The acute angle between the lines \(x \cos 30^{\circ}+y \sin 30^{\circ}=3\) and \(x \cos 60^{\circ}+y \sin 60^{\circ}=5\) is

Given equations of lines, \(x \cos 30^{\circ}+y \sin 30^{\circ}=3\) and \(x \cos 60^{\circ}+y \sin 60^{\circ}=5\) \(\Rightarrow \sqrt{3} x+y=6\) and \(x+\sqrt{3} y=10\)
\(\therefore \quad\) Slope of line \(\sqrt{3} x+y=6=m_1=-\sqrt{3}\)
Slope of line \(x+\sqrt{3} y=10=\mathrm{m}_2=\frac{-1}{\sqrt{3}}\)
\(\begin{aligned} \therefore \quad \tan \theta & =\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ & =\left|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3}) \times\left(\frac{-1}{\sqrt{3}}\right)}\right| \\ & =\left|\frac{-3+1}{2 \sqrt{3}}\right| \\ & =\left|\frac{-2}{2 \sqrt{3}}\right| \\ \therefore \quad \tan \theta & =\frac{1}{\sqrt{3}} \\ \therefore \quad \Rightarrow \theta & =30^{\circ}\end{aligned}\)