Question

求解题：若A=60°，b=1，c=4，则（a+b+c）/（sinA+sinB+sinC）=？

请告诉我过程

Answer 1

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

三角形面积S=bc*sinA/2=根号3*c/4=根号3
所以c=4

a^2=b^2+c^2-2bc*cosA=13
a=根号13

所以a/sinA=根号13/(根号3/2)=2根号39/3=2R

所以(a+b+c)/(sinA+sinB+sinC)=2根号39/3

Answer 2

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 3

a=√(1+16-8*cos60)=√13

（a+b+c）/（sinA+sinB+sinC）=a/siaA=√13/（√3/2）=(2√39)/3