A median of a triangle is $k=5 \mathrm{dm}$; this median forms angles of $\varphi=47^{\circ} 16^{\prime}$ and $\phi=25^{\circ} 38^{\prime}$ with the sides; what is the area of the triangle?
Let's extend the median $A D=k$ of triangle $A B C$ so that $D E=D A$. The quadrilateral $B A C E$ is a parallelogram because its diagonals ($A E$ and $B D$) bisect each other. Therefore, the area of triangle $A B E$ is equal to the area of triangle $A B C$. But

$$
A B=2 k \frac{\sin A E B}{\sin A B E}
$$

and

$$
B E=2 k \frac{\sin B A E}{\sin A B E}
$$

where

$$
A E B \angle=\psi, B A E \angle=\varphi, A B E \angle=180^{\circ}-(\varphi+\psi)
$$

Thus, the area of triangle $A B E$, and consequently of triangle $A B C$, is:

$$
t=\frac{2 k^{2} \sin \varphi \sin \psi}{\sin (\varphi+\psi)}=16.623 \mathrm{dm}^{2}
$$

(Odön Spitzer.)

The problem was also solved by: Dénes A., Détshy K., Devecis M., Dolowschiák M., Freibauer E., Goldziher K., Laczkó E., Orlowszky F., Porkoláb J., Probst E., Szabó I., Szabó K., Weisz J.