In a quadrilateral, $a: b: c: d=9: 8: 7: 6$. The angle $\alpha=72^{\circ} 36^{\prime}$ is formed by the sides $a$ and $d$, and the area of the quadrilateral is $225 \mathrm{~cm}^{2}$. What are the lengths of the sides and the angles of the quadrilateral?
The figure shows the labeling.

![](https://cdn.mathpix.com/cropped/2024_05_02_cf1f0f89c46dade9c574g-1.jpg?height=423&width=458&top_left_y=192&top_left_x=823)

The diagonal $BD=f$ can be expressed using the cosine rule from $\triangle ABD$ and $\triangle BCD$ as follows:

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

from which

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

Expressing the sides with a proportional factor $x$:

$$
a=9 x, \quad b=8 x, \quad c=7 x, \quad d=6 x
$$

Substituting these values into (1):

$$
\cos \gamma=\frac{64 x^{2}+49 x^{2}-81 x^{2}-36 x^{2}+108 x^{2} \cos \alpha}{112 x^{2}}=\frac{27 \cos \alpha-1}{28}=\frac{7.086}{28}
$$

from which

$$
\gamma=75^{\circ} 22^{\prime}
$$

(Another value of $\gamma$: $284^{\circ} 38^{\prime}$, which leads to a concave quadrilateral in this case.)

The double area of the quadrilateral is:

$$
a d \sin \alpha+b c \sin \gamma=450
$$

Thus,

$$
54 x^{2} \sin \alpha+56 x^{2} \sin \gamma=450
$$

from which

$$
x^{2}=\frac{450}{54 \sin 72^{\circ} 36^{\prime}+56 \sin 75^{\circ} 22^{\prime}}=\frac{450}{105.72}
$$

and thus

$$
x=2.063
$$

Therefore,

$$
a=18.567 \text{ cm}, \quad b=16.504 \text{ cm}, \quad c=14.441 \text{ cm} \text{ and } \quad d=12.378 \text{ cm}.
$$

The calculation of the two missing angles was mostly done by solving for the diagonal $f$ using the cumbersome cosine rule (especially cumbersome if we do not know the logarithmic form) and then finding the angles $\beta_{1}$ and $\beta_{2}$ using the sine rule twice.

Using the tangent rule, we can avoid the cumbersome calculation of the diagonal $f$ and directly find the angles $\beta_{1}$ and $\delta_{1}$ as well as $\beta_{2}$ and $\delta_{2}$. The tangent rule can be derived from the sine rule in a short step:

$$
a: b=\sin \alpha: \sin \beta
$$

from which

$$
(a+b):(a-b)=(\sin \alpha+\sin \beta):(\sin \alpha-\sin \beta)
$$

or

$$
\frac{a+b}{a-b}=\frac{2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}}{2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}=\operatorname{tg} \frac{\alpha+\beta}{2} \cdot \operatorname{cotg} \frac{\alpha-\beta}{2}=\frac{\operatorname{tg} \frac{\alpha+\beta}{2}}{\operatorname{tg} \frac{\alpha-\beta}{2}}
$$

This is the tangent rule.

Since $\frac{\alpha+\beta}{2}=\frac{180^{\circ}-\gamma}{2}=90^{\circ}-\frac{\gamma}{2}$ and $\operatorname{tg}\left(90-\frac{\gamma}{2}\right)=\operatorname{cotg} \frac{\gamma}{2}$, from (2) we get

$$
\operatorname{tg} \frac{\alpha-\beta}{2}=\frac{(a-b) \operatorname{cotg} \frac{\gamma}{2}}{a+b}
$$

Applying (3) to $\triangle ABD$:

$$
\operatorname{tg} \frac{\delta_{1}-\beta_{1}}{2}=\frac{(a-d) \operatorname{cotg} \frac{\alpha}{2}}{a+d}=\frac{6.189 \operatorname{cotg} 36^{\circ} 18^{\prime}}{30.945}
$$

from which

$$
\frac{\delta_{1}-\beta_{1}}{2}=15^{\circ} 15^{\prime}
$$

or

$$
\delta_{1}-\beta_{1}=30^{\circ} 30^{\prime}
$$

On the other hand,

$$
\delta_{1}+\beta_{1}=180^{\circ}-\alpha=107^{\circ} 24^{\prime}
$$

and thus from (4) and (5):

$$
\delta_{1}=68^{\circ} 57^{\prime} \quad \text { and } \quad \beta_{1}=38^{\circ} 27^{\prime}
$$

Similarly, from $\triangle BCD$:

$$
\operatorname{tg} \frac{\delta_{2}-\beta_{2}}{2}=\frac{(b-c) \operatorname{cotg} \frac{\gamma}{2}}{b+c}=\frac{2.063 \operatorname{cotg} 37^{\circ} 41^{\prime}}{30.945}
$$

from which

$$
\frac{\delta_{2}-\beta_{2}}{2}=4^{\circ} 56^{\prime}
$$

or

$$
\delta_{2}-\beta_{2}=9^{\circ} 52^{\prime}
$$

and since

$$
\delta_{2}+\beta_{2}=180^{\circ}-\gamma=104^{\circ} 38^{\prime}
$$

therefore

$$
\delta_{2}=57^{\circ} 15^{\prime} \quad \text { and } \quad \beta_{2}=47^{\circ} 23^{\prime}
$$

Thus,

$$
\beta=\beta_{1}+\beta_{2}=85^{\circ} 50^{\prime}
$$

and

$$
\delta=\delta_{1}+\delta_{2}=126^{\circ} 12^{\prime}
$$