$解：如图，设直线l交A_1A_2于点E，交A_3A_4于点D.$
$∵六边形A_1A_2A_3A_4A_5A_6的每个内角都相等，$
$∴∠A_2=∠A_3= \frac{(6−2)×180}{6}=120°.$
$∵五边形B_1B_2B_3B_4B_5的每个内角都相等，$
$∴∠B_2B_3B_4=\frac{(5−2)×180}{5}=108°,$
$∴∠B_4B_3D=180°−108°=72°.$
$∵A_3A_4//B_3B_4，$
$∴ ∠EDA_3=∠B_4B_3D=72°.$
$∵四边形A_2A_3DE的内角和为(4−2)×180°=360°，$
$∴∠α=∠AED=360°−∠A_2−∠A_3−∠EDA_3=360°−120°−120°−72°=48°\ $
