In the convex quadrilateral $A B C D$, the consecutive angles are: $\alpha=79^{\circ}, \beta=63^{\circ}, \gamma=131^{\circ}$, and the side $A B$ is seen from the intersection point $E$ of the diagonals at an angle $\varepsilon=123.5^{\circ}$. Calculate the angle $C A B$.

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The consecutive angles of the convex quadrilateral $A B C D$ are: $\alpha=79^{\circ}, \beta=63^{\circ}, \gamma=131^{\circ}$, and the side $A B$ is seen from the intersection point $E$ of the diagonals at an angle $\varepsilon=123.5^{\circ}$. Calculate the angle $C A B$.
Let the angle $C A B$ be denoted by $\varphi$. Using the known angles and the fact that the sum of the angles in a triangle is $180^{\circ}$, the angle values shown on the diagram are obtained.

![](https://cdn.mathpix.com/cropped/2024_05_02_077745275892e3857ab7g-1.jpg?height=521&width=849&top_left_y=260&top_left_x=627)

It is evident that the equality $\frac{A C}{C B} \cdot \frac{C B}{C D}=\frac{A C}{C D}$ holds. Using the Law of Sines in triangles $A B C$, $B C D$, and $A C D$, we get

$$
\begin{gathered}
\frac{A C}{C B}=\frac{\sin 63^{\circ}}{\sin \varphi}, \quad \frac{C B}{D C}=\frac{\sin \left(42.5^{\circ}-\varphi\right)}{\sin \left(6.5^{\circ}+\varphi\right)} \\
\frac{A C}{D C}=\frac{\sin 87^{\circ}}{\sin \left(79^{\circ}-\varphi\right)}
\end{gathered}
$$

If we substitute these expressions into the previous equality, we obtain the equation

$$
\sin 63^{\circ} \cdot \sin \left(42.5^{\circ}-\varphi\right) \cdot \sin \left(79^{\circ}-\varphi\right)=\sin 87^{\circ} \cdot \sin \varphi \cdot \sin \left(6.5^{\circ}+\varphi\right)
$$

From this, using the angle sum identities repeatedly and rearranging, we get

$$
\begin{gathered}
\left(\sin 87^{\circ} \cos 6.5^{\circ}-\sin 63^{\circ} \cos 42.5^{\circ} \cos 79^{\circ}\right) \sin ^{2} \varphi+ \\
+\left(\sin 63^{\circ} \sin 121.5^{\circ}+\sin 87^{\circ} \sin 6.5^{\circ}\right) \sin \varphi \cos \varphi- \\
-\sin 63^{\circ} \sin 42.5^{\circ} \sin 79^{\circ} \cos ^{2} \varphi=0
\end{gathered}
$$

Using trigonometric tables: $0.8669 \operatorname{tg}^{2} \varphi+0.8728 \operatorname{tg} \varphi-0.599=0$. Solving this quadratic equation, we get $\operatorname{tg} \varphi_{1}=0.4636, \operatorname{tg} \varphi_{2}=-1.4703$. Considering that $0^{\circ}<\varphi<79^{\circ}$, the first root gives $\varphi=24^{\circ} 52^{\prime}$.

(P. T.)