在三角形abc中,角ABC所对的边a,b,c已知3acosA=ccosB+bcosc.求cosA的值、若a=1,cosB

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满意的话请及时点下采纳哟.:

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