Find 3rd point of triangle given two points and one angle I have the following problem. I want to find the position of a point $(x, y)$ on a 2D Plane, by knowing three fixed points ($A$, $B$ and $C$) and two angles ($\alpha$, $\theta$). The angles are between the unknown point ($U$) and the known points.

My first intuition was to set up a system with two formulas, one for each of the angles, and then solving it. I am pretty sure this will result in two possible results, but I don't need to worry about that. I have tried solving it, but had no idea how I could solve it.
My second thought was more of a guess, if it was possible to solve it by using circles, that would always create a specific angle along the edge (shown in the lower diagram on picture 2). picture 2 I would then just find the intersection of the circles and have the missing 4th point. Is that a valid option as well?
Is there an easier way to calculate the point ($U$) other, than how I wanted to do it?
 A: 
Let $O_1$ is center of circumcircle around $ABO$ and $O_2$ is center of circumcircle around $CBO$. And let $D$ is middle of $AB$ and $E$ is middle of $BC$. Then $\angle AO_1D=\alpha$, $\angle BO_2E=\theta$.
Vector $DO_1$ is vector $AD$ rotated 90$^\circ$ clockwise and scaled by $\cot \alpha$. Then $O_1$ has following coordinates
$$x_{O_1}=\frac{x_A+x_B}{2}+\frac{y_B-y_A}{2}\cot\alpha, y_{O_1}=\frac{y_A+y_B}{2}+\frac{x_A-x_B}{2}\cot\alpha$$
The same consideration for $O_2$ gives:
$$x_{O_2}=\frac{x_B+x_C}{2}+\frac{y_C-y_B}{2}\cot\theta, y_{O_2}=\frac{y_B+y_C}{2}+\frac{x_B-x_C}{2}\cot\theta$$
Triangle $BO_1O_2$ is congruent to $OO_1O_2$ then $O$ is symmetric to $B$ with respect to $O_1O_2$.
Then $$\overrightarrow{O_1O}=2\left(\frac{\overrightarrow{O_1B}\cdot \overrightarrow{O_1O_2}}{\overrightarrow{O_1O_2}\cdot \overrightarrow{O_1O_2}}\right)\overrightarrow{O_1O_2}-\overrightarrow{O_1B}$$
$$x_O=2x_{O_1}+2k(x_{O_2}-x_{O_1})-x_{B},y_O=2y_{O_1}+2k(y_{O_2}-y_{O_1})-y_{B}$$
$$k=\frac{(x_B-x_{O_1})(x_{O_2}-x_{O_1})+(y_B-y_{O_1})(y_{O_2}-y_{O_1})}{(x_{O_2}-x_{O_1})^2+(y_{O_2}-y_{O_1})^2}$$
A: Let
$$ x_1 = \overline{AU} , x_2 = \overline{BU} , x_3 = \overline{CU} $$
Then from the law of cosines applied to $\triangle ABU, \triangle BCU, \triangle ACU $, we have
$ x_1^2 + x_2^2 - 2 x_1 x_2 \cos(\alpha) = \overline{AB}^2 $
$ x_2^2 + x_3^2 - 2 x_2 x_3 \cos(\theta) = \overline{BC}^2 $
$ x_1^2 + x_3^2 - 2 x_1 x_3 \cos(\alpha + \theta) = \overline{AC}^2 $
This system can be solved easily using an iterative numerical method like the standard multivariate Newton-Raphson method, and will converge to the solution with high accuracy in very few iterations ($\approx 5$ iterations).  Or one can use a solver from such math packages like MATLAB or Wolframalpha.
Now that $x_1, x_2, x_3$ are found, the problem is to intersect two circles, the first with center $A$ and radius $x_1$ and the other centered at $B$ with radius $x_2$.  To choose the correct intersection point, calculate the distance from the intersection point and point $C$, choose the intersection point that is $x_3$ away from $C$.
