三角形ABC中,内角A,B,C的对边分别是a,b,c,已知a,b,c成等比数列,且cosB=3/4.

1个回答

  • 1.a,b,c成等比数列,所以a*c=b^2

    根据正弦定理,a/sinA=b/sinB=c/sinC

    所以sinA=a/b*sinB,sinC=c/b*sinC

    cotA+cotC=cosA/sinA+cosC/sinC

    =(cosA*sinC+sinA*cosC)/sinA*sinC

    =sin(A+C)/[(a/b*sinB)*(c/b*sinC)]

    =sinB/[(a/b*sinB)*(c/b*sinC)]

    =1/sinB

    =4/(根号7)

    2.a,b,c成等比数列,设公比为q,

    则b=a*q,c=a*q^2

    cosB=(a^2+c^2-b^2)/2*a*c

    =(a^2+a^2*q^4-a^2*q^2)/2*a*a*q^2

    =(1+q^4-q^2)/2*q^2

    =3/4

    化简为:2*q^4-5*q^2+2=0

    解得:q=1/(根号2),或者q=根号2

    向量BA点乘向量BC=a*c*cosB

    =a*a*q^2*cosB

    =3/2

    将cosB和q代入,解得:a=2,此时q=1/(根号2),c=1,a+c=3

    或者a=1,此时q=根号2,c=2,a+c=3