Question

在△ABC中,设a/c=(√3)-1,cotC/cotB=(2a-c)/c,求A,B,C

Answer 1

cotC/cotB=cosCsinB/(cosBsinC).所以

2a/c=1+cosCsinB/(cosBsinC)=(cosCsinB+cosBsinC)/(cosBsinC)=sin(B+C)/(cosBsinC)=sinA/(sinCcosB)=a/(c*cosB)

所以得到cosB=1/2.B=60°

(√3+1)/2=c/a=sinC/sinA=sin(120°-A)/sinA=(√3/2)cotA+1/2.由此得cotA=1.A=45°

所以C=75°