已知在三角形ABC中,向量m=(-1,根号3),向量n(cosA,sinA),且向量m×向量n=1.（1）求角A；

m·n
=-cosA+√3sinA=1
-(1/2)cosA+(√3/2)sinA=1/2
cos(2π/3)cosA+sin(2π/3)sinA=1/2
cos(2π/3-A)=cos(π/3)
2π/3-A=π/3
A=π/3
tanB
=tan（π/4 +B -π/4 )
=(tan（π/4 +B)-1)/(tan（π/4 +B)+1)
=(-4)/(-2)
=2
tanA=√3
tanC
=-tan(A+B)
=-(tanA+tanB)/(1-tanAtanB)
=-(2+√3)/(1-2√3)
=(2+√3)(1+2√3)/11
=(8+5√3)/11