应该是这样的吧:
(根号3sinB-cosB)(根号3sinC-cosC)=4cosBcosC.
3sinBsinC-根号3sinBcosC-根号3cosBsinC+cosBcosC=4cosBcosC
3cos(B+C)=-根号3sin(B+C)
即有tan(B+C)=-根号3
即有-tanA=-根号3,tanA=根号3
故有角A=60度.
应该是这样的吧:
(根号3sinB-cosB)(根号3sinC-cosC)=4cosBcosC.
3sinBsinC-根号3sinBcosC-根号3cosBsinC+cosBcosC=4cosBcosC
3cos(B+C)=-根号3sin(B+C)
即有tan(B+C)=-根号3
即有-tanA=-根号3,tanA=根号3
故有角A=60度.