如图圆o内切于△ABC,AB=9,AC=2+2√3,CB=7+2√3,则角C为多少

题目

如图圆o内切于△ABC,AB=9,AC=2+2√3,CB=7+2√3,则角C为多少

答案

根据余弦定理,
AB^2=AC^2+BC^2-2*AC*BC*cosC,
81=4+8√3+12+49+28√3+12-2*(2+2√3)*(7+2√3)*cosC,
cosC=(128-63√3)/37,