已知abc分别为三角形内角ABC的对边bcosC+根号3bsinC-a-c=0.1求证ABC成等差数列2若b=根号3,求2a+c的最大值

Question

已知abc分别为三角形内角ABC的对边bcosC+根号3bsinC-a-c=0.1求证ABC成等差数列 2若b=根号3,求2a+c的最大值

Answer 1

1. a/sinA=b/sinB=c/sinC=k
a=ksinA, b=ksinB, c=ksinC
ksinBcosC+√3ksinBsinC-ksinA-ksinC=0
2sinB(sinCsinπ/3+cosCcosπ/3)=sinA+sinC
2sinBcos(C-π/3)=sinA+sinC
sin(B+C-π/3)+sin(B-C+π/3)=sinA+sinC
sin(π-A-π/3)+sin(B-C+π/3)=sinA+sinC
sin(2π/3-A)-sinA+sin(B-C+π/3)=sinC
2cosπ/3sin(π/3-A)+sin(B-C+π/3)=sinC
sin(π/3-A)+sin(B-C+π/3)=sinC
2sin(2π/3+B-A-C)/2cos(-A-B+C)/2=sinC
2sin(B-π/6)cos(C-π/2)=sinC
2sin(B-π/6)sinC=sinC
sinC0, sin(B-π/6)=1/2, B-π/6=π/6, B=π/3
2.b=√3, a/sinA=c/sinC=√3/sinπ/3
a=2sinA, c=2sinC
2a+c=2(2sinA+sinC)