If two angles of \(\triangle \mathrm{ABC}\) are \(\frac{\pi}{4}\) and \(\frac{\pi}{3}\), then the ratio of the smallest and greatest side is

Two angles of triangle are \(\frac{\pi}{4}\) and \(\frac{\pi}{3}\) Let the third angle be \(\alpha\).
\(\begin{array}{l}
\therefore \frac{\pi}{4}+\frac{\pi}{3}+\alpha=\pi \\
\therefore 45^{\circ}+60^{\circ}+\alpha=180^{\circ} \Rightarrow \alpha=75^{\circ}
\end{array}\)
We know that side opposite to smallest angle is the smallest side and side opposite to largest angle is the largest side.
\(\therefore \frac{\mathrm{c}}{\sin 75^{\circ}}=\frac{\mathrm{a}}{\sin 45^{\circ}}\)
We know that \(\sin 75^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
\(\therefore \frac{\mathrm{c}(2 \sqrt{2})}{\sqrt{3}+1}=\mathrm{a} \sqrt{2} \Rightarrow \frac{\mathrm{a}}{\mathrm{c}}=\frac{2}{\sqrt{3}+1}\)
By rationalizing, we get \(\frac{a}{c}=\frac{\sqrt{3}-1}{1}\)
Note : \(\sin 75^{\circ}\) can be calculated as follows :
\(\sin 75^{\circ} =\sin \left(45^{\circ}+30^{\circ}\right)\)
\(=\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)