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Homework Help: Napier's analogy

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data

    In a triangle ABC, a=6 b=3 cos(A-B)=4/5. Find the angle C.

    2. Relevant equations



    3. The attempt at a solution

    here we need to find tan(A-B/2)
    I used the formula tan2x=2tanx/(1/tan^2x)
    and got 2 values of tan(A-B/2) as -3 and 1/3
    On what explanation do I reject one of them?
    -90<A-B/2 <90
    so tanA-B/2 can be both positive and negative.
    Please explain in detail
     
  2. jcsd
  3. Sep 14, 2010 #2

    LCKurtz

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    Using the following formula with [itex]\theta = A - B[/itex]

    [tex]\tan\frac \theta 2 = \frac {1-\cos(\theta)}{\sin(\theta)}[/tex]

    I get [itex]A - B = \pm 1/3[/itex]

    Check what this gives for C using Napier's identity and I think your question will be answered.
     
  4. Sep 14, 2010 #3
    I assume u mean tan(A-B/2) = +/- 1/3
    How did the negative sign come? You took two values for sin(theta) ?
    After solving I got c= -90 or c=90
    both can be correct
     
  5. Sep 14, 2010 #4

    LCKurtz

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    Yes.

    Yes

    A triangle with -90 degrees? I don't think so. And you can check, using the fact that it is a right triangle, that the numbers all work.
     
  6. Sep 14, 2010 #5
    Thanks
     
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