2. Inside the area of $\measuredangle A O B$, a ray $O C$ is drawn such that $\measuredangle A O C$ is $40^{\circ}$ less than $\measuredangle C O B$ and is equal to one third of $\measuredangle A O B$. Determine $\measuredangle A O B$.

翻译完成，保留了原文的换行和格式。
Solution. Let $\measuredangle A O C=x$ and $\measuredangle C O B=y$. Then $\measuredangle A O B=x+y$ and it holds that $x=y-40^{\circ}$ and $x=\frac{1}{3}(x+y)$. If we substitute the first equation into the second, we get $y-40^{\circ}=\frac{1}{3}\left(y-40^{\circ}+y\right)$, from which it follows that $3 y-120^{\circ}=2 y-40^{\circ}$, so $y=80^{\circ}$. Finally, $x=80^{\circ}-40^{\circ}=40^{\circ}$.