在△ABC中,角A=60°,b=1,其面积为根号3,则(a+b+c)\(sinA+sinB+sinC)=?



两种办法:
在△ABC中 A=60° b=1 S△ABC=√3,求:(a+b+c)/(sinA+sinB+sinC)=?

(a+b+c)/(sinA+sinB+sinC)=2R
SΔ=(1/2)bcsinA--->c=2SΔ/(bsinA)=4
--->a2=b2+c2-2bccosA=13
--->(a+b+c)/(sinA+sinB+sinC) = 2R = a/sinA = √13/(√3/2) =2√39/3

因为S△ABC=bcsinA/2=[1*c*(√3/2)]/2=√3
所以,c=4
根据余弦定理有:a^=b^+c^-2bccosA=1+16-2*1*4*(1/2)=13
所以,a=√13
根据正弦定理a/sinA=b/sinB=c/sinC,则:
(a+b+c)/(sinA+sinB+sinC)=a/sinA=(√13)/(√3/2)=2√39/3
S=1/2*b*csinA=√3
c=√3/(bsinA)=√3/(1*√3/2)=2
cosA=(b^2+c^2-a^2)/2bc
=(1+4-a^2)/4=1/2
5-a^2=2
a^2=3
a=√3

在三角形ABC中有
a/sinA=b/sinB=c/sinC=√3/√3/2=2
所以
(a+b+c)\(sinA+sinB+sinC)=a/sinA=2
2√39/3


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