Let $ABCD$ be the quadrilateral in question, with $AB=a=33 \, \text{m}$, $BC=b=63 \, \text{m}$, $CD=c=16 \, \text{m}$, and finally $DA=d=56 \, \text{m}$. Therefore, the square of the diagonal $BD$ is

$$
f^{2}=b^{2}+c^{2}-2bc \cos \beta=d^{2}+a^{2}-2da \cos \beta
$$

From this,

$$
\begin{gathered}
b^{2}+c^{2}-d^{2}-a^{2}=2(bc-da) \cos \beta \\
\cos \beta=\frac{b^{2}+c^{2}-d^{2}-a^{2}}{2(bc-da)} \\
=\frac{3969+256-3136-1089}{2(bc-da)}=0 \\
\beta=90^{\circ}
\end{gathered}
$$

Thus,

$$
\begin{gathered}
f^{2}=4225 \\
f=65 \, \text{m}
\end{gathered}
$$

The second diagonal $e$, since we are dealing with a cyclic quadrilateral,

$$
e=\frac{ac+bd}{f}=\frac{4056}{65}=62.4 \, \text{m}
$$

Therefore,

$$
\begin{gathered}
\cos \alpha=\frac{a^{2}+b^{2}-c^{2}}{2ab}=\frac{1164 \cdot 24}{4158} \\
\log \cos \alpha=3 \cdot 06603-3 \cdot 61888=9 \cdot 44715-10 \\
\alpha=73^{\circ} 44' 25'' \\
\gamma=106'' 15' 35''
\end{gathered}
$$

The area

$$
S=\frac{ad+bc}{2}=\frac{1848+1008}{2}=1428 \, \text{m}^{2}
$$

(Pál Holbay, senior high school 8th grade student, Nyitra).
A quadrilateral has sides of $33 \mathrm{~m}$, $63 \mathrm{~m}$, $16 \mathrm{~m}$, and $56 \mathrm{~m}$. The angle between the first and fourth sides is equal to the angle between the second and third sides. What are the angles, the diagonals, and the area?