解：如图，设直线$l$交$A_{1}A_{2}$于点$E，$交$A_{3}A_{4}$于点$D$
∵六边形$A_{1}A_{2}A_{3}A_{4}A_{5}A_{6}$的每个内角都相等
$∴∠A_{2} = ∠A_{3} = (6-2)×180°÷6 = 120°$
∵五边形$B_{1}B_{2}B_{3}B_{4}B_{5}$的每个内角都相等
$∴∠B_{2}B_{3}B_{4}=(5-2)×180°÷5=108°$
$∴∠B_{4}B_{3}D =180°-108°=72°$
$∵A_{3}A_{4}//B_{3}B_{4}$
$∴∠EDA_{3}=∠B_{4}B_{3}D =72°$
∵四边形$A_{2}A_{3}DE $的内角和为$(4-2) × 180°= 360°$
$∴α=∠A_{2}ED= 360°- ∠A_{2}- ∠A_{3} - ∠EDA_{3}$
$=360°- 120°-120°-72°=48°$
