题目如图,在Rt△ABCRt\triangle ABCRt△ABC中,∠ACB=90∘\angle ACB=90^{\circ}∠ACB=90∘,将△ABC\triangle ABC△ABC绕点CCC逆时针方向旋转58∘58^{\circ}58∘得到△A1B1C\triangle A_{1}B_{1}C△A1B1C.若点A1A_{1}A1恰好落在ABABAB边上,则∠B1\angle B_{1}∠B1度数为( )A.29∘29^{\circ}29∘B.32∘32^{\circ}32∘C.58∘58^{\circ}58∘D.61∘61^{\circ}61∘知识点:勾股定理章节:图形的性质 / 三角形