Problem 8. Given an isosceles triangle $A B C(A B=B C)$ with the angle at the vertex equal to $102^{\circ}$. Point $O$ is located inside triangle $A B C$ such that $\angle O C A=30^{\circ}$, and $\angle O A C=21^{\circ}$. Find the measure of angle $\angle B O A$.
Answer: $81^{\circ}$.

Solution. Let $B H$ be the height/median/bisector of the triangle. Let $S$ be the intersection of ray $C O$ and segment $B H$. Note that $A S=S C$. For example, in triangle $A S C$, the median $S H$ coincides with the height.

![](https://cdn.mathpix.com/cropped/2024_05_06_5f8906e33c6403142659g-17.jpg?height=562&width=1236&top_left_y=1084&top_left_x=410)

Let's calculate the angles: