已知三角形ABC的三边a>b>c且a+c=2b,A-C=90,求a:b:c

1个回答

  • ∵ a+c=2b,A-C=90°,

    由正弦定理得

    sinA+sinC=2sinB

    sinC=sin(A-90° )=-cosA

    cosC=cos(A-90°)=sinA

    ∵ A+B+C=180°

    sinA+sinC=2sinB=2sin(A+C)=2sinAcosC+2cosAsinC

    sinA+sinC=2sinAcosC+2cosAsinC

    sinA+sinC-2sinAcosC-2cosAsinC=0

    sinA-cosA -2sinA sinA +2cosAcosA =0

    2 sin^2A+2 sinAcosA –sinA-2 sinAcosA-2cos^2A+cosA=0

    (sinA-cosA)(2sinA+2cosA-1)=0

    ∵A-C=90°,A=C+90°A≠45°

    sinA-cosA=0 不成立.

    ∴2sinA+2cosA-1=0

    sinA+cosA=1/2

    sin^2A+cos^2A+2sinAcosA=1/4

    2sinAcosA=(1/4)-1=-3/4

    2sinAsinC=3/4

    sinAsinC=3/8

    4sin^2B=(sinA+sinC)^2=sinA^2+sinC^2+3/4=sinA^2+cosA^2+3/4=7/4

    sinB=√7/4

    sinA+sinC=2sinB

    sinA+sinC=√7/2

    sinAsinC=3/8

    sinA,sinC是方程x^2-√7/2x+3/8=0 两根

    sinC=(2√7-1)/8

    sinA=√7/2-(2√7-1)/8 =(2√7+1)/8

    sinB=√7/4

    a:b:c= sinA:sinB:sinC=(2√7+1)/8

    :√7/4:(2√7-1)/8=2√7+1:2√7:2√7-1

    a:b:c=2√7+1:2√7:2√7-1

    希望给你有所帮助.

    吉林 汪清