Find the slope of a line given a point and an angle I'm trying to figure out this problem and feel like it's something that must be so simple that I could've done in high school no problem, but for some reason my brain is frozen this morning.  I would really appreciate any help, and want to say thanks in advance. I tried to draw a picture below; I want to find the slope of a line given a point $(x,y)$ and $\theta$.

 A: Another method:
If you know the exterior angle theorem, 
you know that The exterior angle is the sum of remote interior angles
thus:
$ \tan^{-1}\frac{y}{x} = \theta +  $ unknown angle
thus,
$ \tan^{-1}\frac{y}{x} - \theta =  $ unknown angle
$ \tan(\tan^{-1}\frac{y}{x} - \theta )=   \tan( $unknown angle)

$$ \tan(\tan^{-1}\frac{y}{x} - \theta )=   Slope$$
A: $\tan \left( \tan^{-1}\left(\frac{y}{x}\right) - \theta\right)$ is the slope $m$. 
Then use "point slope formula" (if you want an equation of the line, that is...)
$y-y_1 = m(x - x_1)$

For variety, I'll explain.
Labeling the origin "$O$" and the point $(x, y)$ "$P$", the segment $\overline{OP}$ makes an angle of $\tan^{-1} \left(\frac{y}{x}\right)$ with the positive x-axis. But this is the sum of $\theta$ and the angle $\phi$ that your line makes with the positive x-axis (since we have opposite interior angles). 
So $\tan^{-1}\left(\frac{y}{x}\right) - \theta = \phi$. 
Finally, $\tan \phi = m$.
A: The slope is given by the change in y for a given change in x. From trig, you have that the tangent of an angle in a right triangle is the measure of the side of the triangle opposite the angle divided by the measure of the side adjacent to the angle (not the hypotenuse).
See diagram.
Then you have
$$
\tan\theta = \frac{\Delta y}{\Delta x}
$$
which is the slope of the line.
A: find slope of the line you know as m1. m2 will be the slope of the unknown line and Ө the angle. The eqn tan(Ө) = (m2 - m1)/ (1 + (m1*m2))
