在三角形ABC中,角A=60度,b=1,三角形ABC的面积=根号3,则 a+b+c/sinA+sinB+sinC=?

a =60°, b=1, S=sqr3 =bccosA/2 得c=4
cosA =(b^2 + c^2 -a^2)/(2bc) =1/2
a =sqrt13 (sqrt为根号）
sinA = sqrt(3) /2
sinB =b*sinA/a
sinC =c*sinA/a
(a+b+c)/(sinA+sinB+sinC) =(a+b+c)/[sinA*(1+b/a +c/a)]
=a*(a+b+c)/[sinA*(a+b+c)] =a/sinA =2*sqrt(13)/sqrt3