可以不?
2b cosC=2a
==> cosC=a/b=sinA/sinB
==> -cos(A+B)=sinA/sinB
==> -(cosAcosB-sinAsinB)=sinA/sinB
==> -cosAcosBsinB+sinAsinB²=sinA
==> -cosAsinBcosB=sinA(1-sin²B)
==> cosAsinBcosB=sinAcos²B
==> sinAcos²B+cosAsinBcosB=0
==> cosB(sinAcosB+cosAsinB)=0
==> cosBsin(A+B)=0
==> cosB=0
==> B=π/2
如图
由正弦定理得:
2sinBcosC=2sinA=2sin(B+C)=2sinBcosC+2cosBsinC
cosBsinC=0
C为三角形内角,sinC恒>0
因此只有cosB=0
B为三角形内角,B=π/2
初中学的早忘了