解：$\because \triangle ABM\cong\triangle ACN，$$\angle B = 20^{\circ}，$$\angle CAN = 30^{\circ}，$
$\therefore \angle BAM=\angle CAN = 30^{\circ}，$$\angle B=\angle C = 20^{\circ}。$
又$\because \angle B+\angle C+\angle BAC = 180^{\circ}，$
$\therefore \angle BAC = 180^{\circ}-\angle B-\angle C = 180^{\circ}-20^{\circ}-20^{\circ}=140^{\circ}。$
$\therefore \angle MAN=\angle BAC-\angle BAM-\angle CAN = 140^{\circ}-30^{\circ}-30^{\circ}=80^{\circ}$
