在三角形ABC中,角A,B,C,的对边分别为a,b,c,已知向量M=(a+c,b-c),N=(a-c,b),且M向

(1)∵向量M⊥向量N∴向量M*向量N=0∴(a+c)(a-c)+(b-a)*b=0∴a^2-c^2+b^2-ab=0∴a^2+b^2-c^2=ab∴cosC=(a^2+b^2-c^2)/(2ab)=ab/(2ab)=1/2∴C=π/3。(2)∵△ABC∴A+B+C=π。∵C=π/3∴A+B=2π/3∴B=2π/3-A∴sinB=sin(2π/3-A)=sin(2π/3)cosA-cos(2π/3)sinA=[(√3)/2]cosA+(1/2)sinA∵sinA+sinB=(√6)/2∴sinA+[(√3)/2]cosA+(1/2)sinA=(√6)/2∴(3/2)sinA+[(√3)/2]cosA=(√6)/2∴(√3)sin(A+π/6)=(√6)/2∴sin(A+π/6)=(√2)/2∴A+π/6=π/4或A+π/6=3π/4∴A=π/12或A=7π/12。楼主,