是3acosC=2ccosA吧?
3acosC=2ccosA→3sinAcosC=2sinCcosA→sinA/cosA=2sinC/3cosC→tanA=2sinC/3cosC=1/3
所以tanC=1/2
tanB=tan(π-A-C)=-tan(A+C)=-[(tanA+tanC)/(1-tanAtanC)]=-1
因为B∈[0,π],所以B=3π/4
是3acosC=2ccosA吧?
3acosC=2ccosA→3sinAcosC=2sinCcosA→sinA/cosA=2sinC/3cosC→tanA=2sinC/3cosC=1/3
所以tanC=1/2
tanB=tan(π-A-C)=-tan(A+C)=-[(tanA+tanC)/(1-tanAtanC)]=-1
因为B∈[0,π],所以B=3π/4