三角形ABC中,a,b,c分别为角A,B,C的对边,已知b-c=2a*cos(π/3 +C),求角A

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三角形ABC中,a,b,c分别为角A,B,C的对边,已知b-c=2a*cos(π/3 +C),求角A
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三角形ABC中,a,b,c分别为角A,B,C的对边,已知b-c=2a*cos(π/3 +C),求角A
三角形ABC中,a,b,c分别为角A,B,C的对边,已知b-c=2a*cos(π/3 +C),求角A

三角形ABC中,a,b,c分别为角A,B,C的对边,已知b-c=2a*cos(π/3 +C),求角A
已知b-c=2a*cos(π/3 +C),
结合正弦定理 a/sinA=b/sinB=c/sinC
得:sinB-sinC=2sinA·cos(π/3 +C),
→sin(A+C)-sinC=2sinA·[cos(π/3)cosC-sin(π/3)sinC],
→sinAcosC+cosAsinC-sinC=2sinA(1/2·cosC-√3/2·sinC),
→sinAcosC+cosAsinC-sinC=sinAcosC-√3sinAsinC,
→cosAsinC-sinC=-√3sinAsinC,
→cosA-1=-√3sinA,
→√3sinA+cosA=1,[根据asinα+bcosα=√(a²+b²)sin(α+φ),(tanφ=b/a)]
→2sin(A+30°)=1,
→sin(A+30°)=1/2,
得:A=120°.