Please help me find a formula to find the 3rd point in a right triangle I'm trying to figure out how to plot a 3rd point on a graph
Given the following line segments and angles 

Is there a formula for the 3rd point?
Note: This image is just for an example.  The base line of the triangle will not always be parallel to the x-axis.
Thanks!
 A: Name the 90° vertex A, the 30° vertex B and the unknown vertex C
$$ \begin{pmatrix} x_C \\ y_C \end{pmatrix} = \begin{pmatrix} x_A \\ y_A \end{pmatrix} + \frac{1}{\sqrt{3}}  \begin{pmatrix} -(y_B-y_A) \\ (x_B-x_A) \end{pmatrix}  $$
Example: A $=(1,1)$ B $=(5,2)$ C $= (1,1)+\frac{1}{\sqrt{3}} (-1,4)$

Wolfram Alpha Link
A: There are various ways of finding it. The one I would favour in this particular scenario is:


*

*Work out the length of the edge from the right angle to the point, using trig.

*Find a unit vector from the right angle to the point.

*Multiply and add.


The interesting one in the non-axis-aligned case is point 2, but it has a simple trick. You can easily find a unit vector from the right angle to the other point (vector subtraction), and then a 90 degree rotation takes $(x, y)$ to $(\pm y, \mp x)$ with the sign depending on the direction of rotation.
A: Seeing as it is a right angled triangle, we know that the angle at the point $(?,?)$ is going to be $180 - (90 + 30) = 60$. Using the sine rule, we can then work out the length of that vertical line, which we get to be:
$$\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$$
$$\frac{a}{\sin(30)} = \frac{10}{\sin(60)} \implies a = \frac{10}{\sin(60)} \cdot \sin(30) = \frac{10}{\sqrt3}.$$
This is your $y$ co-ordinate, and as you know that it is in a straight line up, you're $x$ co-ordinate will remain the same and so your point will be
$$\left( 10, \frac{10}{\sqrt3} \right).$$
