circlescoordinate systemsgeometry
I have a circle with $A$ as a center, $B$ and $C$ two points on the circle.
I have the coordinates of $A$ (the center) and $B$ (the point on the circle). How can I find the coordinates of $C$ (another point on the circle knowing that $\angle BAC$ is $0.384$ rad (or $22^\circ$).
The Y axis is upside down (see picture http://i.stack.imgur.com/4Vp4W.jpg)
So I have $(X_a, Y_a)$ and $(X_b, Y_b)$ and I need to find $(X_c, Y_c)$ respecting the given axis.
Thanks!
$\def\A{{\bf A}} \def\B{{\bf B}} \def\C{{\bf C}} \def\R{{\bf R}} \def\D{{\bf D}} \def\f{\phi}$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,
$$\begin{eqnarray*}
A_x &=& 2 \\
A_y &=& 1 \\
c &=& r = 1 \\
b &=& 2.61 \\
A &=& 72^\circ \\
\f &=& 30^\circ.
\end{eqnarray*}$$
Let's look at a simpler problem. Suppose you have the situation depicted in the figure below:
Then, given the angle $\alpha$, the coordinates of the point $C''$ are:
$$ C''_x = r\cos\alpha \qquad\mbox{and}\qquad C''_y = r\sin\alpha $$
where $r$ is the radius of the circle.
Now let's look at a slightly more complicated problem, depicted below:
This is very similar to the situation above. In fact,
$$ C'_x = r\cos(\alpha+\beta) \qquad\mbox{and}\qquad C'_y = r\sin(\alpha+\beta) $$
By using the trigonometric relations $\sin(\alpha+\beta) = \sin\alpha\cos\beta + \sin\beta\cos\alpha$ and $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$, we can write the above as follows:
$$ C'_x = r\cos\alpha\cos\beta - r\sin\alpha\sin\beta \qquad\mbox{and}\qquad C'_y = r\sin\alpha\cos\beta + r\sin\beta\cos\alpha $$
But, wait... By looking at the previous situation and replacing $C''$ with $B'$ and $\alpha$ with $\beta$, we see that
$$ B'_x = r\cos\beta \qquad\mbox{and}\qquad B'_y = r\sin\beta $$
Therefore, we can write
$$ C'_x = B'_x\cos\alpha - B'_y\sin\alpha \qquad\mbox{and}\qquad C'_y = B'_x\sin\alpha + B'_y\cos\alpha $$
But what you want is this, instead:
Well, we can just move everything rigidly by the vector $-\vec{OA}$ so that $A$ is now the origin of the coordinate system and we get the situation just above. This amounts to subtracting $A$ from both $B$ and $C$ to get $B'$ and $C'$ in the above, and we find
$$ C_x - A_x = (B_x-A_x)\cos\alpha - (B_y-A_y)\sin\alpha $$ $$ C_y - A_y = (B_x-A_x)\sin\alpha + (B_y-A_y)\cos\alpha $$
Then, finally,
$$ C_x = A_x + (B_x-A_x)\cos\alpha - (B_y-A_y)\sin\alpha $$ $$ C_y = A_y + (B_x-A_x)\sin\alpha + (B_y-A_y)\cos\alpha $$
Best Answer
Consider the angle formed between line ($AC$) and the x-axis. The picture you provided shows point $B$ on the x-axis which is not necessarily the case. The x-distance from point $A$ would be $\text{radius}\times \sin(\text{angle})$. The y-distance from point $A$ would be $\text{radius}\times \cos(\text{angle})$