在三角形ABC中,已知（sinA+sinB+sinc)(sinA+sinB-sinC)=3sinAsinB,a

在三角形ABC中,已知（sinA+sinB+sinc)(sinA+sinB-sinC)=3sinAsinB,a
在三角形ABC中,已知（sinA+sinB+sinc)(sinA+sinB-sinC)=3sinAsinB,a

在三角形ABC中,已知（sinA+sinB+sinc)(sinA+sinB-sinC)=3sinAsinB,a
（sinA+sinB+sinc)(sinA+sinB-sinC)=3sinAsinB
由正弦定理化成边得
a²+2ab+b²-c²=3ab
c²=a²-ab+b²
cosC=(a²+b²-c²)/(2ab)=(ab)/(2ab)=1/2
C=π/3
acosA+bcosB=ccosC
sinAcosA+sinBcosB=(1/2)sinC
(1/2)(sin2A+sin2B)=(1/2)sinC
2sin(A+B)cos(A-B)=sinC
cos(A-B)=1/2
因a

根据正弦定理a/sinA=b/sinB=c/sinC
把(sinA+sinB+sinc)(sinA+sinB-sinC)=3sinAsinB化为
(sinA+sinB)² - sin²C=3sinAsinB
sin²A+sin²B+2sinAsinB-sin²C=3sinAsinB
sin²...

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根据正弦定理a/sinA=b/sinB=c/sinC
把(sinA+sinB+sinc)(sinA+sinB-sinC)=3sinAsinB化为
(sinA+sinB)² - sin²C=3sinAsinB
sin²A+sin²B+2sinAsinB-sin²C=3sinAsinB
sin²A+sin²B-sin²C=sinAsinB
a²+b²-c²=ab
根据余弦定理cosC=(a²+b²-c²)/(2ab)=(ab)/(2ab)=1/2
则∠C=π/3或写成∠C=60°
∵acosA+bcosB=ccosC
∴根据正弦定理a/sinA=b/sinB=c/sinC化为
sinAcosA+sinBcosB=(½)sinC
根据sin(2α)=2sinα·cosα继续化为
(½) sin2A+ （½）sin2B=(½)sinC
sin2A+ sin2B= sinC
sin[ (A+B)+(A -B)] +sin[ (A+B)-(A-B)]=sinC
sin(A+B)cos(A- B)+cos(A+B)sin(A-B) +sin(A+B)cos(A- B)-cos(A+B)sin(A-B) =sinC
2sin(A+B)cos(A- B)=sinC
2sin(π-C)cos(A-B)=sinC
2sinCcos(A-B)=sinC
2cos（A-B)=1
cos（A-B)=½
∵a＜b
∴A＜B
∴A-B= -π/3
又∵A+B=π-C=π- π/3= (2/3) π
解得A=π/6 B=π/2
∴A=π/6 B=π/2 ∠C=π/3

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