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The angle of a curve

  1. Mar 4, 2013 #1
    Hello everyone, I've been googling how to find the angle of a curve but the results are not the kind I'm looking for.

    Let's say I have a shape that has a curve in it at some point. Something like this.


    I'm curious what I need to be reading in order to find the angle of the curve. what information do I need to know about the top part in order to find the angle at which the line curves?
  2. jcsd
  3. Mar 4, 2013 #2


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    First you will have to tell us what you mean by "angle of a curve". In this picture, does the circle meet the triangle "smoothly"?
  4. Mar 4, 2013 #3
    I've labled the shape a bit better.


    I want to measure the angle of [itex]a[/itex] and [itex]b[/itex] and then I want to find out the degree at which the line curves.

    Is is a smooth 70 degree curve? Maybe a smooth 60 degree curve?

    This is basically what I'm asking myself. Sorry I'm not being that clear, I'm not even sure if what I'm asking can be solved. I'm just playing with what I'm learnt so far.
  5. Mar 4, 2013 #4


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    No, this is not going to have any simple answer like '60' or '70'. The angles will depend both on the length of ab and the radius of the circle.

    I am going to assume ab is less than the diameter of the circle. In particular, drawing lines from a and b to the center of the circle, call it O, gives a triangle with two sides of length Oa= Ob= r, the radius of the circle and one side of length ab. If we call the angle Oa and Ob make, then, by the cosine law, [itex]ab^2= 2r^2- 2r^2cos(\theta)= 2r^2(1- cos(\theta))[/itex]. From that, [itex]cos(\theta)= (2r^2- ab^2)/2r^2[/itex] so that the angle between Oa and ab is [itex]\theta= arccos(2r^2- ab^2)/2r^2[/itex]. Since the angle between a tangent to a circle and a radius is 90 degrees, to find the angle between ab and the tangent, add 90 degrees to that:
    [itex]arccos(2r^2- ab^2)/2r^2)+ 90[/itex] degrees.
  6. Mar 4, 2013 #5

    Thanks so much. Quite a bit more work than I thought but thanks for making the effort to help :)
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