(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.
樓主,