Solve the triangle if we know the median $k_{3}$ corresponding to side $AB$, and the angles $(\alpha$ and $\beta)$ at the base $AB$.
First solution. Let $CD$ be the median to the side $AB$, and let $\angle ACD = \gamma_1$, $\angle DCB = \gamma_2$. Then,

$$
k_3: \frac{c}{2} = \sin \alpha : \sin \gamma_1
$$

and

$$
k_3: \frac{c}{2} = \sin \beta : \sin \gamma_2
$$

from which:

$$
\frac{\sin \alpha}{\sin \beta} = \frac{\sin \gamma_1}{\sin \gamma_2}
$$

and thus

$$
\frac{\sin \alpha + \sin \beta}{\sin \alpha - \sin \beta} = \frac{\sin \gamma_1 + \sin \gamma_2}{\sin \gamma_1 - \sin \gamma_2}
$$

or

$$
\frac{\tan \frac{\gamma_1 + \gamma_2}{2}}{\tan \frac{\gamma_1 - \gamma_2}{2}} = \frac{\tan \frac{\alpha + \beta}{2}}{\tan \frac{\alpha - \beta}{2}}
$$

therefore

$$
\tan \frac{\gamma_1 - \gamma_2}{2} = \frac{\tan \frac{\alpha - \beta}{2}}{\tan \frac{\alpha + \beta}{2}} \cdot \tan \frac{\gamma_1 + \gamma_2}{2}
$$

or finally

$$
\tan \frac{\gamma_1 - \gamma_2}{2} = \frac{\tan \frac{\alpha - \beta}{2}}{\tan \frac{\alpha + \beta}{2}} \cdot \cot \frac{\alpha + \beta}{2}
$$

From this equation, we can calculate $\gamma_1 - \gamma_2$; but $\gamma_1 + \gamma_2$ is known, so $\gamma_1$ and $\gamma_2$ can be determined. In the triangles $ADC$ and $BDC$, one side and two angles are known, so the missing elements can be calculated.

Second solution. If $\angle CDB = \epsilon$, then

$$
a^2 = \frac{c^2}{4} + k_3^2 + c k_3 \cos \epsilon
$$

and

$$
b^2 = \frac{c^2}{4} + k_3^2 - c k_3 \cos \epsilon
$$

from which

$$
2a^2 + 2b^2 - c^2 = 4k_3^2
$$

But

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

so

$$
c^2 = \frac{4k_3^2 \sin^2 \gamma}{2 \sin^2 \alpha + 2 \sin^2 \beta - \sin^2 \gamma}
$$

(Nándor Bayer, Losonc.)