a=2√3,tan2分之(A+B)+tan2分之C=4,2sinBcosC=sinA,求A B b c

a=2√3,tan2分之(A+B)+tan2分之C=4,2sinBcosC=sinA,求A B b c
a=2√3,tan2分之(A+B)+tan2分之C=4,2sinBcosC=sinA,求A B b c

a=2√3,tan2分之(A+B)+tan2分之C=4,2sinBcosC=sinA,求A B b c
tan[（A+B）/2]+tanC/2=4,tan(A+B)=tan(π-C）
tan[（A+B）/2]+tanC/2=tan[π/2-C/2]+tanC/2=cot(C/2)+tan(C/2)
=cos(C/2)/sin(C/2)+sin(C/2)/cos(C/2)=[sin(C/2)^2+cos(C/2)^2]/(sinC/2)(cosC/2)=2/sinC=4,
sinC=1/2,C=π/6或5π/6
2sinBcosC=sinA,sinA= sin[π-(B+C)] =sin(B+C),
所以2sinBcosC= sin(B+C),
2sinBcosC= sinBcosC+ cosBsinC,
sinBcosC- cosBsinC=0,
sin (B-C)=0,
B-C=0.
所以B=C=π/6,5π/6应舍去（不能有二个钝角）,
∠A=2π/3,根据正弦定理,a/sinA=b/sinB,可得b=2,c=2.