# 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!

-
An informal sketch should convince you that there are many $C$ that work. Did the problem say anything else about $C$? Typical in this sort of game, for example, would be to say that $C$ is on the $x$-axis. –  André Nicolas Jul 25 '11 at 18:59
Post a link to the image (upload it to imgur if necessary), and someone will edit it into your question. –  Rahul Jul 25 '11 at 18:59

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:

-
There's also another half-line on which C can lie, on the other side of AB. –  Ilmari Karonen Jul 25 '11 at 19:31
@Ilmari: Yes, I thought adding both may lead to more confusion, so I arbitrarily picked one. :-) –  ShreevatsaR Jul 25 '11 at 19:37

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.

-

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.

-

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.

-