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Find Lambda from this arbitrary triangle

  1. Nov 18, 2015 #1
    Folks,

    I am puzzled how one obtains equation 3.2.31 based on the schematic as attached! Can you help?
    Is there an online source I can refer to to learn how to obtain angles and magnitudes of complex schematics?

    Thanks
     

    Attached Files:

  2. jcsd
  3. Nov 18, 2015 #2

    RUber

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    Homework Helper

    I am still working on this one...
    However, I feel like the secret will be in manipulating the equation you have.
    Edited*
    ##\frac{\sin \psi }{ m - \cos \psi } = \frac{\sin( \pi - \psi) }{ a/p +\cos( \pi - \psi) } = \frac{p\sin( \pi - \psi) }{ a + p \cos ( \pi - \psi) }##
    This should help you visualize the triangle you should be working with for ##\tan \lambda##.
     
    Last edited: Nov 18, 2015
  4. Nov 18, 2015 #3
    Hi, thanks....
    This is as far as I got....see attached.....
     

    Attached Files:

  5. Nov 18, 2015 #4

    RUber

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    Homework Helper

    That looks good. You extension works to give you a solution for lambda. But it doesn't help get to that tan lambda formula you were looking for.

    Try to extend the side length a to make a right triangle with hypotenuse of p. Use sin and cos to determine the side lengths.
    Then you can directly get the formula for tangent using the opposite/adjacent of the appropriate right trangle which includes angle lambda.
     
  6. Nov 19, 2015 #5
    Hi RUber,

    I have it. The extended segment of "a" which forms a right angled triangle with P is Pcos(pi-psi). The tan of lambda is obvious. Thank you very much.
    I have learned a new way of tacking triangles!
     
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