Question

已知a b c为三角形abc的三内角,向量m=(-2,-1)向量n=(cos(A+π/6),sin(A-π/3))且向量

Answer 1

（1）由向量m⊥向量n可得：

-2cos(A+π/6)-sin(A-π/3)=0,

∵cos(A+π/6)=cos[π/2+(A-π/3)]=-sin(A-π/3)

∴sin(A-π/3)=0,

∵A∈(0,π),∴A-π/3=0,即A=π/3.

（2）(sin²C-cos²C)/(1-sin2C)=-2

即(sin²C-cos²C)/(sinc-cosc)²=-2,

∴(sinC+cosC)/(sinC-cosC)=-2,

∴(tanC+1)/(tanC-1)=-2,

解得：tanC=1/3.

tanB=tan[π-(A+C)]=-tan(A+C)=-(tanA+tanc)/(1-tanAtanC)=-(6+5√3)/3