Is it possible to find the angle measure of two lines following y=mx+b format? I didn't think so because it is impossible to determine the slope of a vertical line, but I also could be wrong. I tried to relate two slopes to their definite angle measures. I used slopes 1 and -1 for 90 degrees, and 1 and 0 for 45 degrees. I can't find any relationship after using multiple slopes, so I don't know if there is already a way.

I'm confused about what you're actually asking. Perhaps a diagram would be of assistance?

Perhaps you could use the [math]y = mx +b[/math] to find a vector for each line and then use the scalar product?

The acute angle θ between two straight lines is given by:

 

[math]\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|[/math]

 

Where m₁ and m₂ are the gradients of the lines.

Let's make a right-angled triangle with [math]y=mx+c[/math] as it's hypotenuse. If it's horizontal base is 2 units long, then it's hieght will be 2m.

 

Now on to some easy trig:

[math]\tan\theta=\frac{opp}{adj}[/math]

Factor in the sides of our triangle.

[math]\tan\theta=\frac{2m}{2}[/math]

Re-arange.

[math]\theta=\tan^{-1}\frac{2m}{2}[/math]

 

If I got that right (no garuntees) then the 2 can be replaced with whatever makes life easy.