21.2. In triangle $A B C$, the bisectors of angles $B A C$ and $A B C$ intersect at point $O$. Find angle $A C B$ if angle $A O B$ is $125^{\circ}$.

$(7-8$ grade $)$
21.2. In triangle $A B C$ (Fig. 15), we have: $\angle O A B + \angle O B A = 180^{\circ} - 125^{\circ} = 55^{\circ}$. Since $A O$ and $B O$ are the bisectors of angles $C A B$ and $C B A$, the sum of these angles is $2 \cdot 55^{\circ} = 110^{\circ}$. Then $\angle A C B = 180^{\circ} - (\angle C A B + \angle C B A) = 180^{\circ} - 110^{\circ} = 70^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_f0d089db3d527e92a992g-29.jpg?height=445&width=488&top_left_y=677&top_left_x=790)

Fig. 15