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Trigonometry problem

  1. Jan 5, 2006 #1
    I have reduced a problem I am trying to solve to something simpler though rather difficult: Four side lengths and an angle uniquely describe a quadrilateral, solve one of the angles adjacent to the one given. No matter what algebraic / trigonometric manipulations I apply I cannot get a simplified relation between the angle and the five given variables.
    I would appreciate a pointer or two, thankyou.
     
  2. jcsd
  3. Jan 5, 2006 #2
    I don't think you will get any simlification there unless the angles and the given sides determine a special property to the quadirateral. For example you understand that it is a rhombhus the the things are easy, or similar is the case for kite. If you are very good in algebra I would recommend vector analysis here for simplification. Write one of the sides adjacent to the given angle in vector (a)form by taking that direction as X-axis. Then you know the direction of the vector adjacent b to it as well as its length. so write it in vector form. You get its end point. Now find a point using equation such that the satisfy the given lenghts from endpoint of second vector b as well as from origin. Now from the angle made by the position vector to this point c along X-axis you can find the adjacent angle. And if you want the other adjacent angle then find the angle made by the vector drawn from the endpoint of b to this calculated point c.
     
  4. Jan 6, 2006 #3

    VietDao29

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    Homework Helper

    Let's say you have a quadrilateral ABCD, AB = a, BC = b, CD = c, DA = d, and you have an angle B.
    Here's my way to tackle the problem. Using cosine law, we have
    [tex]a ^ 2 + b ^ 2 - 2ab \cos B = c ^ 2 + d ^ 2 - 2cd \cos D[/tex]
    [tex]\Leftrightarrow \cos D = \frac{c ^ 2 + d ^ 2 - a ^ 2 - b ^ 2 + 2ab \cos B}{2cd}[/tex]
    [tex]\Leftrightarrow D = \arccos \left( \frac{c ^ 2 + d ^ 2 - a ^ 2 - b ^ 2 + 2ab \cos B}{2cd} \right)[/tex]
    Now you have angle B, and D. So
    A + C = 3600 - (B + D).
    Let [itex]\alpha = A + C[/itex].
    Again, use the cosine law, we have:
    [tex]b ^ 2 + c ^ 2 - 2bc \cos C = a ^ 2 + d ^ 2 - 2ad \cos A[/tex]
    [tex]\Leftrightarrow b ^ 2 + c ^ 2 - 2bc \cos C = a ^ 2 + d ^ 2 - 2ad \cos (\alpha - C)[/tex]
    [tex]\Leftrightarrow b ^ 2 + c ^ 2 - 2bc \cos C = a ^ 2 + d ^ 2 - 2ad (\cos \alpha \cos C + \sin \alpha \sin C)[/tex]
    [tex]\Leftrightarrow a ^ 2 + d ^ 2 - b ^ 2 - c ^ 2 - 2ad (\cos \alpha \cos C + \sin \alpha \sin C) + 2bc \cos C = 0[/tex]
    [tex]\Leftrightarrow a ^ 2 + d ^ 2 - b ^ 2 - c ^ 2 - 2ad \sin \alpha \sin C + 2bc \cos C - 2ad \cos \alpha \cos C = 0[/tex]
    [tex]\Leftrightarrow a ^ 2 + d ^ 2 - b ^ 2 - c ^ 2 - 2ad \sin \alpha \sin C + (2bc - 2ad \cos \alpha) \cos C = 0[/tex]
    [tex]\Leftrightarrow (2bc - 2ad \cos \alpha) \cos C - 2ad \sin \alpha \sin C = b ^ 2 + c ^ 2 - a ^ 2 - d ^ 2[/tex]
    This equation has the form a sin x + b cos x = c, which can be solved by dividing both sides by [itex]\sqrt{a ^ 2 + b ^ 2}[/itex]
    Can you go from here?
    My way is a real mess... Someone may come up with something else simplier.
     
    Last edited: Jan 6, 2006
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