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Known: $r$, coordinates of $A$, angle $BAC$=72°
Task: find coordinates of B and C.

So, I have 4 unknown parameters to compute, but only 3 equations.

$r^2$=$(x_a-x_b)^2+(y_a-y_b)^2 $
$(2.49r)^2 = (x_a-x_c)^2 + (y_a - y_c)^2$
$(2.28r)^2 = (x_b-x_c)^2 + (y_b - y_c)^2$

How to get the fourth equation? Or maybie is there a simplier algorithm? Working in my way may be very time-consuming and wearisome.

It's not a homework, this problem has naturally occured when making an OpenGL program.

enter image description here

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    There was a mistake and I only corrected the picture , but no equations. Question again corrected and I apologise for it.2012-07-10

2 Answers 2

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$\def\A{{\bf A}} \def\B{{\bf B}} \def\C{{\bf C}} \def\R{{\bf R}} \def\D{{\bf D}} \def\f{\phi}$As others have mentioned, if the task is to find the coordinates of $B$ and $C$, this problem is underdetermined.

Let's first consider the triangle with point $A$ located at the origin and point $C$ lying along the positive $x$-axis. Then the coordinates of the points can be found with simple trigonometry, $\begin{eqnarray*} \A_0 &=& (0,0) \\ \B_0 &=& (c \cos A, c\sin A) \\ \C_0 &=& (b,0). \end{eqnarray*}$ The angle $A$, and the sides $b$ and $c$ have been given. This is an SAS triangle. The triangle can be solved by finding $a$ with the law of cosines and one of the other angles with the law of sines.

(Added: For completeness, $A = 72^\circ$, $b = 2.61r$, and $c=r$. The law of cosines gives $a = \sqrt{b^2+c^2-2bc\cos A} = 2.49r$. Then $\frac{\sin A}{a} = \frac{\sin B}{b}$ implies $B = 86^\circ$. Lastly, $A+B+C = 180^\circ$ implies $C = 22^\circ$.)

The collection of triangles you are interested in have vertices of the form $\begin{equation*} \D = \A + \R(\f)\D_0 \tag{1} \end{equation*}$ where $\A = (A_x,A_y)$ is the given location of $A$, and where $\R(\f)$ is a rotation matrix. The transformation (1) is a counterclockwise rotation by the angle $\f$, followed by a shift so the point $A$ has the given coordinates. In components, $\begin{eqnarray*} A_x &=& A_x \\ A_y &=& A_y \\ B_x &=& A_x + c\cos A\cos\f - c\sin A\sin\f \\ &=& A_x + c\cos(A+\f) \\ B_y &=& A_y + c\cos A\sin\f + c\sin A\cos\f \\ &=& A_y + c\sin(A+\f) \\ C_x &=& A_x + b\cos \f \\ C_y &=& A_y + b\sin \f. \end{eqnarray*}$

Below we plot the triangle before rotation and translation in black. (We set $r = 1$ in the figure.) The dotted triangle has been rotated counterclockwise by $\f = 30^\circ$, and then translated so the new location of point $A$ is $(2,1)$. For reference,
$\begin{eqnarray*} A_x &=& 2 \\ A_y &=& 1 \\ c &=& r = 1 \\ b &=& 2.61 \\ A &=& 72^\circ \\ \f &=& 30^\circ. \end{eqnarray*}$

enter image description here

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    @kowalt: Thanks for the bounty. Cheers!2012-07-10
4

B and C will not have any fixed co-ordinate.
Locus of B,C will be a circle with centre at A.

Now we have only one information about point B($x,y$) that is, it is at a distance $r$ from
A($x_a,y_a$).Hence,we have $(x-x_a)^2+(y-y_a)^2=r^2$ which is indeed an equation of circle, centered at A with radius $r$.And if you fix B on that circle you will get corresponding point C.

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    In other words to find a fixed location of B or C you need one more piece of information like slope of AC or something like that. As you have mentioned in your question that you have more variables than equations is because you donot have any other information about B other than it is at distance $r$ from A2012-07-07