2. As shown in Figure 3, the diagonals of quadrilateral $ABCD$ intersect at point $O, \angle BAD=\angle BCD=60^{\circ}$, $\angle CBD=55^{\circ}, \angle ADB=$ $50^{\circ}$. Then the degree measure of $\angle AOB$ is
2. $80^{\circ}$.

As shown in Figure 6, draw perpendiculars from point $C$ to $A B$, $B D$, and $A D$, with the feet of the perpendiculars being $E$, $F$, and $G$ respectively.
It is easy to see that $\angle A B D=70^{\circ}$, $\angle C B D=\angle C B E=55^{\circ}$, $\angle B D C=\angle C D G=65^{\circ}$.
Therefore, $B C$ is the angle bisector of $\angle D B E$, and $C D$ is the angle bisector of $\angle B D G$.
Thus, $C E=C F=C G$.
Hence, $A C$ is the angle bisector of $\angle E A G$.
Therefore, $\angle A O B=\angle C A D+\angle A D B=80^{\circ}$.