Equation solving

The angle that the arc subtends with the centre can be found using the definition of angle. Let this angle be $\theta$ . Then:

$\theta = \frac{16}{r}$

Now, if one were to draw a line from O that intersects AB at C such that $OC \perp AB$ . Then:

$AC = 5$ $\angle AOC = \frac{\theta}{2}$

Now, by using the definition of sine on $\triangle AOC$ :

$\sin\left(\frac{\theta}{2}\right) = \frac{5}{r}$ $\implies \sin\left(\frac{8}{r}\right) = \frac{5}{r}$

$\implies r \sin\left(\frac{8}{r}\right) - 5=0$

The equation above must be solved to obtain $r$ . This is a nonlinear equation. This can be solved in an iterative method as follows. Assume the root is:

$r = r_o$

Let:

$f(r) = r \sin\left(\frac{8}{r}\right) - 5$

It's derivative is:

$f'(r) = \sin\left(\frac{8}{r}\right) - \frac{8}{r} \cos\left(\frac{8}{r}\right)$

Use the following formula to evaluate the new estimate of the root:

$r_n = r_o - \frac{f(r_o)}{f'(r_o)}$

After getting $r_n$ , use that as the new $r_o$ and substitute it in the formula above to again obtain $r_n$ . Keep repeating this until the results of $r_n$ show repeatability. The final $r_n$ would be the answer. It comes out to be $r_n \approx 5.002$ . This is called the Newton Raphson method of solving for the roots of an equation.