Solve the triangle if the sum of two heights $\left(m_{1}+m_{3}=s\right)$, one side (b), and the angle opposite to it $(\beta)$ are given.
$$
m_{1}=b \sin \gamma, m_{3}=b \sin \alpha
$$

from which

$$
m_{1}+m_{3}=s=b(\sin \alpha+\sin \gamma)=2 b \sin \frac{\alpha+\gamma}{2} \cos \frac{\alpha-\gamma}{2}
$$

or

$$
s=2 b \cos \frac{\beta}{2} \cos \frac{\alpha-\gamma}{2}
$$

thus

$$
\cos \frac{\alpha-\gamma}{2}=\frac{s}{2 b \cos \frac{\beta}{2}}
$$

from which the value of $\frac{\alpha-\gamma}{2}$ can be calculated, and thus $\alpha, \beta, \gamma$ and $b$ are known.

Then, using the sine rule, $a, c$ and $t$ can be determined.

(Preisich Gusztáv, Beszterczebánya.)

The problem was also solved by: Bartók I., Deutsch E., Deutsch I., Eckstein J., Enyedi B., Haar A., Hirschfeld Gy., Kertész G., König D., Kürti I., Pám M., Pivnyik I., Riesz K., Sonnenfeld I., Szücs A.