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

Let three angles of the triangle be given as \(\mathrm{A}=\frac{\pi}{4}, \mathrm{~B}=\frac{\pi}{3}\) and \(\mathrm{C}=\frac{\pi}{4}+\frac{\pi}{6}\)
Let a, b, c be the opposite to angles A, B, C respectively.
As \(\frac{\sin A}{a}=\frac{\sin C}{c}\), we get
\(
\text { Required Ratio }=\frac{\mathrm{a}}{\mathrm{c}}=\frac{\sin \mathrm{A}}{\sin \mathrm{C}}
\)
\(
=\frac{\sin \left(\frac{\pi}{4}\right)}{\sin \left(\frac{\pi}{4}+\frac{\pi}{6}\right)}
\)
\(\begin{aligned} & =\frac{\sin \frac{\pi}{4}}{\sin \frac{\pi}{4} \cos \frac{\pi}{6}+\cos \frac{\pi}{4} \sin \frac{\pi}{6}} \\ & =\frac{2}{\sqrt{3}+1} \\ & =\frac{2(\sqrt{3}-1)}{2} \\ & =(\sqrt{3}-1): 1\end{aligned}\)