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Law of cosine

  1. Jan 28, 2007 #1

    disregardthat

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    1. The problem statement, all variables and given/known data

    In triangle ABC where you only know the sides: a, b and c I must find angle B.
    a=8
    b=6
    c=12

    2. Relevant equations

    Law of cosines: c^2 = a^2 + b^2 -2ab*cos(C)
    When angle C is at the opposite of side c, (same for a and b)

    3. The attempt at a solution

    12^2 = 6^2 + 8^2 -2*6*8*cos(C)

    144 = 36+64-96cos(C)
    44=-96cos(C)
    -(44/96) = cos(C)
    - (11/24) = cos(C)

    (I know the answer is supposed to be 117.4)
    And cos(117.4)=-(11/24)

    The problem is:
    How do I find the angle? The explanation says: "Use calculator (degree mode)"
    How do I do that?

    I have an TI-84 Plus
     
  2. jcsd
  3. Jan 28, 2007 #2

    Dick

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    Look for 'arccos' or 'cos^(-1)'.
     
  4. Jan 28, 2007 #3
    Are you sure about that? Check your formula again. The largest angle must be opposite the largest side, and side b is not the largest side
     
  5. Jan 29, 2007 #4

    disregardthat

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    Thanks, I found it, and I filled in the number, and I got the answer.
     
  6. Jan 29, 2007 #5

    disregardthat

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    Thanks, I found it, and I filled in the number, and I got the answer.
     
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