In a quadrilateral, $a: b: c: d=9: 8: 7: 6$. One of the angles $\alpha=72^{\circ} 36^{\prime} 18^{\prime \prime}$; the area is $225 \mathrm{~cm}^{2}$. What are the lengths of the sides and the angles of the quadrilateral?
According to the condition,

$$
a=\frac{3}{2} d, \quad b=\frac{4}{3} d \quad \text { and } \quad c=\frac{7}{6} d
$$

On the other hand, by Carnot's theorem,

$$
a^{2}+d^{2}-2 a d \cos \alpha=b^{2}+c^{2}-2 b c \cos \gamma
$$

or

$$
\cos \gamma=\frac{b^{2}+c^{2}-a^{2}-d^{2}+2 a d \cos \alpha}{2 b c}=
$$

thus, in this problem,

$$
\cos \gamma=\frac{27 \cos \alpha-1}{28}=0.25240
$$

or

$$
\gamma=75^{\circ} 27^{\prime} 6^{\prime \prime}
$$

Furthermore,

$$
a d \sin \alpha+b c \sin \gamma=2 t
$$

or

$$
\begin{gathered}
d^{2}\left(\frac{3 \sin \alpha}{2}+\frac{7 \sin \gamma}{4}\right)=2 t \\
d=\sqrt{\frac{8 t}{6 \sin \alpha+7 \sin \gamma}}=12.05 \mathrm{~cm}
\end{gathered}
$$

thus,

$$
a=18.07 \mathrm{~cm}, b=16.06 \mathrm{~cm} \text {, and } c=14.05 \mathrm{~cm} \text {. }
$$

If in the quadrilateral $A B C D$ the diagonal $B D$ divides the angles $\beta$ and $\delta$ into $\beta_{1}, \beta_{2}$ and $\delta_{1}, \delta_{2}$, then in the triangle $A B D$ by the tangent theorem,

$$
\frac{a+d}{a-d}=\frac{\operatorname{tg} \frac{\beta_{1}+\delta_{1}}{2}}{\operatorname{tg} \frac{\beta_{1}-\delta_{1}}{2}}=\frac{\operatorname{tg}\left(90^{\circ}-\frac{\alpha}{2}\right)}{\operatorname{tg} \frac{\beta_{1}-\delta_{1}}{2}}
$$

or

$$
\operatorname{tg} \frac{\beta_{1}-\delta_{1}}{2}=\frac{1}{5} \operatorname{ctg} \frac{\alpha}{2}
$$

thus $\frac{\beta_{1}-\delta_{1}}{2}$ can be calculated. Since

$$
\beta_{1}+\delta_{1}=180^{\circ}-\alpha
$$

therefore, both $\beta_{1}$ and $\delta_{1}$ can be determined. We find that

$$
\beta_{1}=68^{\circ} 49^{\prime} 1^{\prime \prime}, \delta_{1}=38^{\circ} 24^{\prime} 41^{\prime \prime}
$$

Similarly, from the triangle $B C D$, we get

$$
\beta_{2}=61^{\circ} 51^{\prime} 39^{\prime \prime}, \delta_{2}=42^{\circ} 51^{\prime} 15^{\prime \prime}
$$

or

$$
\beta=\beta_{1}+\beta_{2}=130^{\circ} 40^{\prime} 40^{\prime \prime}, \text { and } \delta=\delta_{1}+\delta_{2}=81^{\circ} 15^{\prime} 56^{\prime \prime}
$$

The problem was solved by: Jenő Silbermann.