已知a,b,c分别为三角形ABC三个内角A、B、C的三边,acosC+根号3asinC-b-c=0.若a=2,三角形ABC的面积为根号3,求b

一问：sinAcosC＋√3sinAsinC－sinB－sinC=0sinAcosC＋√3sinAsinC－sin(A＋C)－sinC=0sinAcosC＋√3sinAsinC－sinAcosC－cosAsinC－sinC=0√3sinAsinC－cosAsinC－sinC=0√3sinA=1＋cosA因tan(A/2)=(sinA)/(1＋cosA...