已知三角形abc的内角abc的对边为abc,且b+c/a=cosC+∫3sin C求A的大小
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(b+c)/a=cosC+√3sin C
正弦定理代入:
(sinB+sinC)/sinA=cosC+√3sin C
sin(A+C)+sinC=sinA(cosC+√3sin C)
sinAcosC+cosAsinC+sinC=sinAcosC+√3sinAsinC
cosAsinC+sinC=√3sinAsinC
sinC≠0,约去:
cosA+1=√3sinA
√3sinA-cosA=1
√3/2.sinA-1/2.cosA=1/2
sinAcosΠ/6-cosAsinΠ/6=1/2
sin(A-Π/6)=1/2
A-Π/6=Π/6,或A-Π/6=5Π/6
A=Π/3,或A=Π
后式不成立,所以A=Π/3
正弦定理代入:
(sinB+sinC)/sinA=cosC+√3sin C
sin(A+C)+sinC=sinA(cosC+√3sin C)
sinAcosC+cosAsinC+sinC=sinAcosC+√3sinAsinC
cosAsinC+sinC=√3sinAsinC
sinC≠0,约去:
cosA+1=√3sinA
√3sinA-cosA=1
√3/2.sinA-1/2.cosA=1/2
sinAcosΠ/6-cosAsinΠ/6=1/2
sin(A-Π/6)=1/2
A-Π/6=Π/6,或A-Π/6=5Π/6
A=Π/3,或A=Π
后式不成立,所以A=Π/3
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