With usual notations if the angles of a triangle are in the ratio \(1: 2: 3\), then their corresponding sides are in the ratio.

Let the angles be \(\mathrm{x}, 2 \mathrm{x} 3 \mathrm{x}\)
\(
\therefore \mathrm{x}+2 \mathrm{x}+3 \mathrm{x}=180^{\circ} \Rightarrow 6 \mathrm{x}=180^{\circ} \Rightarrow \mathrm{x}=30^{\circ}
\)
Thus angles of the triangle are \(30^{\circ}, 60^{\circ}, 90^{\circ}\).
\(\text { Now, } \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \)
\( \therefore \frac{\mathrm{a}}{\sin 30^{\circ}}=\frac{\mathrm{b}}{\sin 60^{\circ}}=\frac{\mathrm{c}}{\sin 90^{\circ}} \Rightarrow \frac{\mathrm{a}}{\left(\frac{1}{2}\right)}\) \(=\frac{\mathrm{b}}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{\mathrm{c}}{(1)} \)
\( \therefore 2 \mathrm{a}=\frac{2 \mathrm{~b}}{\sqrt{3}}=\mathrm{c} \Rightarrow=\frac{\mathrm{c}}{2} \text { and } \mathrm{b}=\frac{\sqrt{3} \mathrm{c}}{2} \)
\( \therefore \mathrm{a}: \mathrm{b}: \mathrm{c}=\frac{\mathrm{c}}{2}: \frac{\sqrt{3} \mathrm{c}}{2}: \mathrm{c}=1: \sqrt{3}: 2\)