三角形ABC中，sin2A = sin2B+sin2C+sinB*sinC ,则角A等于

sin2a=sin2b+sin2c+sinb*sinc
-sin(2b+2c)=2sin(b+c)cos(b-c)+sinb*sinc
-2sin(b+c)cos(b+c)=2sin(b+c)cos(b-c)+sinb*sinc
2sin(b+c)(cos(b-c)+cos(b+c))=-sinbsinc
4sin(b+c)*cosb*cosc=-sinbsinc
sin(b+c)=-1/4*tanb*tanc

三角形ABC中，sin2A=sin2B+sinB*sinC+sin2C=-2sin(A)cos(B+C)=2sin(B+C)cos(B-C)+sinB*sinC，4sinAcosBcosC+sinB*sinC=0，cosBcosC＜0，B或C为钝角，三角形ABC是钝角三角形 不好意思啊，只能想到这了