Given lines are $m x^2+2 h x+n y^2=0$
Let $\mathrm{m}_1$ and $\mathrm{m}_2$ be the slopes of the lines
$$
\begin{aligned}
& \therefore \mathrm{m}_1+\mathrm{m}_2=\frac{-2 \mathrm{~h}}{\mathrm{n}} \text { and } \mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{m}}{\mathrm{n}} \\
& \text { Now }\left(\mathrm{m}_1-\mathrm{m}_2\right)^2=\left(\mathrm{m}_1+\mathrm{m}_2\right)^2-4 \mathrm{~m}_1 \mathrm{~m}_2 \\
& =\left(\frac{-2 \mathrm{~h}}{\mathrm{n}}\right)^2-\frac{4 \mathrm{~m}}{\mathrm{n}}=\frac{4 \mathrm{~h}^2-4 \mathrm{mn}}{\mathrm{n}^2}=\frac{3 \mathrm{~m}^2+6 \mathrm{mn}+3 \mathrm{n}^2}{\mathrm{n}^2} \ldots[\text { From (1)] } \\
& =\frac{3(\mathrm{~m}+\mathrm{n})^2}{\mathrm{n}^2} \\
& \therefore \mathrm{m}_1-\mathrm{m}_2=\frac{\sqrt{3}(\mathrm{~m}+\mathrm{n})}{\mathrm{n}}
\end{aligned}
$$
$\therefore$ Let $\theta$ be the required angle.
$$
\begin{aligned}
& \text { Now } \tan \theta=\frac{\left|\mathrm{m}_1-\mathrm{m}_2\right|}{1+\mathrm{m}_1 \mathrm{~m}_2} \\
& =\frac{\left(\frac{\sqrt{3}(\mathrm{~m}+\mathrm{n})}{\mathrm{n}}\right)}{\left(1+\frac{\mathrm{m}}{\mathrm{n}}\right)}=\frac{\sqrt{3}(\mathrm{~m}+\mathrm{n})}{\mathrm{n}} \times \frac{\mathrm{n}}{\mathrm{m}+\mathrm{n}}=\sqrt{3} \Rightarrow \theta=60^{\circ}=\frac{\pi^{\mathrm{c}}}{3}
\end{aligned}
$$