The medians of a triangle are individually: $s_{1}=\frac{1}{2} \sqrt{673}, s_{2}=\frac{1}{2} \sqrt{592}$ and $s_{3}=\frac{1}{2} \sqrt{505}$; solve the triangle using trigonometric methods.
If the midpoints of the sides of triangle $ABC$ are denoted by $A_{1}, B_{1}, C_{1}$, and the lengths of the medians are denoted by $s_{1}, s_{2}, s_{3}$, then from the Stewart's theorem (K. M. L. IX. year, problem 948):

$$
A B^{2} \cdot A_{1} C + A C^{2} \cdot A_{1} B = A A_{1}^{2} \cdot B C + B C \cdot B A_{1} \cdot A_{1} C
$$

If the lengths of the sides are denoted by $a, b, c$ respectively, and considering that

$$
A_{1} C = A_{1} B = \frac{B C}{2} = \frac{a}{2}
$$

then

$$
c^{2} \cdot \frac{a}{2} + b^{2} \cdot \frac{a}{2} = s_{1}^{2} \cdot a + a \cdot \frac{a}{2} \cdot \frac{a}{2}
$$

thus

$$
2 b^{2} + 2 c^{2} - a^{2} = 4 s_{1}^{2}
$$

and similarly

$$
\begin{aligned}
& 2 c^{2} + 2 a^{2} - b^{2} = 4 s_{2}^{2} \\
& 2 a^{2} + 2 b^{2} - c^{2} = 4 s_{3}^{2}
\end{aligned}
$$

from which we can calculate the values of $a^{2}, b^{2}, c^{2}$, and then

$$
\begin{aligned}
& a = \frac{2}{3} \sqrt{2 s_{2}^{2} + 2 s_{3}^{2} - s_{1}^{2}} \\
& b = \frac{2}{3} \sqrt{2 s_{3}^{2} + 2 s_{1}^{2} - s_{2}^{2}} \\
& c = \frac{2}{3} \sqrt{2 s_{1}^{2} + 2 s_{2}^{2} - s_{3}^{2}}
\end{aligned}
$$

Given our data,

$$
a = 13 ; b = 14 ; c = 15
$$

and since

$$
\begin{aligned}
& \cos \frac{\alpha}{2} = \sqrt{\frac{s(s-a)}{b c}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} \\
& \cos \frac{\beta}{2} = \sqrt{\frac{s(s-b)}{c a}} = \sqrt{\frac{49}{65}} = \frac{7}{\sqrt{65}} \\
& \cos \frac{\gamma}{2} = \sqrt{\frac{s(s-c)}{a b}} = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}}
\end{aligned}
$$

therefore

$$
\begin{gathered}
\alpha = 53^{\circ} 7^{\prime} 50^{\prime \prime} \\
\beta = 59^{\circ} 29^{\prime} 26^{\prime \prime} \\
\gamma = 67^{\circ} 22^{\prime} 44^{\prime \prime}
\end{gathered}
$$

(Imre Dömény, Budapest.)

The problem was also solved by: Ádámffy E., Bartók I., Dömény E., Eckhart F., Enyedi B., Glück J., Haar A., Heimlich P., Hirschfeld Gy., Jánosy Gy., Kertész G., Kiss J., Kürti I., Liehner A., Neidenbach E., Pám M., Pichler S., Pivnyik I., Popoviciu M., Rássy P., Riesz K., Rosenberg J., Schwarz Gy., Schuster Gy., Schwemmer I., Szücs A.