Given a triangle with two known vertices and the angle, get the coordinates of the last vertex I have tried attaching an image of the triangle I am working with but since I am a new user this site will not let me post images (kind of defeats the purpose, but anyways).
I have the following triangle:
Point A = (x:40, y:100)
Point B = (x:50, y:50)
Point C = ??
d = 20 degrees (which is the angle between vectors BA and CA.
I am tring to find out the coordinates of Point C.  I have tried using the law of cosines and scoured the net looking for a close example that I can learn from and figure out why I can't get the correct formula for this.
Can any one please lend a hand in figuring out this formula.
Thank you!
 A: Just to repeat what everyone has told you: what you have stated in the question is not enough information about C. Here's a picture:

A: I don't think you have enough information to find a unique solution, or even to narrow the solution down to a finite number of possibilities.  A triangle in the plane has six degrees of freedom, i.e. you need six real numbers to uniquely specify the locations of its vertices.  You only have five known parameters.
A: From  the information you've given point C can be any point along the vector AC such that the Y coordinate is less than the Y coordinate of A. This gives an infinite number of solutions.
A: Those numbers are unpleasant and irrational, so I'm going to do a simpler example. The idea is that we know one side length and one angle of a triangle. But we can quickly construct two such triangles - for example, if AB = 5, angle ABC is 30 degrees, and angle BCA is a right angle, then we quickly get that AC is $5/2$ and BC is $\frac{5}{2} \sqrt 3$.
On the other hand, if AB = 5, ABC is 30 degrees, and angle BAC is right, then we get that AC is $\dfrac{5}{\sqrt 3}$ and BC is $\dfrac{10}{\sqrt 3}$. 
So two different triangles, but each with one side and one angle the same. So it's not unique.
