + 12}\right|$
i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\dfrac{49}{4} - 12}}{13}\right| = \left|\dfrac{1}{13}\right| = \dfrac{1}{13}$
$\implies$ $\theta = \tan^{-1} \left(\dfrac{1}{13}\right)$