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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.
Re: Determining XY coordinates from quadrilateral measurements
Because you don't have right angles you should define where xy axes are located.
Re: Determining XY coordinates from quadrilateral measurements
Sorry...D is the Y axis with C (roughly) being the X axis.
Re: Determining XY coordinates from quadrilateral measurements
Hello, bstones!
Quote:
I'm trying to determine the xycoordinates 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 $\displaystyle AB,\,BC,\,CD,\,AC,\,BD.$
We are given: .$\displaystyle \begin{Bmatrix}AB\:=\: 2.69 & D\!A \:=\:2.67 \\ BC \:=\: 2.79 & AC \:=\:3.93 \\ C\!D \:=\: 2.71 & B\!D \:=\:3.75 \end{Bmatrix}$
$\displaystyle \text{Let }A = \angle D\!AB,\;\theta = \angle B\!AX$
$\displaystyle \cos A \:=\:\frac{2.67^2 + 2.69^2  3.75^2}{2(2.67)(2.69)} \:=\:0.021058714$
$\displaystyle A \:=\:88.79335538^o \quad\Rightarrow\quad A \:\approx\:88.79^o$
$\displaystyle \text{Then: }\:\theta \:=\:1.21^o$
$\displaystyle \begin{array}{ccccccc}x_{_B} &=& 2.69\cos\theta &=& 2.689410039 &\approx& 2.69\\ y_{_B} &=&2.69\sin\theta &=&0.056335109 &\approx&0.06\end{array}$
$\displaystyle \text{Therefore: }\:\boxed{B(2.69,\,0.06)}$
$\displaystyle \text{Let }D = \angle CDA,\;\phi = D  90^o$
$\displaystyle \cos D \:=\:\frac{2.67^2 + 2.71^2  3.93^2}{2(2.67)(2.71)} \:=\:0.067160054$
$\displaystyle D \:=\:93.85088622^o \quad\Rightarrow\quad D \:\approx\:93.85^o$
$\displaystyle \text{Then: }\:\phi \:=\:3.85^o$
$\displaystyle \begin{array}{ccccccc}x_{\tiny {C}} &=& 2.71\cos\phi &=& 2.703884217 &\approx& 2.70 \\ y_{\tiny{C}} &=& 2.71\sin\phi\:{\color{red}+\:2.67} &=& 0.181961923 + 2.67 & \approx & 2.85 \end{array}$
$\displaystyle \text{Therefore: }\:\boxed{C(2.70,\,2.85)}$
Re: Determining XY coordinates from quadrilateral measurements
Thank you very much everyone, especially Soroban. That was very helpful and informative.