Question

半径为R的圆外接于△ABC，且2R(sin^2A-sin^2C)=(根号3*a-b)sinB,求角C

Answer 1

2R(sinA+sinC)(sinA-sinC)=(√3a-b)sinB
有正弦定理
2RsinA=a,2RsinC=c
所以(a+c)(sinA-sinC)=(√3a-b)sinB
sinA=a/2R,sinB=b/2R,sinC=c/2R
所以(a+c)(a-c)=(√3a-b)b
a^2-c^2=√3ab-b^2
a^2+b^2-c^2=√3ab

cosC=(a^2+b^2-c^2)/2ab=√3/2
C=30度

Answer 2

用弦化边来做∵2R(sin^2A-sin^2c)=(根号3*a-b)sinB asinA-csinC=(a/sinA=c/sinC=2R a/sin(7π/12)=2R a=(根号2 根号6）R S△abc

追问

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