(2a+c)BC*BA+cCA*CB=0,
(2a+c)ac*cosB+cba*cosC=0,
(2a+c)cosB+b*cosC=0,(1)
由正弦定理,设a=ksinA,b=ksinB,c=ksinC(k>0)代入(1)得
(2sinA+sinC)cosB+sinBcosC=0,
1).2sinAcosB+sinCcosB+sinBcosC=0,
2sinAcosB+sin(C+B)=0,
2sinAcosB+sin(180度-A)=0,
2sinAcosB+sinA=0,
(2cosB+1)sinA=0,
2cosB+1=0(sinA不等于0),
cosB=-1/2,B=120度
2).求什么?
由余弦定理得b^2=a^2+c^2-2ac*cosB,
4=a^2+c^2-2ac*(-1/2),
4=a^2+c^2+ac,
AB^2+BC^2+2AB*BC=4,