题目如图,OAOAOA,OBOBOB是⊙O\odot O⊙O的半径,点CCC在劣弧AB^\widehat {AB}AB上,连接ABABAB,ACACAC,BCBCBC.若∠AOB=120∘\angle AOB=120^{\circ}∠AOB=120∘,∠BAC=21∘\angle BAC=21^{\circ}∠BAC=21∘,则∠ABC\angle ABC∠ABC的度数是( )A.42∘42^{\circ}42∘B.39∘39^{\circ}39∘C.37∘37^{\circ}37∘D.35∘35^{\circ}35∘知识点:圆周角定理I章节:图形的性质 / 圆