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+sinC
sinAcosB-sinBcosA=sinB+sin(A+B)
sinAcosB-sinBcosA=sinB+sinAcosB+cosAsinB
sinB+2cosAsinB=0
sinB(2cosA+1)=0
B为三角形内角,sinB>0,要等式成立,只有2cosA+1=0
cosA=-1/2
A=2π/3
2.
由正弦定理得a/sinA=b/sinB=c/sinC
bc/(sinBsinC)=(a/sinA)^2
bc=(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

A=120°,面积为根号3，如何输入图片？