Question

已知三角形ABC,A=2C,2b=a+c,求A:B:C的值

Answer 1

A=2C,∴C是锐角,sinC>0,cosC>0.

B=180°-3C,∴sinB=sin3C=3sinC-4(sinC)^3,

2b=a+c,由正弦定理,2sinB=sinA+sinC,

∴2[3sinC-4(sinc)^3]=2sinCcosC+sinC,

两边都除以sinC,得2[3-4(sinC)^2]=2cosC+1,

∴8(cosC)^2-2cosC-3=0,

解得cosC=3/4,

∴sinC=√7/4,sinA=2sinCcosC=3√7/8,

sinB=3sinC-4(sinC)^3=5√7/16,

∴a:b:c=sinA:sinB:sinC=6:5:4.