This is similar to this question, except I am finding the radius now using the Bisection method and then Newton's method for finding a zero.
This is a computer science for a Numerical Methods course. In the question the arc length is 4, and the chord length is 3.
I derived a function such that $f(r) = 0$ when $r$ is the radius like this:
$L = r\theta$ $4 = r\theta$ Using the Law of Cosines:
$3^2 = 2r^2(1 - \cos\theta)$ $ \Rightarrow \theta = \cos^{-1}(1 - 9/2r^2)$ $ \Rightarrow 4 = r\cos^{-1}(1 - 9/2r^2)$
So $f(r) = 4 - r\,\cos^{-1}(1 - 9/2r^2)$ will be $0$ when $r$ is the radius.
This all works well so far, but for the two methods I need an initial approximation. For bisection I need a left and a right $x$, and for newtons method I should use the average of my two $x$ values from the bisection. It's easy to show that a lower bound for the radius is $1.5$ since the chord length is 3. If the radius was less than $1.5$ the chord length would be greater than the diameter, which is a contradiction. I am having a lot of difficulty finding an upper bound though. If I just pick a value like $1.7$ or $\pi/2$ then I get the correct answer, but although I'm not sure that it's required for my assignment, I'd like to know how I can find an upper bound that I can show is actually provably greater than the radius.
The actually radius assuming my program and equation is correct is $1.567769039908$