Solve the triangle if $a: b=1: 2, \quad \alpha: \beta=1: 3$ and $c=5 \text{ cm}$.
Given

$$
a: b=\sin \alpha: \sin \beta
$$

therefore

$$
\frac{\sin \alpha}{\sin \beta}=2
$$

or

$$
\frac{\sin 3 \alpha}{\sin \alpha}=\frac{3 \sin \alpha-4 \sin ^{3} \alpha}{\sin \alpha}=3-4 \sin ^{2} \alpha=2
$$

from which the usable value is

$$
\sin \alpha=\frac{1}{2}
$$

thus

$$
\alpha=30^{\circ}
$$

and

$$
\beta=90^{\circ}, \gamma=60^{\circ}
$$

Since the triangle is a right triangle, therefore

$$
a^{2}+c^{2}=b^{2}=4 a^{2},
$$

or

$$
a=c \sqrt{\frac{1}{3}}=\frac{5}{3} \sqrt{3} \quad \mathrm{~cm} \text { and } \quad b=\frac{10}{3} \sqrt{3} \mathrm{~cm} .
$$

(Neumann Frida, Budapest.)