Question

在△ABC中,∠A=60°,b=1,c=4。则（a+b+c）/(sinA+sinB+sinC)=

Answer 1

根据正弦定理,
2RsinA=a,
2RsinB=b,
2RsinC=c,
因此,
2R(sinA+sinB+sinC)=a+b+c,
(a+b+c)/(sinA+sinB+sinC)=2R,
由余弦定理,
A=60°,b=1,c=4,
a^2=b^2+c^2-2bc*cosA=13,
a=√13,
2R=a/sinA=√13/(√3/2)=2√39/3

Answer 2

由正弦定理：a/sinA=b/sinB=c/sinC
故而：（a+b+c）/(sinA+sinB+sinC)=a/sinA
由余弦定理：a^2=b^2+c^2-2bc*cosA
得：a=√13，
故而：a/sinA=√13/(√3/2)=2√39/3