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Procrastinater needs help withv Law of Cosines

  1. Feb 8, 2006 #1
    Ok I did this thing on law of cosines and here is what i came up with, but i am pretty sure it is wrong so help me change it to solve for cos(C)

    c^2=a^2+b^2-2abcos(C)

    so...

    Cos(C)=____c^2____
    . a^2+b^2-2ab

    is that right?

    if not how do i make it to equal cos(C)?
    (need this by like right now cause i am a procrastinator!!! :P)
     
  2. jcsd
  3. Feb 8, 2006 #2
    wouldnt cos(x) = (a^2 + b^2 - c^2) / 2ab
     
  4. Feb 8, 2006 #3
    probably... i will use that for my hw then, thanks
     
  5. Feb 9, 2006 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No!! You both are apparently thinking of

    c2= (a2+ b2- 2ab) cos(C)
    which it is NOT!! (Notice the parentheses on the right side.)
    The cosine law says c2= a2+ b2- 2ab cos(C). (Notice lack of parentheses!)

    The problem is not the "cosine", it is simple algebra:

    Suppose the problem were c= a+ bx. How would you solve for x?
     
  6. Feb 9, 2006 #5
    I don't see how mathmike's is wrong, but I may just be stupid.

    Isn't

    [tex]cos (C) = \frac{a^2 + b^2 - c^2}{2ab}[/tex]

    the same as

    [tex]cos (C) = \frac{c^2 - a^2 - b^2}{-2ab}[/tex]

    ?

    You could take a -1 out of the numerator and then cancel out the negatives, right?
     
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