Solve the triangle if the perimeter and the two angles at the base are given. For example, $k=1 \mathrm{dm}, \alpha=35^{\circ} 40^{\prime} 15^{\prime \prime}, \quad \beta=$ $48^{\circ} 20^{\prime} 26^{\prime \prime}$.
Minthogy

$$
b=a \frac{\sin \beta}{\sin \alpha} \quad \text{and} \quad c=a \frac{\sin \gamma}{\sin \alpha}
$$

therefore

$$
k=a+b+c=a\left(1+\frac{\sin \beta}{\sin \alpha}+\frac{\sin \gamma}{\sin \alpha}\right)=a \frac{\sin \alpha+\sin \beta+\sin \gamma}{\sin \alpha}
$$

which means

$$
a=\frac{k}{\sin \alpha+\sin \beta+\sin \gamma} \sin \alpha
$$

thus

$$
b=\frac{k}{\sin \alpha+\sin \beta+\sin \gamma} \sin \beta
$$

and

$$
c=\frac{k}{\sin \alpha+\sin \beta+\sin \gamma} \sin \gamma .
$$

The area of the triangle

$$
t=\frac{a b}{2} \sin \gamma=\left(\frac{k}{\sin \alpha+\sin \beta+\sin \gamma}\right)^{2} \frac{\sin \alpha \sin \beta \sin \gamma}{2}
$$

Numerical values in this problem: $a=2.508 \mathrm{~cm}, b=3.214 \mathrm{~cm}, c=4.278 \mathrm{~cm}$ and $t=4 \mathrm{~cm}^{2}$.

(Dezső Grünhut, Budapest.)