已知三角形ABC sinA(sinB+cosB)-sinA=0,sinB+cos2C=0 求A B C的值?

sina(sinb+cosb)-sinc=0
sinasinb+sinacosb-sin(a+b)=0
sinasinb+sinacosb-(sinacosb+cosasinb)=0
sinasinb-cosasinb=0
(sina-cosa)sinb=0
sina=cosa
a=45度
sinb+cos2c=0
sinb+cos2(180-a-b)=0
sinb+cos(360-2a-2b)=0
sinb+cos(2a+2b)=0
sinb+cos(90+2b)=0
sinb-sin2b=0
sinb-2sinbcosb=0
1-2cosb=0
cosb=1/2
b=60度
c=180-60-45=75度