tanB/tanC=cotC/cotB=(2a-c)/c=(2sinA-sinC)/sinC
去分母得tanBcosC=2sinA-sinC
整理得sinBcosC=2sinAcosB-sinCcosB
即sinBcosC+sinCcosB=2sinAcosB
sin(B+C)=2sinAcosB
sinA=2sinAcosB
sinA(cosB-1/2)=0
在三角形中,正弦值为正
所以cosB=1/2
B=π/3
则C=2π/3-A
a/c=sinA/sinC=sinA/sin(2π/3-A)=√3-1
即sin(2π/3-A)/sinA=(√3+1)/2
展开整理得√3/2+cotA/2=(√3+1)/2
即cotA=1
解得A=π/4
则C=5π/12