题目如图,△ABC\triangle ABC△ABC内接于⊙O\odot O⊙O,∠B=65∘\angle B=65^{\circ}∠B=65∘,∠C=70∘\angle C=70^{\circ}∠C=70∘.若BC=32BC=3\sqrt{2}BC=32,则半径的长为( )A.666B.333C.323\sqrt{2}32D.2\sqrt{2}2知识点:垂径定理章节:图形的性质 / 圆