Find point given a line and two angles Let's say I have two points $p_1=(x_1, y_1)$ and $p_2=(x_2, y_2)$, which are given as two points of a triangle $T$. And let's say I know the angles of $T$ at $p_1$ and $p_2$.  How do I find the third point $p_3=(x_3, y_3)$ of $T$?
 A: I believe that this is what you want.  We can use basic linear algebra.
We have the two sides of the triangle.  I write these sides as lines:
$$f(x)=m_1x+b_1=\frac{y_1-y_3}{x_1-x_3}x+b_1$$
$$g(x)=m_2x+b_2=\frac{y_2-y_3}{x_2-x_3}x+b_2$$
$m$ is the slope of the line.  I assume that the angles you are talking about are the angles of elevation or depression of the lines $f$ and $g$.  To turn an angle into slope, we use $\tan \theta$ where $\theta$ is the angle.  Thus
$$f(x)=\tan (\theta_1)x+b_1$$
$$g(x)=\tan (\theta_2)x+b_2$$
To find $b_1$ and $b_2$, solve for them in $f(x_1)=y_1$ and $g(x_2)=y_2$ respectively.
$$f(x_1)=y_1=\tan (\theta_1)x_1+b_1\implies b_1=y_1-\tan (\theta_1)x_1$$
$$g(x_2)=y_2=\tan (\theta_2)x_2+b_2\implies b_2=y_2-\tan (\theta_2)x_2$$
To find $x_3$, we know this is the unique point such that $f(x_3)=g(x_3)$. Thus
$$
f(x_3)=g(x_3)=\tan (\theta_1)x_3+y_1-\tan (\theta_1)x_1=
\tan (\theta_2)x_3+y_2-\tan (\theta_2)x_2 \implies
x_3=
\frac{\tan (\theta_1)x_1-\tan (\theta_2)x_2+y_2-y_1}{\tan (\theta_1)-\tan (\theta_2)}$$
I'm sure there are other ways to do this, but this is the first thing that comes to mind.
