在△ABC中,若A=60°,a=√3,则a+b+c/sinA+sinB+sinC等于

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在△ABC中,若A=60°,a=√3,则a+b+c/sinA+sinB+sinC等于

由正弦定理
a/sinA=b/sinB=c/sinC=2Ra
a/sinA=√3/(√3/2)=2
∴a=2sinA,b=2sinB,c=2sinC
(a+b+c)/(sinA+sinB+sinC)
=2(sinA+sinB+sinC)/(sinA+sinB+sinC)
=2

根据正弦定理
(a+b+c)/(sinA+sinB+sinC)
=2R(sinA+sinB+sinC)/(sinA+sinB+sinC)
=2R
=a/sinA
=√3/(√3/2)
=2

a/SinA=b/SinB=c/SinC=a+b+c/SinA+SinB+SinC
解得2

2