4A. Given a convex quadrilateral $A B C D$ for which: $\angle A B D=50^{\circ}$, $\measuredangle A D B=80^{\circ}, \measuredangle A C B=40^{\circ}$ and $\measuredangle D B C=\measuredangle B D C+30^{\circ}$. Find $\measuredangle D B C$.
Solution. According to the conditions of the problem, we get that $\measuredangle B A D=50^{\circ}$, which means triangle $A B D$ is isosceles. Let $k$ be the circle with center $D$ and radius $\overline{D A}$. Clearly, $B \in k$. Since $\measuredangle A D B=80^{\circ}, \measuredangle A C B=40^{\circ}$, and $\measuredangle A D B$ is a central angle over the arc $A B$, it follows that $C \in k$. Therefore, triangle $B C D$ is also isosceles. From the conditions of the problem, we have:

$$
\begin{aligned}
180^{\circ} & =\measuredangle D B C+\measuredangle D C B+\measuredangle B D C \\
& =\measuredangle B D C+30^{\circ}+\measuredangle B D C+30^{\circ}+\measuredangle B D C
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_84a5520b68919fd51f4cg-05.jpg?height=438&width=440&top_left_y=1139&top_left_x=1030)

from which $\measuredangle B D C=40^{\circ}$, hence $\measuredangle D B C=70^{\circ}$.

ZB. Calculate $\frac{\sin x+\cos x}{\sin x-\cos x}$, if $\sin x \cdot \cos x=\frac{2}{5}$.