brad89 Posted December 7, 2005 Share Posted December 7, 2005 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. Link to comment Share on other sites More sharing options...

Dave Posted December 7, 2005 Share Posted December 7, 2005 I'm confused about what you're actually asking. Perhaps a diagram would be of assistance? Link to comment Share on other sites More sharing options...

Lyssia Posted December 7, 2005 Share Posted December 7, 2005 Perhaps you could use the [math]y = mx +b[/math] to find a vector for each line and then use the scalar product? Link to comment Share on other sites More sharing options...

BigMoosie Posted December 8, 2005 Share Posted December 8, 2005 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 m_{1} and m_{2} are the gradients of the lines. Link to comment Share on other sites More sharing options...

the tree Posted December 8, 2005 Share Posted December 8, 2005 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. Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now