三角。。。。

AB/AC=4/3

由正弦定理 AB/AC=sinC/sinB=sinC/sin(180-60-C)=4/3

sinC/sin(120-C)=4/3

3sinC=4sin(120-C)=4(sin120cosC-cos120sinC)
=4(sqrt(3)cosC/2+sinC/2)
sinC=2sqrt(3)cosC

因为sinC^2+cosC^2=1

sinC^2+sinC^2/12=1

sinC^2=12/13

sinC=sqrt(12/13)

注：sqrt是根号

不妨设AB=4a，则AC=3a，BC^2=16a^2+9a^2-2*4a*3a*cos60=13*a^2
BC=根号13*a
然后sinC/sinA=AB/BC
可得sinc