题目如图,射线ABABAB与⊙O\odot O⊙O相切于点BBB,经过圆心OOO的射线ACACAC与⊙O\odot O⊙O相交于点DDD、CCC,连接BCBCBC,若∠A=42∘\angle A=42^{\circ}∠A=42∘,则∠ACB=______∘.\angle ACB= \_\_\_\_\_\_^{\circ}.∠ACB=______∘.知识点:切线的判定章节:图形的性质 / 圆