The area of a triangle $t=234 \mathrm{~cm}^{2}$, perimeter $k=108 \mathrm{~cm}$, one of the angles $\alpha=130^{\circ} 26^{\prime} 59^{\prime \prime}$. What are the lengths of the sides and the angles?
Let $s$ denote the semi-perimeter of the triangle, thus:

$$
\tan \frac{1}{2} \alpha=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\frac{t}{s(s-a)}
$$

from which

$$
s-a=\frac{t}{s} \cot \frac{1}{2} \alpha
$$

or

$$
s-a=\frac{2 t}{k} \cot \frac{1}{2} \alpha
$$

Substituting the given values, we obtain:

$$
s-a=2, \text { hence } a=52 \text{ cm and } b+c=56 \text{ cm. }
$$

According to Mollweide's first equation:

$$
b+c: a=\cos \frac{1}{2}(\beta-\gamma): \sin \frac{1}{2} \alpha
$$

from which

$$
\cos \frac{1}{2}(\beta-\gamma)=\frac{b+c}{a} \sin \frac{1}{2} \alpha
$$

$$
\frac{1}{2}(\beta-\gamma)=12^{\circ} 5^{\prime} 20^{\prime \prime}
$$

but

$$
\frac{1}{2}(\beta+\gamma)=90^{\circ}-\frac{1}{2} \alpha=24^{\circ} 46^{\prime} 30.5^{\prime \prime}
$$

Thus, from (4) and (5) we get:

$$
\left.\beta=36^{\circ} 51^{\prime} 50.5^{\prime \prime} \text { and } \gamma\right)=12^{\circ} 41^{\prime} 10.5^{\prime \prime}
$$

The sides $b$ and $c$ are determined using the sine rule:

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

Substituting the values, we find that $b=41 \text{ cm}$ and $c=15 \text{ cm}$.

(Bernát Friedmann, 8th grade, S.-A.-Ujhely.)

The problem was also solved by:

Mór Feuer, Károly Fröhlich, János Galter, Emil Geist, Zsigmond Goldstein, Béla Grünhut, Ervin Hofbauer, Nándor Kántor, Jenő Reif, Béla Schneider, István Szabó.