正弦定理题目一道：A=60,a=12,则a+b+c/sinA+sinB+sinC=?

a/sinA=b/sinB=c/sinC=2R
2R=a/sinA=12/sin60°=24/√3

(a+b+c)/(sinA+sinB+sinC)
=(2RsinA+2RsinB+2RsinC)/(sinA+sinB+sinC)
=2R
=24/√3
=8√3

正弦定理：设a/sinA=b/sinB=c/sin/C=k，
a=k*sinA,b=k*sinB,c=k*sinC,
代入a+b+c/sinA+sinB+sinC=k=12/sin60

既然是通用解
不妨设A=B=C=60度，a=b=c=12
答案就是(12+12+12)/(sqrt(3)/2+sqrt(3)/2+sqrt(3)/2)=12/(sqrt(3)/2)
=24/sqrt(3)
=8*sqrt(3)

8根3