△ABC的三个内角为A,B,C的对边分别为a,b,c，a=2根号3，tan(A+B)/2+tanC/2=4,2sinBcosC=sinA,求A,B及b,c

Question

如题所述

Answer 1

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,C=π/6或5π/6
cos(A/2)=cos[π-(B+C)]/2=sin[(B+C)/2]
cos(A/2)^2=sin[(B+C)/2]^2=[1-cos(B+C)}/2=sinBsinC,1-cosBcosC+sinBsinC=2sinBsinC
cos(B-C)=1,B-C=0(π不符合题意）
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,b=2,c=2