brad89 Posted December 7, 2005 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.
Dave Posted December 7, 2005 Posted December 7, 2005 I'm confused about what you're actually asking. Perhaps a diagram would be of assistance?
Lyssia Posted December 7, 2005 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?
BigMoosie Posted December 8, 2005 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 m1 and m2 are the gradients of the lines.
the tree Posted December 8, 2005 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.
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