Question

在三角形ABC中,a,b,c分别是角A,B,C对边的长,且满足cosB/cosC=-b/（2a+c）,求角B的值.

Answer 1

由正弦定理,设a=ksinA,b=ksinB,c=ksinC(k>0),则b/(2a+c)=sinB/(2sinA+sinC),

cosB/cosC=-sinB/(2sinA+sinC),

2cosBsinA+cosBsinC+sinBcosC=0,

2cosBsinA+sin(B+C)=0,

2cosBsinA+sinA=0,

sinA(2cosB+1)=0,sinA>0,

2cosB+1=0,

cosB=-1/2,B=2π/3