若在三角形ABC中A=60°,b=1,SABC=√3,则(a+b+c)/(sinA+sinB+sinC)=

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若在三角形ABC中A=60°,b=1,SABC=√3,则(a+b+c)/(sinA+sinB+sinC)=

若在三角形ABC中A=60°,b=1,SABC=√3,则(a+b+c)/(sinA+sinB+sinC)=
若在三角形ABC中A=60°,b=1,SABC=√3,则(a+b+c)/(sinA+sinB+sinC)=

若在三角形ABC中A=60°,b=1,SABC=√3,则(a+b+c)/(sinA+sinB+sinC)=
计算得c=4
b=根号13
(a+b+c)/(sinA+sinB+sinC)=b/sinB=2根号39/3

SABC=bc×sinA/2=√3c/4=√3
c=4
a^2=b^2+c^2-2bc×cosA=13
a=√13
(a+b+c)/(sinA+sinB+sinC)
= (2RsinA+2RsinB+2RsinC)/(sinA+sinB+sinC)
= 2R
= a/sinA
= √13/(√3/2)
= 2√39/3

bcsinA=2S
c=4
a^2=b^2+c^2-2bccosA=13
a=√13
由合比定理和正弦定理
a/sinA=b/sinB=c/sinC=(a+b+c)/(sinA+sinB+sinC)
所以(a+b+c)/(sinA+sinB+sinC)
=a/sinA
=√13/(√3/2)
=2√39/3