Problem 7.2.1. In the figure, rays $O A, O B, O C, O D, O E, O F$ are such that:

- $O B$ is the bisector of angle $A O C$
- $O E$ is the bisector of angle $D O F$;
- $\angle A O F=146^{\circ}, \angle C O D=42^{\circ}$.

How many degrees does angle $B O E$ measure?

![](https://cdn.mathpix.com/cropped/2024_05_06_ce96a31ee4d4d7b89169g-02.jpg?height=368&width=719&top_left_y=1140&top_left_x=369)
Answer: $94^{\circ}$.

Solution. Let's find the sum of angles $A O B$ and $E O F$:

$$
\angle A O B+\angle E O F=\frac{\angle A O C}{2}+\frac{\angle D O F}{2}=\frac{\angle A O F-\angle C O D}{2}=\frac{146^{\circ}-42^{\circ}}{2}=52^{\circ}
$$

Now it is not difficult to find the required angle:

$$
\angle B O E=\angle A O F-(\angle A O B+\angle E O F)=146^{\circ}-52^{\circ}=94^{\circ} .
$$