Solve the triangle if one side $a$, the angle $\alpha$ opposite to it, and the difference between the other two sides $b-c$ are given. \left(a=631.28 \mathrm{dm}, \alpha=63^{\circ} 35^{\prime} 30^{\prime \prime}, b-c=373 \mathrm{dm}.\right)
According to Mollweide's equation,

$$
\frac{b-c}{a}=\frac{\sin \frac{\beta-\gamma}{2}}{\sin \frac{\beta+\gamma}{2}}=\frac{\sin \frac{\beta-\gamma}{2}}{\cos \frac{\alpha}{2}}
$$

thus

$$
\sin \left(\frac{\beta-\gamma}{2}\right)=\frac{b-c}{a} \cos \frac{\alpha}{2}
$$

Therefore, $\beta$ and $\gamma$ can be calculated since we know $(\beta+\gamma)$ and $(\beta-\gamma)$, without knowing $\beta$ and $\gamma$ themselves. $b$ and $c$ can be calculated based on Carnot's theorem; indeed,

$$
\begin{gathered}
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha=(b-c)^{2}+2 b c[1-\cos \alpha]= \\
=(b-c)^{2}+4 b c \sin ^{2} \frac{\alpha}{2}
\end{gathered}
$$

thus

$$
b c=\frac{a^{2}-(b-c)^{2}}{4 \sin ^{2} \frac{\alpha}{2}}
$$

Knowing $b c$ and $(b-c)$, $b$ and $c$ can be calculated. The numerical results in this problem are:

$$
\beta=88^{\circ} 20^{\prime} 55^{\prime \prime}, \quad \gamma=28^{\circ} 3^{\prime} 35^{\prime \prime}
$$

and

$$
b=704.55 \mathrm{dm}, \quad c=331.55 \mathrm{dm}
$$

(Pál Füstös, Eger.)

Number of solutions: 38.