Question

已知三角形abc分别是三角形三个内角ABC的对边，acosC+根号3asinC-b-c=0，求A

Answer 1

acosC+√3asinB-b-c=0
利用正弦定理 a/sinA=b/sinB=c/sinC
sinAcosC+√3sinAsinC-sinB-sinC=0
∵ sinB=sin(A+C),
sinAcosC+√3sinAsinC-sin(A+C)-sinC=0
sinAcosC+√3sinAsinC-sinAcosC-cosAsinC-sinC=0
√3sinAsinC=sinC+cosAsinC
√3sinA=1+cosA
2√3sin(A/2)cos(A/2)=2cos²(A/2)
√3tan(A/2)=1
tan(A/2)=√3/3
∵0<A/2<π/2
∴ A/2=π/6
∴ A=π/3

Answer 2

一问：sinAcosC＋√3sinAsinC－sinB－sinC=0
sinAcosC＋√3sinAsinC－sin(A＋C)－sinC=0
sinAcosC＋√3sinAsinC－sinAcosC－cosAsinC－sinC=0
√3sinAsinC－cosAsinC－sinC=0
√3sinA=1＋cosA
因tan(A/2)=(sinA)/(1＋cosA)=√3/3
得：A/2=30°，即A=60°

Answer 3

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