Angle between 3 points I have three points $(x_1, y_1), (x_c, y_c), (x_3, y_3)$, where I know $(x_1, y_1), (x_c, y_c)$, the angle $\theta$, and $c$ on the dash line in the following figure. How to calculate the point $(x_3, y_3)$?
I think of this form:
$$
\theta = arccos\left(\frac{a\cdot b}{||a||\cdot ||b||}\right)
$$
where
$$
a = (x_1 - x_c, y_1 - y_c)\\
b = (x_1 - x_3, y_1 - y_3)
$$
More information:

 A: So, since $(x_1,y_1)$ and $(x_c,y_c)$ are known, vector $a$ is known and we are basically looking for vector $b$, such that $\angle(a,b)=\theta$.
If you draw this situation, you will find that, for given $a$, there is a whole line of vectors for $b$ that satisfy this angle property.
However, to find one, we can add one more constraint, e.g. $|b|=1$ (or fix one of the coordinates of $b$).
Once you found (all possible) $b=(x_b,y_b)$, you can get $x_3=x_1-x_b$ and $y_3=y_1-y_b$.
A: Let's approach the problem via a simplified case, where $(x_c,y_c)$ is at origin, and $(x_1, y_1)$ is on the $x$ axis at $(a, 0)$:

Obviously, we can calculate $a$ from the original coordinates,
$$a = \sqrt{\left(x_1 - x_c\right)^2 + \left(y_1 - y_c\right)^2}$$
We have three unknowns, $x$, $y$, and $b$ (the distance from origin to $(x,y)$), and three equations:
$$\begin{cases}
x = b \cos \theta \\
y = b \sin \theta \\
(x - a)^2 + y^2 = c^2
\end{cases}$$
There is a pair of solutions:
$$\begin{cases}
x = a \left(\cos\theta\right)^2 \pm \cos\theta \sqrt{c^2 - a^2 \left(\sin\theta\right)^2} \\
y = a \sin\theta \cos\theta \pm \sin\theta \sqrt{c^2 - a^2 \left(\sin\theta\right)^2} \\
b = a \cos\theta \pm \sqrt{c^2 - a^2\left(\sin\theta\right)^2}
\end{cases}$$
Pick either the upper or the lower signs for all three, but note that only the triplet for which $b \ge 0$ is actually valid. Indeed, for my illustration above, I've shown the "-" solution; the "+" solution would have $(x,y)$ somewhere near $(a,c)$, making the lower right corner angle somewhat near ninety degrees.
However, now that we do know (possibly two valid values of) $b$, we can go back to looking at the situation in the original coordinates.
Numerical solutions, using atan2():
The simplest way is to use the atan2() function available in most programming languages. ($\operatorname{atan2}(y,x) = \tan(y/x)$, except the former also takes account the quadrant, too.)
With it, in original coordinates,
$$\begin{cases}
b = a \cos\theta \pm \sqrt{c^2 - a^2\left(\sin\theta\right)^2}, & b \ge 0 \\
\theta_0 = \operatorname{atan2}(y_1 - y_c, x_1 - x_c) \\
x_3 = x_c + b \cos \left ( \theta_0 + \theta \right ) \\
y_3 = y_c + b \sin \left ( \theta_0 + \theta \right )
\end{cases}$$
If you want positive $\theta$ to be clockwise, use $(\theta_0 - \theta)$ instead in the formulas for $x_3$ and $y_3$, above.
Symbolic solutions, via coordinate system transformation:
A two-dimensional rotation matrix is defined as
$$\mathbf{R} = \left[\begin{matrix}\cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi\end{matrix}\right] = \left[\begin{matrix}C&-S\\S&C\end{matrix}\right]$$
and rotating a point $\vec{p} = (x, y)$ by matrix $\mathbf{R}$ is
$$\mathbf{R} \vec{p} = \left[\begin{matrix}C&-S\\S&C\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right]$$
i.e.
$$\begin{cases}
x' = C x - S y \\
y' = S x + C y
\end{cases}$$
In this particular case, we need to rotate our simplified case solutions using a matrix which rotates point $(a,0)$ to $(x_1-x_c, y_1-y_c)$:
$$\begin{cases}
x_1 - x_c = C a \\
y_1 - y_c = S a
\end{cases} \iff \begin{cases}
C = \frac{x_1 - x_c}{a} \\
S = \frac{y_1 - y_c}{a}
\end{cases}$$
Applying the above rotation to our simplified case results, and a translation to move $(x_c, y_c)$ back to its proper place from origin, we get:
$$\begin{cases}
b = a \cos\theta \pm \sqrt{c^2 - a^2\left(\sin\theta\right)^2}, & b \ge 0 \\
x = b \cos\theta \\
y = b \sin\theta \\
C = \frac{x_1 - x_c}{a} \\
S = \frac{y_1 - y_c}{a} \\
x_3 = x_c + C x - S y \\
y_3 = y_c + S x + C y
\end{cases}$$
or equivalently
$$\begin{cases}
b = a \cos\theta \pm \sqrt{c^2 - a^2\left(\sin\theta\right)^2}, & b \ge 0 \\
x_3 = x_c + \frac{b}{a}(x_1 - x_c)\cos\theta - \frac{b}{a}(y_1 - y_c)\sin\theta \\
y_3 = y_c + \frac{b}{a}(y_1 - y_c)\cos\theta + \frac{b}{a}(x_1 - x_c)\sin\theta
\end{cases}$$
or equivalently
$$\begin{cases}
z = \frac{b}{a} = \cos\theta \pm \sqrt{\frac{c^2}{a^2} - \left(\sin\theta\right)^2}, & z \ge 0 \\
x_3 = x_c + z(x_1 - x_c)\cos\theta - z(y_1 - y_c)\sin\theta \\
y_3 = y_c + z(y_1 - y_c)\cos\theta + z(x_1 - x_c)\sin\theta
\end{cases}$$
.
A: If you know $c$ then you can apply the cosine rule to the triangle:
$$
c^2=a^2+b^2-2ab\cos\theta
$$
and solve for $b$.
Notice that you could find one, two or no solutions, depending on the known values.
