已知在三角形ABC中,角ABC的对边分别为abc.且4cos^2*C/2-cos2C=7/2.a+b=5.c=根号7 求

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  • ∵ab≤(a+b)^2/4

    =25/4

    ab最大值 25/4<9

    CosC

    =(a^2+b^2-c^2)/2ab

    =[(a+b)^2-c^2-2ab]/2ab

    =(25-7-2ab)/2ab

    =(18-2ab)2ab

    =9/ab-1>0 (9/25/4=36/25>1)

    ∴0°<C<90°

    ∵4cos^2C/2-cos2C

    =2(2cos^2C/2-1)+2-cos2C

    =2cosC+2-cos2C

    =2cosC+2-2cos^2C+1

    =-2cos^2C+2cosC+3

    =7/2

    ∴cos^2C-cosC+1/4=0

    (cosC-1/2) ^2=0

    CosC=±1/2

    C=60°

    C=120°舍掉(∵0°<C<90°上边已确定)

    SinC=sin60°=√3/2

    ∵.a+b=5

    ∴(a+b) ^2

    =a^2+b^2+2ab

    =25

    A^2+b^2=25-2ab

    c ^2=√7 ^2=7

    C^2=a^2+b^2-2abcosC

    ab=(a^2+b^2-c^2)/2cosC

    =(25-2ab-7)/2*1/2

    =18-2ab

    3ab=18

    ab=6

    S=1/2absinC

    =1/2*6*√3/2

    =3√3/2

    C=60° S=3√3/2

    吉林 汪清LLX