Question

在△ABC中,2sin*2（A/2）=√3sinA,sin(B-C)=2cosBsinC,则AC/AB=?

Answer 1

2sin²(A/2)=1-cosA=√3sinA

√3sinA+cosA=1

2(√3/2sinA+½cosA)=1

√3/2sinA+½cosA=½

sinA·cos(π/6)+sin(π/6)·cosA=½

sin(A+π/6)=½

∴A+π/6=5π/6

A=2π/3

sinBcosC=3cosBsinC

正弦定理和余弦定理得a^2=b^2+c^2+bc,a^2=2(b^2-c^2)

所以b^2-3c^2-bc=0,

b/c==(1+√13)/2