∴cos[π-(b+c)]+cosbcosc-√3c osbsinc = 0
-cos(B+C)+cosBcosC-√3c osbsinc = 0
cosBcosC - √3cosBsinC=cos(B+C)
cosBcosC-√3c OSB sinc = cosBcosC-sinb sinc
∴√3cosBsinC=sinBsinC
∫sinC≠0
∴√3cosB=sinB, then tanB=√3.
∴B=π/3
According to sine theorem: 2r = b/sinb =1/sin (π/3) = 2/√ 3.
Then a+c=2RsinA+2RsinC=2R(sinA+sinC).
=2R[sinA + sin(π - A - π/3)]
=2R[sinA + sin(2π/3 - A)]
= 2R[sinA+sin(2π/3)cosA-cos(2π/3)sinA]
= 2R[ Sina+(√3/2)cosA+(1/2) Sina]
=2R[(3/2)sinA + (√3/2)cosA]
=(2/√3)? (√3)[(√3/2) Sina+(1/2)cosA]
=2sin(A + π/6)
∫△ABC is an acute triangle
When b