(2b-c)cosA-acosC=0
则利用正弦定理得到:
(2sinB-sinC)cosA-sinA*cosC=0
2sinBcosA-(sinCcosA+sinAcosC)=0
2sinBcosA-sin(A+C)=0
2sinBcosA-sinB=0
所以cosA=1/2
所以A=60°
(2b-c)cosA-acosC=0
则利用正弦定理得到:
(2sinB-sinC)cosA-sinA*cosC=0
2sinBcosA-(sinCcosA+sinAcosC)=0
2sinBcosA-sin(A+C)=0
2sinBcosA-sinB=0
所以cosA=1/2
所以A=60°