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Trigonometry Obscurity!

  1. Sep 5, 2004 #1
    Hey everyone,
    I was looking over some old pre-calculus exams and I found this rather obscure looking question.. It's about trigonometry.

    You're given a triangle ABC, and the legs are a (BC),b (AC), c (AB).
    You're given the lenghts of a=5, b-8, and the angle C between them is 140.

    The question is, what's the length of the leg "c" and what are the other 2 degrees? :yuck:

    Is that even possible?? :bugeye:
    Thx,
    --Xeno
     
  2. jcsd
  3. Sep 5, 2004 #2
    Are you familiar with the sine and cosine rules?
     
  4. Sep 5, 2004 #3
    Of course I am!! :grumpy:
    But they can only be applied to a triangle with a right angle, and I tried to divide this triangle to 2 with right angles but it just didnt seem to work out.. Any ideas?
     
  5. Sep 5, 2004 #4

    Hurkyl

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    devious is referring to the law of sines and the law of cosines...

    Given any triangle with sides A, B, and C with angles [itex]\alpha, \beta, \gamma[/itex] (with A opposite [itex]\alpha[/itex], etc):


    The law of sines:
    [tex]
    \frac{\sin \alpha}{A} = \frac{\sin \beta}{B} = \frac{\sin \gamma}{C}
    [/tex]

    The law of cosines:
    [tex]
    C^2 = A^2 + B^2 - 2AB\cos \gamma
    [/tex]

    (and, of course, similar formulae for the other choices of angle)
     
    Last edited: Sep 5, 2004
  6. Sep 5, 2004 #5

    Alkatran

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    Notice that since cos(90) = 0, the law of cosine turns into pythagoras' (sp?) theorem with right triangles. That's the rule I assume you were talking about.
     
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