在△ABC中,向量m=(cosC/2,sinC/2),向量n=(cosC/2,-sinC/2),且向量m与向量n的夹角为π/3,求∠C,已知c=7/2

m*n=cos^2(C/2)-sin^2(C/2)=cosC=|m|*|n|*cos(π/3)
|m|=|n|=cos^2(C/2)+sin^(C/2)=1
cosC=cos(π/3)=1/2
∵C∈(0,π)
∴C=π/3三角型面积S=(3倍根3)/2，求a+b.设AB=c=7/2,AC=b,BC=a (absinC)/2=3√3/2 absin(π/3)=3√3ab=6(1) c^2=a^2+b^2-2abcosC=a^2+b^2-ab (a+b)^2=c^2+3ab=(49/4)+18=121/4 a+b=11/2周长为7/2+11/2=9