The angle of a curve

by uperkurk
Tags: angle, curve
 P: 159 I've labled the shape a bit better. I want to measure the angle of $a$ and $b$ 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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,345 The angle of a curve 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, $ab^2= 2r^2- 2r^2cos(\theta)= 2r^2(1- cos(\theta))$. From that, $cos(\theta)= (2r^2- ab^2)/2r^2$ so that the angle between Oa and ab is $\theta= arccos(2r^2- ab^2)/2r^2$. 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: $arccos(2r^2- ab^2)/2r^2)+ 90$ degrees.
 Quote by HallsofIvy 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, $ab^2= 2r^2- 2r^2cos(\theta)= 2r^2(1- cos(\theta))$. From that, $cos(\theta)= (2r^2- ab^2)/2r^2$ so that the angle between Oa and ab is $\theta= arccos(2r^2- ab^2)/2r^2$. 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: $arccos(2r^2- ab^2)/2r^2)+ 90$ degrees.