5.7. In an isosceles triangle \(ABC\) with lateral sides \(AB = BC\), the angle \(ABC\) is \(80^\circ\). Inside the triangle, a point \(O\) is taken such that the angle \(OAC\) is \(10^\circ\) and the angle \(OCA\) is \(30^\circ\). Find the angle \(AOB\).
5.7. Answer. $70^{\circ}$.

Let $K$ be the point of intersection of the altitude dropped from vertex $B$ and the bisector of angle $OAB$ (see figure). First, show that point $K$ lies on the extension of $OC$. Then prove that triangles

![](https://cdn.mathpix.com/cropped/2024_05_21_20eb9f6ef2c568605a63g-46.jpg?height=378&width=526&top_left_y=1012&top_left_x=310)

Fig. 5.7.

$ABK$ and $AOK$ are congruent (by side and two adjacent angles) and from the isosceles triangle $OAB$, find the required angle.