正弦定理问题在△ABC中,若A=π/3,b=12,c=π/6,则a+b+c/sinA+sinB+sinC=?

Question

正弦定理问题
在△ABC中,若A=π/3,b=12,c=π/6,则a+b+c/sinA+sinB+sinC=?

Answer 1

A=π/3,
B=π/6
C=π/2
a/sinA=c/sinC=b/sinB=12/(1/2)=24
a=24 * sinA=24 * √3/2=12√3
c=24 * sinC=24
(a+b+c)=(12+24+12√3) = 12(3+√3)
(sinA+sinB+sinC) = (1/2 + √3/2 + 1) = (3+√3)/2
(a+b+c)/(sinA+sinB+sinC) = 12(3+√3) /[ (3+√3)/2] = 24