Question

在△ABC中,A,B,C的对边分别为a,b,c,已知3acosA=ccosA+bcosC。若a=1,cosB+cosC=2根

Answer 1

1. 根据正弦定理，a/sinA=b/sinB=c/sinC=2R
将已知条件两边除以2R（外接圆半径）
=》3sinAcosA=sinCcosB+cosCsinB=sin(B+C)=sin(180-A)=sinA
=》cosA=1/3
2. cosA=1/3 =》sinA=2根号2/3
cosB=cos(180-A-C)=-cosAcosC+sinAsinC=-cosC/3+2根号2/3sinC
cosB+cosC=2cosC/3+2根号2/3sinC=2根号3/3(1/根号3cosC+根号2/根号3sinC)=2根号3/3 =》
=》1/根号3cosC+根号2/根号3sinC=1
=》sin(alpha+C)=1 其中cos(alpha)=根号2/根号3
=》alpha+C=PI/2
=》sinC=cos(alpha)=根号2/根号3
根据正弦定理，c/sinC=a/sinA=>c=a/sinA*sinC=1/[2根号2/3]*[根号2/根号3]
=根号3/2 呃

Answer 2

题目略看不懂……