12. (6 points) As shown in the figure, given that lines $A B$ and $C D$ intersect at point $O$, if $\angle A O C=20^{\circ}, \angle E O D=60^{\circ}$, then $\angle A O E$ $=$ $\qquad$ , $\angle B O C=$ $\qquad$ .
【Analysis】From the figure, we know that $\angle A O C=20^{\circ} , \angle E O D=60^{\circ}$, and the sum of $\angle A O E$ with $\angle A O C$ and $\angle E O D$ equals $180^{\circ}$. Therefore, we can find the degree of $\angle A O E$; $\angle B O C$ and $\angle A O C=20^{\circ}$ are supplementary angles, and according to the definition of supplementary angles, we can solve it.

【Solution】Solution: $\angle A O E=180^{\circ}-\angle A O C-\angle E O D=180^{\circ}-20^{\circ}-60^{\circ}=100^{\circ}$. $\angle B O C=180^{\circ}-\angle A O C=180^{\circ}-20^{\circ}=160^{\circ}$.

Therefore, the answers are: $100^{\circ} ; 160^{\circ}$.