已知△ABC的三内角A,B,C与所对的边a,b,c满足2b-c/a=cosC/cosA 1 、求角A的大小 2、如果用p

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  • 1、(2b-c)cosA-acosC=0

    由正弦定理b/sinB=a/sinA=c/sinC=2R

    b=2RsinB

    a=2RsinA

    c=2RsinC

    (2b-c)cosA-acosC=0

    2R(2sinB-sinC)cosA-2RsinAcosC=0

    (2sinB-sinC)cosA-sinAcosC=0

    2sinBcosA-sinCcosA-sinAcosC=0

    2sinBcosA-(sinCcosA+sinAcosC)=0

    2sinBcosA-sin(A+C)=0,

    2sinBcosA-sin(180-B)=0,

    所以:2sinBcosA-sinB=0,

    因为:A、B∈(0,π),sinB≠0

    所以:cosA=1/2,

    所以:A=60度

    2、sin^2 B+sin^2 C=p^2 sin^2 A

    p^2=(sin^2 B+sin^2 C)/0.75

    =(sin^2(120-C)+sin^2 C)/0.75

    =(5-cos2C-根号3的sin2C)/3 (C在0到120之间)

    =【5-2sin(2C+30°)】/3

    ∴P^2∈【1,7/3)

    ∴P∈【1,根号21 /3)