m=(2a-c,cosC),n=(cosB,-b)
∵m垂直n
∴(2a-c)cosB-bcosC=0
根据正弦定理:
(2sinA-sinC)cosB-sinBcosC=0
2sinAcosB-(sinCcosB+sinBcosC)=0
2sinAcosB-sin(C+B)=0
2sinAcosB-sinA=0
∵sinA≠0∴cosB=1/2
∵0
m=(2a-c,cosC),n=(cosB,-b)
∵m垂直n
∴(2a-c)cosB-bcosC=0
根据正弦定理:
(2sinA-sinC)cosB-sinBcosC=0
2sinAcosB-(sinCcosB+sinBcosC)=0
2sinAcosB-sin(C+B)=0
2sinAcosB-sinA=0
∵sinA≠0∴cosB=1/2
∵0