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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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    You can use simple trigonometry. The $x$-coordinate of $B$ for instance is $A_x+r\cos(72^\circ)$ where $A_x$ is the $x$-coordinate of $A$.2012-07-06
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    @DavidMitra That is possible only if AC is parallel to $x$-axis.Isn't it?2012-07-06
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    @SaurabhHota Yes, thanks. I assumed as much.2012-07-06
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    This picture is simplified, hence it can be misleading. |AC| needn't to be parallel to x-axis, so simple geometry is't a correct solution in this case.2012-07-06
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    Your equations seem to assume that the lengths $AC$ (or $BC$) is known; together with $AB$ and $\angle BAC$, you can consider the shape of the triangle to be known. Okay. Now ... You say you only know the coordinates of $A$. Without additional information, you can turn (or even flip) the triangle all over and keep vertex $A$ in place, so your problem is "under-determined". Perhaps your application allows you to add another constraint in the form of some parameter, say, the angle $AB$ (or $AC$) makes with the horizontal. In any case, there is no single answer here.2012-07-08
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    By the way: your figure and your equations don't completely agree on what the side lengths are.2012-07-08
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    @DayLateDon Actually the angle is correct.Then BC comes out to be $2.49r$.See comments in my post.2012-07-09
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    @SaurabhHota: I don't doubt the angle. I'm just pointing-out that kowalt's side lengths don't match in the two places they appear. The figure has $1r$, $2.61r$, $2.49r$, but the equations have $1r$, $2.61r$, $2.28r$. The problem statement says that only the length $r$ is known; given the color-coding of the figure, one *might* infer that the adjacent length $2.61r$ is *also* known, in which case $2.49r$ is the correct length of the opposite side. In any case, kowalt should fix the discrepancy, and should also mention explicitly that we know (if indeed we do) the side length $2.61r$.2012-07-09
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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
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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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    First statement is not revelatory and the second is nonsense.2012-07-06
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    @Kowalt Why do you think so?2012-07-06
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    Sooooooooooooorry! "A" angle is certainty, BC was calculated by myself(therefore it is red), it equals to 2.49r. Question corrected :)2012-07-06
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    @kowalt Ok. But still Locus of point B is a circle.2012-07-07
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    I still can't see where the answer is. Your equation is exactly the same as my first, doesn't it? Also statement "Locus of B,C will be a circle with centre at A" still doesn't make sense for me. Did you mean the same circle?2012-07-07
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    I means if you sketch a circle with center at A and radius $r$ then B can be anywhere on that circle and then correspondingly C can be located.2012-07-07
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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