The angles of a triangle and side $c$ are given. Express the height corresponding to side $c$ using these.
I. solution. We use the usual notation. Let the projection of vertex $C$ onto line $AB$ be $C_1$, so the height we are looking for is: $m_c = CC_1$. We can assume that $\alpha \leq \beta$, so $\alpha$ is definitely an acute angle. From the right-angled triangle $ACC_1$,

$$
m_c = AC \sin \alpha = b \sin \alpha
$$

On the other hand, applying the sine theorem to the sides $b$ and the given $c$,

$$
b = c \frac{\sin \beta}{\sin \gamma}, \quad \text{and thus} \quad (1) \quad m_c = \frac{c \sin \alpha \sin \beta}{\sin \gamma}
$$

Bod Katalin (Miskolc, Herman O. lg. III. o. t.)

II. solution. If $\alpha \leq \beta < 90^\circ$, then $C_1$ is on the extension of side $AB$ beyond $B$, $AB = AC_1 - BC_1$, and for the $\beta'$ angle at $B$ in the right-angled triangle $BCC_1$, $\beta' = 180^\circ - \beta$. Thus, $BC_1 = m_c \operatorname{ctg} \beta' = m_c \operatorname{ctg}(180^\circ - \beta) = -m_c \operatorname{ctg} \beta$. Substituting this into the previous equality, we again get (2). Finally, if $\beta = 90^\circ$, then on the one hand $C_1 \equiv B, AB = AC_1$, and on the other hand $\operatorname{ctg} \beta = 0$, so (2) is also valid in this case.

Simenszky Csilla (Makó, József A. g. III. o. t.)

Remark. Expression (2) can be transformed into (1):

$$
m_c = \frac{c}{\frac{\cos \alpha}{\sin \alpha} + \frac{\cos \beta}{\sin \beta}} = \frac{c \sin \alpha \sin \beta}{\sin (\alpha + \beta)} = \frac{c \sin \alpha \sin \beta}{\sin \gamma}
$$