(tanA-tanB)/(tanA+tanB)=(sinAcosB-cosAsinB)/(sinAcosB+cosAsinB)
=(sinAcosB-cosAsinB)/sin(A+B)
=(sinAcosB-cosAsinB)/sinC…①.
正弦定理(b+c)/c=(sinB+sinC)/sinC…②,
由①,②得
sinAcosB-cosAsinB=sinB+sinC=sinB+sin(A+B)=sinB+sinAcosB+cosAsinB,
∴ sinB(1+2cosA)=0,
∵ sinB≠0,
∴ cosA=-1/2,
A=120°