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Finding angles of a parallelogram.

  1. Oct 2, 2015 #1
    1. The problem statement, all variables and given/known data
    This question has popped up recently and I was completely stumped. How to find the angles of a parallelogram given only area and the length of the diagonals? I'm trying to find a generic solution or formula that works for a non-rhombus.

    2. Relevant equations


    3. The attempt at a solution

    Parallelogram and Triangle Laws
    Setting angle variable equal height(sin(angle variable))
    Vector addition laws
     
  2. jcsd
  3. Oct 2, 2015 #2

    HallsofIvy

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    The area of a parallelogram is "base times height". "Base" is the length of one side, b. The height is the length of perpendicular from a vertex to the base. Given angle [itex]\theta[/itex] the height is the "opposite side" while the length of that side, a, is the hypotenuse, h, so [itex]h= a sin(\theta)[/itex] The area is [itex]ab sin(\theta)[/itex].
     
  4. Oct 3, 2015 #3
    I still don't understand how to obtain sides a and b from the diagonal lengths since there are no right angles to use Pythagoras.
     
  5. Oct 3, 2015 #4
    I got a solution using the Law of Cosines with the angle of intersection of the diagonals to get the lengths of the sides.
     
  6. Oct 3, 2015 #5
    So you were given the angle of intersection? How do you find the sides without that information? I tried substituting A=absinθ into the equation, after working through sine changes, but the solution keeps cancelling out.
     
  7. Oct 3, 2015 #6
    No, you can use some simple trig in one quadrant of the parallelogram to get the angle (in terms of A and diagonals d1 and d2). Remember, each quadrant has 1/4 the total area A.
     
    Last edited: Oct 3, 2015
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