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Math Help - trouble with Cosine Rule

  1. #1
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    trouble with Cosine Rule

    Hi all,

    I had this equation

    x = a + b - c

    these figures fit the equation

    a=10, b=5, c=2 giving x=11

    121 = 100 + 25 - 4

    then it occured to me that the original equation looked very like the Cosine Rule

    c = a + b - 2ab Cos(c)

    so that

    2ab Cos(c) = a + b - c

    so combining this with my original equation I would have thought

    x = 2ab Cos(c) which it doesn't 2 x 10 x 5 x Cos(2) = 99.94......

    what did I miss or misunderstand?

    Many thanks,

    Pro
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  2. #2
    Member anonimnystefy's Avatar
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    Re: trouble with Cosine Rule

    hi procyon

    it shouldn't be cos(c) but in there should be an angle opposite to the side c in the triangle with sides a,b and c.
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  3. #3
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    Re: trouble with Cosine Rule

    Quote Originally Posted by anonimnystefy View Post
    hi procyon

    it shouldn't be cos(c) but in there should be an angle opposite to the side c in the triangle with sides a,b and c.
    Wow, thanks for the quick reply

    Not sure that I understand you though

    I'll put the equations together

    a + b - c = x
    a + b - c = 2ab.Cos(c)

    Surely 2ab.Cos(c) must equal x for any given a, b & c (10, 5 & 2 in this case)

    hmmm...
    just had a thought, a triangle can't have sides 10, 5 & 2. Is this perhaps where it falls down?
    Last edited by procyon; August 14th 2011 at 02:18 PM. Reason: extra thought
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  4. #4
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    Re: trouble with Cosine Rule

    Quote Originally Posted by procyon View Post
    Wow, thanks for the quick reply
    Not sure that I understand you though
    I'll put the equations together
    a + b - c = x
    a + b - c = 2ab.Cos(c)
    Surely 2ab.Cos(c) must equal x for any given a, b & c (10, 5 & 2 in this case)
    It is easy: you are missing the point.
    In that formula: c^2=a^2+b^2-2ab\cos(C) you have different variables, c~\&~C and c\ne C.
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  5. #5
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    Re: trouble with Cosine Rule

    Quote Originally Posted by Plato View Post
    It is easy: you are missing the point.
    In that formula: c^2=a^2+b^2-2ab\cos(C) you have different variables, c~\&~C and c\ne C.
    Hi Plato,

    Many thanks for clearing that up for me. That's the trouble when you suddenly remember a formula through the mists of time

    I had thought 2ab.Cos(c) not 2ab.Cos()

    thanks again
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