设三角形ABC的内角A,B,C所对的边为a,b,c,且acosB-bcosA=b+c 1求A

1。由正弦定理得sinAcosB-sinBcosA=sinB+sinCsinAcosB-sinBcosA=sinB+sin(A+B)sinAcosB-sinBcosA=sinB+sinAcosB+cosAsinBsinB+2cosAsinB=0sinB(2cosA+1)=0B为三角形内角,sinB>0,要等式成立,只有2cosA+1=0cosA=-1/2A=2π/32。由正弦定理得a/sinA=b/sinB=c/sinCbc/(sinBsinC)=(a/sinA)^2bc=(a/sinA)^2(sinBsinC)S△ABC=(1/2)bcsinA=(1/2)a^2(sinBsinC)/sinA=(1/2)(2√3)^2 (1/4)/(√3/2)=√3