Coordinates of third point in a triangle Of triangle ABC, I am given the coordinates of two points (A and B) and the angles between the side AB and each of the two other sides. 
How can I get the coordinates of point C?
 A: The ambiguity mentioned above can be avoided if you consider ordered angles, i.e., encode the orientation of the triangle in your data.  We can then solve the problem
as follows.  Choose coordinates so that $A$ is $(0,1)$, $B$ is $(1,0)$ and $C$ is $(p,q)$.  Then $\tan A=\dfrac q p$ and $\tan B=\dfrac q{1-p}$.  We can solve to get
$$p=\frac{\tan B}{\tan A+\tan B},\quad q=\frac{\tan A \tan B}{\tan A+\tan B}.$$
A: You can't find the coordinates of $C$.
Here's what you can find: since you have the coordinates of $A$ and $B$, you have the length of side $AB$.  You also have all the angles, so you can use the law of sines to find the lengths of all the sides.
However, this is not enough.  An equilateral triangle with points at $(0,0)$ and $(1,0)$ can be drawn so that the third point is above the axis or below the axis; the location of the third point is not uniquely specified from what you have given us.
A: Find an intersection of two lines. Find auxiliary basis vectors for a frame, aligned to the AB line:
$$\vec{e}_1=\frac{B-A}{|B-A|}$$
$$\vec{e}_2=\pm(-e_{1y},e_{1x})$$
They are normalized and orthogonal. The plusminus sign specifies the orientation of the frame (and consequently, will select the solution with appropriate triangle orientation - clockwise for $-$ and counterclockwise for $+$). Now, write the line through AC:
$$C=A+AC(\vec{e}_1\cos\alpha+\vec{e}_2\sin\alpha)$$
and a line through BC:
$$C=B+BC(-\vec{e}_1\cos\beta+\vec{e}_2\sin\beta)$$
If you know the lengths AC and BC (from the law of sines or something similar), any of the two expressions will do just fine. If you want to proceed without assuming the law of sines, equate the two expressions and solve for $AC$ and $BC$. This procedure actually derives the law of sines.
A: Geometrically speaking, this is absolutely feasible because the given conditions meet the ‘ASA’ requirement (See figure 1). 
However, using the same given conditions, we can obtain the second solution. (See figure 2). Hence, we say it is possible but the solution is not unique.
Point C (actually C2 and C1), can be obtained analytically.
By suitable transformations (See figure 3), we can let A = (0, 0) and B = (h, 0) for some non-zero h.
Then, equation of L1 is y = (tan α)x …………(1)
and equation of L2 is y = tan (π-β)(x – h) 
i.e. y =–tan (β)(x – h) …………(2)
Co-ordinates of C1 can be found by solving (1) and (2) since α, β, and h are known quantities.
C2 can be obtained similarly (or by symmetry).
