Finding relative coordinates in triangle 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.

 A: 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.    
A: $\def\A{{\bf A}}
\def\B{{\bf B}}
\def\C{{\bf C}}
\def\R{{\bf R}}
\def\D{{\bf D}}
\def\f{\phi}$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,
$$\begin{eqnarray*}
A_x &=& 2 \\
A_y &=& 1 \\
c   &=& r = 1 \\
b   &=& 2.61 \\
A   &=& 72^\circ \\
\f  &=& 30^\circ.
\end{eqnarray*}$$

