.b*cosA+(a-2c)*cosB=0 由正弦定理得sinBcosA+sinAcosB-2sinCcosB=0 由和角公式得sin(A+B)-2sinCcosB=0 ; sin(180-C)-2sinCcosB=0 ; sinC-2sinCcosB=0 ; cosB=1/2 ;B=60°
（2）b=2根号3,p=a+b+c,由正弦定理a/sinA=c/sinC=b/sinB=4,得a=4sinA,c=4sinC.故p=4sinA+4sinC+2G3=4(sin(120-C)+sinC)+2G3=2G3cosC+6sinC+2G3=4G3(1/2*cosC+G3/2*sinC)+2G3=4G3sin(C+30)+2G3 所以当C=60时,p最大为6G3,p最小为4G3,即4G3