题目如图,△ABC\triangle ABC△ABC是⊙O\odot O⊙O的内接三角形,将劣弧AC^\widehat {AC}AC沿ACACAC折叠后刚好经过弦BCBCBC的中点DDD.若AC=6AC=6AC=6,∠C=60∘\angle C=60^{\circ}∠C=60∘,则⊙O\odot O⊙O的半径长为( )A.137\frac{1}{3}\sqrt{7}317B.237\frac{2}{3}\sqrt{7}327C.1321\frac{1}{3}\sqrt{21}3121D.2321\frac{2}{3}\sqrt{21}3221知识点:解直角三角形章节:图形的变化 / 锐角三角函数