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Math Help - Determining XY coordinates from quadrilateral measurements

  1. #1
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    Determining XY coordinates from quadrilateral measurements

    I'm trying to determine the XY coordinates of 3 corner points of a quadrilateral shape based on known lengths of the sides of that shape. I know the lengths of all 4 sides of the shape as well as the lengths of both cross lengths (effectively making two adjacent triangles). I've attached a picture to indicate the lengths of the triangles / quadrilateral shape as well as the corner of the shape that would serve as the origin of the system. I would like to find the coordinates of the 3 other corners of the shape in respect to the known XY coordinates of the origin. Any assistance would be greatly appreciated.
    Attached Thumbnails Attached Thumbnails Determining XY coordinates from quadrilateral measurements-photo-aug-31-12-15-00-am.jpg  
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  2. #2
    Senior Member MaxJasper's Avatar
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    Re: Determining XY coordinates from quadrilateral measurements

    Because you don't have right angles you should define where x-y axes are located.
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  3. #3
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    Re: Determining XY coordinates from quadrilateral measurements

    Sorry...D is the Y axis with C (roughly) being the X axis.
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  4. #4
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    Re: Determining XY coordinates from quadrilateral measurements

    Hello, bstones!

    I'm trying to determine the xy-coordinates of 3 corner points of a quadrilateral
    based on known lengths of the sides of that shape.
    I know the lengths of all 4 sides of the shape as well as the lengths of both diagonals
    (effectively making two adjacent triangles).

    Code:
          Y
          |
          |               * C 
        D *
          |
          |
          |
          |             * B 
          |
        A * - - - - - - - - - - X
          |
    Draw segments AB,\,BC,\,CD,\,AC,\,BD.

    We are given: . \begin{Bmatrix}AB\:=\: 2.69 & D\!A \:=\:2.67 \\ BC \:=\: 2.79 & AC \:=\:3.93 \\ C\!D \:=\: 2.71 & B\!D \:=\:3.75 \end{Bmatrix}


    \text{Let }A = \angle D\!AB,\;\theta = \angle B\!AX

    \cos A \:=\:\frac{2.67^2 + 2.69^2 - 3.75^2}{2(2.67)(2.69)} \:=\:0.021058714

    A \:=\:88.79335538^o \quad\Rightarrow\quad A \:\approx\:88.79^o

    \text{Then: }\:\theta \:=\:1.21^o


    \begin{array}{ccccccc}x_{_B} &=& 2.69\cos\theta &=& 2.689410039 &\approx& 2.69\\ y_{_B} &=&2.69\sin\theta &=&0.056335109 &\approx&0.06\end{array}

    \text{Therefore: }\:\boxed{B(2.69,\,0.06)}


    \text{Let }D = \angle CDA,\;\phi = D - 90^o

    \cos D \:=\:\frac{2.67^2 + 2.71^2 - 3.93^2}{2(2.67)(2.71)} \:=\:-0.067160054

    D \:=\:93.85088622^o \quad\Rightarrow\quad D \:\approx\:93.85^o

    \text{Then: }\:\phi \:=\:3.85^o


    \begin{array}{ccccccc}x_{\tiny {C}} &=& 2.71\cos\phi &=& 2.703884217 &\approx& 2.70 \\ y_{\tiny{C}} &=& 2.71\sin\phi\:{\color{red}+\:2.67} &=& 0.181961923 + 2.67 & \approx & 2.85 \end{array}

    \text{Therefore: }\:\boxed{C(2.70,\,2.85)}
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  5. #5
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    Re: Determining XY coordinates from quadrilateral measurements

    Thank you very much everyone, especially Soroban. That was very helpful and informative.
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