Multiple Choice Question (MCQ)

If a, b,c are 3 side of △ABC中 such that (b+c):(c+a):(a+b) = 4:5:6, then the maximum internal angle of △ABC is

  1. ×

    90\degree

  2. ×

    150\degree

  3. 120\degree

  4. ×

    75\degree

Collected in the board: Law of Cosines

Steven Zheng posted 1 hour ago

Answer

  1. Let

    b+c=4k
    (1)
    c+a=5k
    (2)
    a+b=6k
    (3)

    Addition of the three equations

    2(a+b+c)=15k, that is,

    a+b+c=7.5k
    (4)

    Substituting (1),(2), (3) to (4) gives

    a=3.5k,b=2.5k,c=1.5k

    Therefore, the maximum internal angle is A,

    Using the law of cosine:

    \cos A = \dfrac{b^2+c^2-a^2}{2bc}

    =\dfrac{6.25k^2+2.25k^2-12.25k^2}{7.5k^2}

    =-1

    \because A∈(0, 180°)

    \therefore A=120°

    Therefore, the maximum internal angle of △ABC is 120°

    Option C is the right choice.

Steven Zheng posted 1 hour ago

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