Given the perimeter $(K)$ and two angles $(\alpha, \beta)$ of a certain triangle; the triangle is to be constructed from these general components. Assuming that $K=62.4 \mathrm{~m}, \alpha=70^{\circ} 20^{\prime} 40^{\prime \prime}, \beta=36^{\circ} 50^{\prime} 30^{\prime \prime}$, the sides $a$ and $b$ of the triangle and its area are to be determined.
We will construct the triangle $ABC$ where $AB=K$, $\angle A=\frac{\alpha}{2}$, and $\angle B=\frac{\beta}{2}$. At point $C$, we will construct angles $\frac{\alpha}{2}$ and $\frac{\beta}{2}$ adjacent to $AC$ and $BC$, respectively. The bisectors of these angles intersect the line $AB$ at points $E$ and $D$. The triangle $EDC$ is the desired triangle.

Proof.

\[
\begin{gathered}
\angle CDE = \angle DCB + \angle DBC = \frac{\beta}{2} + \frac{\beta}{2} = \beta \\
\angle CED = \angle CAE + \angle ACE = \frac{\alpha}{2} + \frac{\alpha}{2} = \alpha \\
ED + DC + CE = ED + DB + EA = AB
\end{gathered}
\]

Calculation of the sides and the area. From triangle $ACB$:

\[
\begin{gathered}
K : CB = \sin \frac{\alpha+\beta}{2} : \sin \frac{\alpha}{2} \\
CB = K \frac{\sin \frac{\alpha}{2}}{\sin \frac{\alpha+\beta}{2}}
\end{gathered}
\]

Similarly,

\[
AC = K \frac{\sin \frac{\beta}{2}}{\sin \frac{\alpha+\beta}{2}}
\]

From the isosceles triangle $CDB$:

\[
CB = 2CD \cos \frac{\beta}{2}
\]

Thus,

\[
CD = \frac{K}{2} \cdot \frac{\sin \frac{\alpha}{2}}{\sin \frac{\alpha+\beta}{2} \cos \frac{\beta}{2}}
\]

From the isosceles triangle $ACE$:

\[
AC = 2EC \cos \frac{\alpha}{2}
\]

Thus,

\[
EC = \frac{K}{2} \cdot \frac{\sin \frac{\beta}{2}}{\sin \frac{\alpha+\beta}{2} \cos \frac{\alpha}{2}}
\]

Finally, the area of the triangle:

\[
\begin{gathered}
T = \frac{EC \cdot DC}{2} \sin (\alpha+\beta) = \\
= \frac{K^2 \sin \frac{\beta}{2} \cdot \sin \frac{\alpha}{2} \cdot 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}}{8 \cdot \left[\sin \frac{\alpha+\beta}{2}\right]^2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2}} = \\
= \frac{K^2}{4} \cdot \tan \frac{\alpha}{2} \tan \frac{\beta}{2} \cot \frac{\alpha+\beta}{2}
\end{gathered}
\]

Substituting the given values: $CD = 23.54 \, \text{m}$, $EC = 14.99 \, \text{m}$, $T = 168.5 \, \text{m}^2$.

(Odön Spitzer.)

The problem was also solved by: Beck F., Dénes A., Devecis M., Friedmann B., Krátky Gy., Misángyi V., Petrogalli G., Porde Gy., Porkoláb J., Schiffer H., Szabó I., Weisz A., Weisz J.