What are the sides of the triangle if the area of the triangle is $t=4920 \mathrm{~cm}^{2}$ and two of its angles are: $\alpha=43^{\circ} 36^{\prime}, \beta=72^{\circ} 23^{\prime}$.
It is known that

$$
2 t=a b \sin \gamma
$$

furthermore, based on the sine theorem,

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

Substituting this value of $b$ into (1),

$$
\begin{gathered}
2 t=\frac{a^{2} \sin \beta \sin \gamma}{\sin \alpha} \\
\text { from which } a=\sqrt{\frac{2 t \sin \alpha}{\sin \beta \sin \gamma}}=\sqrt{\frac{9840 \sin 43^{\circ} 36^{\prime}}{\sin 72^{\circ} 23^{\prime} \sin 64^{\circ} 0^{\prime}}} \approx 89 \mathrm{~cm}
\end{gathered}
$$

With $a$ known, the other two sides can be most simply calculated using the sine theorem:

$$
b=\frac{\alpha \sin \beta}{\sin \alpha} \approx 123 \mathrm{~cm} \quad \text { and } \quad c=\frac{a \sin \gamma}{\sin \alpha} \approx 116 \mathrm{~cm}
$$

Veronika Kollányi (Kaposvár, Munkácsy M. g. IV. o. t.)