利用正弦定理
cosB/cosC=-sinB/(2sinA+sinC)
即 2cosBsinA+cosBsinC=-sinBcosC
即 2cosBsinA=-(cosBsinC+sinBcosC)=-sin(B+C)=-sinA
即 cosB=-1/2
∴ B=2π/3
sinA+sinC
=2sin[(A+C)/2]*cos[(A-C)/2]
=cos[(A-C)/2]
∵ 0
利用正弦定理
cosB/cosC=-sinB/(2sinA+sinC)
即 2cosBsinA+cosBsinC=-sinBcosC
即 2cosBsinA=-(cosBsinC+sinBcosC)=-sin(B+C)=-sinA
即 cosB=-1/2
∴ B=2π/3
sinA+sinC
=2sin[(A+C)/2]*cos[(A-C)/2]
=cos[(A-C)/2]
∵ 0