A+B+C=π
A=60°
则B+C=阿0°
B=120°-C
正弦定理:
AB/AC=sinC/sinB
即sinC/sinB=4/3
sinB=sin(120°-C)=sin120°cosC-cos120°sinC
代入化简
2倍根号3cosC=sinC
显然因为C0.所以cosC>0
两边平方
12(1-(sinC)^2)=(sinC)^2
解得sinC=根号(12/13)
A+B+C=π
A=60°
则B+C=阿0°
B=120°-C
正弦定理:
AB/AC=sinC/sinB
即sinC/sinB=4/3
sinB=sin(120°-C)=sin120°cosC-cos120°sinC
代入化简
2倍根号3cosC=sinC
显然因为C0.所以cosC>0
两边平方
12(1-(sinC)^2)=(sinC)^2
解得sinC=根号(12/13)