在三角形ABC中,角A,B,C的对边分别是a,b,c.已知3acosA=ccosB+bcosC求cosA的值.

2个回答

  • 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