Question

已知三角形ABC,A=2C,2b=a+c,求A:B:C的值

a,b,c为边，A,B,C为三角，求的是a:b:c的值 之前写错了

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.