It is known that in every triangle

$$
m_{b}=a \sin \gamma \quad \text { and } \quad m_{c}=a \sin \beta \text {. }
$$

Therefore,

$$
m_{b}-m_{c}=a(\sin \gamma-\sin \beta)
$$

or

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

However, on the other hand,

$$
a=2 R \sin \alpha
$$

and

$$
\cos \frac{\gamma+\beta}{2}=\sin \left(90^{\circ}-\frac{\gamma+\beta}{2}\right)=\sin \frac{\alpha}{2}
$$

so the equation under (1) can also be written as:

$$
m_{b}-m_{c}=4 R \sin \alpha \sin \frac{\alpha}{2} \sin \frac{\gamma-\beta}{2}
$$

from which

$$
\sin \frac{\gamma-\beta}{2}=\frac{m_{b}-m_{c}}{4 R \sin \alpha \sin \frac{\alpha}{2}}
$$

Substituting the given values, we get:

$$
\frac{\gamma-\beta}{2}=15^{\circ} 48^{\prime} 4^{\prime \prime}
$$

and since

$$
\frac{\gamma+\beta}{2}=90^{\circ}-\frac{\alpha}{2}=66^{\circ} 30^{\prime}
$$

therefore

$$
\gamma=82^{\circ} 18^{\prime} 4^{\prime \prime} \quad \text { and } \quad \beta=50^{\circ} 41^{\prime} 56^{\prime \prime}
$$

The sides are given by the formulas

$$
a=2 R \sin \alpha, \quad b=2 R \sin \beta, \quad c=2 R \sin \gamma
$$

Performing the detailed calculations,

$$
a=12.433 \mathrm{~m}, \quad b=13.156 \mathrm{~m}, \quad c=16.847 \mathrm{~m}
$$

(György Schuster, Budapest.)

Number of solutions: 37.
Let's calculate the sides and angles of the triangle, given that the radius of the circumscribed circle around the triangle is \( R = 8.5 \, \text{m} \), one of the angles of the triangle is \( \alpha = 47^\circ \), and the difference between the heights \( m_b \) and \( m_c \), which are perpendicular to the sides of angle \( \alpha \), is \( m_b - m_c = 2.7 \, \text{m} \).