Solve the triangle if $t=4920 \mathrm{~cm}^{2}, \alpha=43^{\circ} 36^{\prime} 10^{\prime \prime}, \quad \beta=72^{\circ} 23^{\prime} 11^{\prime \prime}$.
It is known that

$$
a b=\frac{2 t}{\sin \gamma}
$$

and

$$
\frac{a}{b}=\frac{\sin \alpha}{\sin \beta}
$$

thus

$$
a^{2}=\frac{2 t \sin \alpha}{\sin \beta \sin \gamma}
$$

or

$$
a=\sqrt{\frac{2 t \sin \alpha}{\sin \beta \sin \gamma}}
$$

Similarly,

$$
b=\sqrt{\frac{2 t \sin \beta}{\sin \gamma \cdot \sin \alpha}} \quad \text { and } \quad c=\sqrt{\frac{2 t \sin \gamma}{\sin \alpha \cdot \sin \beta}}
$$

Substituting the given values, we get

$$
a=89 \text{ cm}, \quad b=123 \text{ cm} \quad c=116 \text{ cm}.
$$

(Viktor Wáhl, Eger.)

The problem was also solved by: Babocsai Gy., Bauer F., Csada I., Ehrenfeld N., Füstös P., Halmi O., Kirchknopf E., Kiss J., Kovács Gy., Kürth R., Lorenc A., Lusztig M., Paunz A., Sárközy P., Spitzer L., Wellis D.