题目如图,ABABAB是⊙O\odot O⊙O的直径,点CCC和点DDD在⊙O\odot O⊙O上,若⊙O\odot O⊙O的半径是444,BD=6BD=6BD=6,则sin∠ACD\sin \angle ACDsin∠ACD的值是( )A.34\frac{3}{4}43B.35\frac{3}{5}53C.73\frac{\sqrt{7}}{3}37D.74\frac{\sqrt{7}}{4}47知识点:切线的性质章节:图形的性质 / 圆