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Angular properties of quadrilaterals

  1. Jul 15, 2007 #1

    Defennder

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    I'm trying to solve a physics problem concerning phasor additions, but unfortunately it seems I've forgotten most of my pre-calc maths, especially geometry.

    Given a irregular quadrilateral with 3 of its sides of known length, and the values of 2 of its interior angles (specifically the angles formed at the junction where 2 of the 3 known sides meet), is there any way where I may determine the length of the unknown side? Or do I require more information?
     
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  3. Jul 15, 2007 #2

    Gib Z

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    Yes, but only if your willing to grind through quite alot of trig. Heres the basic gist of it:

    |_/

    Say that looks something like the 3 sides of the quadrilateral, you know the lengths of all the sides, and the angles |_ and _/ . Draw a line to to top of the line | to the mid point of _, and draw another line from the top of / to the middle of _ as well. Now we have 2 triangles, and in them we know the length of the original sides, plus angles in the right place. Using what you know, for each triangle, use the cosine rule ( [itex] a^2 = b^2+c^2 - 2bc \cos A[/itex] where A is the angle opposite side a) to find the lengths of the sides that you just drew.

    Now, we know all the sides and 1 of the angles, for each of the triangles. use the sine rule ([itex] a/\sin A = b/ \sin B = c / \sin C[/itex]) to find the angle between the 2 lines that you drew. With the known information you the cosine rule again to find the length of the side opposite the angle you just worked out, and thats the unknown side.
     
  4. Jul 15, 2007 #3

    Defennder

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    Ahh, I see it now, thanks Gib!
     
  5. Jul 15, 2007 #4

    Gib Z

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    No worries :)
     
  6. Jul 16, 2007 #5

    symbolipoint

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    Gib Z, you are brilliant. Have you taught Geometry?
     
  7. Jul 17, 2007 #6

    Gib Z

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    Well I do tutor all mathematics to students my own age (for a very petty sum may i add), but thats about all my teaching experience. I don't even get minimum wage...
     
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