# Determining XY coordinates from quadrilateral measurements

• August 30th 2012, 11:30 PM
bstones
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.
• August 31st 2012, 12:23 AM
MaxJasper
Re: Determining XY coordinates from quadrilateral measurements
Because you don't have right angles you should define where x-y axes are located.
• September 1st 2012, 12:20 AM
bstones
Re: Determining XY coordinates from quadrilateral measurements
Sorry...D is the Y axis with C (roughly) being the X axis.
• September 1st 2012, 06:28 AM
Soroban
Re: Determining XY coordinates from quadrilateral measurements
Hello, bstones!

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

$\text{Therefore: }\:\boxed{C(2.70,\,2.85)}$
• September 1st 2012, 02:00 PM
bstones
Re: Determining XY coordinates from quadrilateral measurements
Thank you very much everyone, especially Soroban. That was very helpful and informative.