题目如图,在矩形ABCDABCDABCD中,AB=3AB=3AB=3,BC=2BC=2BC=2,以BCBCBC为直径在矩形内作半圆,自点AAA作半圆的切线AEAEAE,则sin∠CBE=( )\sin \angle CBE=\left(\ \ \right)sin∠CBE=( )A.63\frac{{\sqrt{6}}}{3}36B.23\frac{2}{3}32C.13\frac{1}{3}31D.1010\frac{{\sqrt{10}}}{10}1010知识点:切线长定理章节:图形的性质 / 圆