Let's solve and trigonometrically analyze the triangle, given $m_{a}=10.769, m_{b}=24.706$ and $c=28$.
The construction is as follows: $A B=c$, and a circle is drawn over it as a diameter. The circle $\mathrm{E}$ is drawn from $A$ and $B$ as centers, with radii $m_{a}$ and $m_{b}$, intersecting at points $D, D_{1}$ and $E, E_{1}$. If the lines $A E$ and $B D$, as well as $A E_{1}$ and $B D_{1}$, $A E$ and $B D_{1}$, $A E_{1}$ and $B D$ intersect at points $C, C^{\prime}, C_{1}$, and $C_{1}^{\prime}$, then $A B C, A B C^{\prime}, A B C_{1}$, and $A B C_{1}^{\prime}$ are the sought triangles.

Since $\triangle A B C \cong \triangle A B C^{\prime}$ and $\triangle A B C_{1} \cong \triangle A B C_{1}^{\prime}$, it is sufficient to solve the triangles $\triangle A B C$ and $\triangle A B C_{1}$ trigonometrically. If the angles at vertices $A, B$, and $C$ are $\alpha, \beta$, and $\gamma$, then

$$
\sin \alpha=\frac{m_{b}}{c}, \sin \beta=\frac{m_{a}}{c}
$$

and

$$
\begin{aligned}
\sin \gamma & =\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta= \\
& =\frac{1}{c^{2}}\left(m_{b} \sqrt{c^{2}-m_{a}^{2}}+m_{a} \sqrt{c^{2}-m_{b}^{2}}\right)
\end{aligned}
$$

Substituting the given numerical values, we obtain

$$
\alpha=61^{\circ} 55^{\prime} 33^{\prime \prime}, \beta=22^{\circ} 37^{\prime} 10^{\prime \prime}, \gamma=95^{\circ} 27^{\prime} 17^{\prime \prime}
$$

Furthermore,

$$
a=\frac{m_{b}}{\sin \gamma}=\frac{m_{b} c^{2}}{m_{b} \sqrt{c^{2}-m_{a}^{2}}+m_{a} \sqrt{c^{2}-m_{b}^{2}}}=24.82
$$

and

$$
b=\frac{m_{a}}{\sin \gamma}=\frac{m_{a} c^{2}}{m_{b} \sqrt{c^{2}-m_{a}^{2}}+m_{a} \sqrt{c^{2}-m_{b}^{2}}}=10.818
$$

If the angles at vertices $A, B$, and $C_{1}$ in $\triangle A B C_{1}$ are $\alpha_{1}$, $\beta_{1}$, and $\gamma_{1}$, then

$$
\alpha_{1}=180^{\circ}-\alpha, \beta_{1}=\beta \text {, and thus } \gamma_{1}=180^{\circ}-\left(\alpha_{1}+\beta_{1}\right)=\alpha-\beta
$$

Therefore,

$$
\alpha_{1}=118^{\circ} 04^{\prime} 27^{\prime \prime}, \beta_{1}=22^{\circ} 37^{\prime} 10^{\prime \prime}, \text { and } \gamma_{1}=39^{\circ} 18^{\prime} 23^{\prime \prime}
$$

and

$$
a_{1}=\frac{m_{b}}{\sin \gamma_{1}}=39, \quad b_{1}=\frac{m_{a}}{\sin \gamma_{1}}=17
$$

(Imre Kürti, Eger.)

The problem was also solved by: Ádámffy E., Bartók I., Braun I., Dömény E., Dömény I., Enyedi B., Eckhardt F., Fekete M., Friedländer H., Glück I., Haar A., Hirschfeld Gy., Jánosy Gy., Kertész G., Liebner A., Messer P., Neidenbach P., Pám M., Pichler S., Pazsiczky G., Pető L., Pivnyik I., Popoviciu A., Ragány R., Rássy P., Riesz K., Rosenberg J., Schwemmer I., Schwarz Gy., Söpkéz Gy., Szúcs A.