假设ABC外接圆半径R,
有a=2RsinA b=2RsinB c=2RsinC
c=2S/(b*sinA)=6
a^2=b^2+c^2-2bc*cosA=180-72√3
a=6√(5-2√3)
(sinA+sinB+sinC)/(a+b+c)
=(sinA+sinB+sinC)/[2R(sinA+sinB+sinC)]
=1/（2R）
2R*sinA=a
(sinA+sinB+sinC)/(a+b+c)
=1/2R
=sinA/a
=1/[12√(5-2√3) ]