问题补充:
在三角形ABC中,角A=30°,b=12,三角形的面积是18,则(sinA+sinB+sinC)除以(a+b+c)等于多少?
答案:
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) ]