连接OA、OB,过点O作OC⊥AB于C,
则OC是等腰△OAB底边上的高,可得：
OA = OB = 4 ,
AC = BC = AB/2 = 2√3 ,
OC = √(OA²-AC²) = 2 ,
∠AOC = 60° ,
∠AOB = 2∠AOC = 120° ,
所以,劣弧AB的度数为 120° ,优弧AB的度数为 360-120 = 240° .