在三角形ABC中,内角A,B,C的对边a,b,c.已知b^2=ac,且cosB=3/4.求.(1)cotA+cotC.(2)设向量BA*BC=3/2.求a+c的值..注用正弦和余弦做..高二程度...谢谢
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![在三角形ABC中,内角A,B,C的对边a,b,c.已知b^2=ac,且cosB=3/4.求.(1)cotA+cotC.(2)设向量BA*BC=3/2.求a+c的值..注用正弦和余弦做..高二程度...谢谢](/uploads/image/z/1095880-40-0.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E5%86%85%E8%A7%92A%2CB%2CC%E7%9A%84%E5%AF%B9%E8%BE%B9a%2Cb%2Cc.%E5%B7%B2%E7%9F%A5b%5E2%3Dac%2C%E4%B8%94cosB%3D3%2F4.%E6%B1%82.%EF%BC%881%EF%BC%89cotA%2BcotC.%282%29%E8%AE%BE%E5%90%91%E9%87%8FBA%2ABC%3D3%2F2.%E6%B1%82a%2Bc%E7%9A%84%E5%80%BC..%E6%B3%A8%E7%94%A8%E6%AD%A3%E5%BC%A6%E5%92%8C%E4%BD%99%E5%BC%A6%E5%81%9A..%E9%AB%98%E4%BA%8C%E7%A8%8B%E5%BA%A6...%E8%B0%A2%E8%B0%A2)
在三角形ABC中,内角A,B,C的对边a,b,c.已知b^2=ac,且cosB=3/4.求.(1)cotA+cotC.(2)设向量BA*BC=3/2.求a+c的值..注用正弦和余弦做..高二程度...谢谢
在三角形ABC中,内角A,B,C的对边a,b,c.已知b^2=ac,且cosB=3/4.求.
(1)cotA+cotC.
(2)设向量BA*BC=3/2.求a+c的值..
注用正弦和余弦做..高二程度...
谢谢
在三角形ABC中,内角A,B,C的对边a,b,c.已知b^2=ac,且cosB=3/4.求.(1)cotA+cotC.(2)设向量BA*BC=3/2.求a+c的值..注用正弦和余弦做..高二程度...谢谢
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,
则a+c=3