Solve the triangle if the angle $\gamma=104^{\circ} 15^{\prime} 0.4^{\prime \prime}$ is given, the ratio of the sides enclosing this angle is $\frac{a}{b}=2.88$, and the height corresponding to the third side is $m_{c}=20 \mathrm{~cm}$.
Minthogy

$$
\frac{a}{b}=2.88
$$

therefore

$$
\frac{a+b}{a-b}=\frac{3.88}{1.88}
$$

thus, applying the tangent theorem:

$$
\operatorname{tg} \frac{\alpha-\beta}{2}=\frac{1.88}{3.88} \operatorname{tg}\left(90^{\circ}-\frac{\gamma}{2}\right)
$$

substituting the value of $\gamma$, we get

$$
\alpha-\beta=41^{\circ} 18'
$$

and thus

$$
\alpha=58^{\circ} 31' 29.8'', \beta=17^{\circ} 13' 29.8''
$$

Furthermore

$$
b=\frac{m_{c}}{\sin \alpha}, a=2.88 b, c=\frac{a \sin \gamma}{\sin \alpha}, t=\frac{m_{c} \cdot c}{2}
$$

substituting the values:

$$
b=23.45 \mathrm{~cm}, a=67.54 \mathrm{~cm}, c=76.75 \mathrm{~cm}, t=767.5 \mathrm{~cm}^{2}
$$

(Károly Burján, Kaposvár.)

The problem was also solved by: Bayer B., Engel D., Faith F., Filkorn J., Goldstein A., Hein I., Holzmann M., Keesz J., Kerekes T., Krisztián Gy., Kürth A., Lindtner M., Lukhaub Gy., Lupsa Gy., Perl Gy., Sasvári G., Smodics K.