Question

在△ABC中,满足sin2A-sin2C+sin2B=sinAsinB.那么C的大小为?

Answer 1

(sinA+sinB)^2-sin^2C=3sinAsinB
sin²A+2sinAsinB+sin²B-3sinAsinB=sin²C
sin²A-sinAsinB+sin²B=sin²(A+B)
sin²A(sin²B+cos²B)-sinAsinB+sin²B(sin²A+cos²A)=(sinAcosB+sinBcosA)²
展开得雀野乱
sin²A(sin²B+cos²B)-sinAsinB+sin²B(sin²A+cos²A)
=(sinAcosB+sinBcosA)²脊镇
=sin²Acos²B+2sinAsinBcosAcosB+sin²Bcos²A
两顷档边减去相同项得
sin²Asin²B-sinAsinB+sin²Bsin²A=2sinAsinBcosAcosB
2sin²Asin²B-sinAsinB-2sinAsinBcosAcosB=0
sinAsinB(2sinAsinB-1-2cosAcosB)=0
2sinAsinB-1-2cosAcosB=0
sinAsinB-cosAcosB=1/2
-cos(A+B)=1/2
cos(A+B)=-1/2
A+B=120°
则角C=60度