Answer
θ≈2.2622 radians, or 130°
Work Step by Step
Recall from way back when they first introduced "radians", the equation
s=rθ, or θ = $\frac{s}{r}$ = central angle
In this problem we have θ = $\frac{5}{r}$, or r= $\frac{5}{θ}$
So with radius r ( the distance from the center to point A and to point B), we use the Law of Cosines on the triangle in the book's diagram
$4^{2}$ = $r^{2}$ + $r^{2}$ - 2 (r)(r) cos θ
$16^{}$ = $2r^{2}$ + $2r^{2}$ - 2 $r^{2}$ cos θ
Substitute for r
8 = ($\frac{5}{θ})^{2}$ (1-cos θ)
8 = $\frac{25}{θ^{2}}$ (1-cos θ)
$\frac{8θ^{2}}{25}$ = 1 - cos θ
Get one side of the equation to zero, then call it f(x) for use in Newton's method. And find f'.
0 = - $\frac{8θ^{2}}{25}$ + 1 - cos θ
f(x) = - $\frac{8θ^{2}}{25}$ + 1 - cos θ
f'(x) = - $\frac{16}{25}$θ + 1 +sin θ
It looks like there are 3 zeros, but we can disregard x = 0 and the negative one. We can use initial approximation $x_{1}$ = 2.3
Iterate
$θ_{n+1}$ = $θ_{n}$ - $\frac{f(θ_{n})}{f'(θ_{n})}$
until the numbers in the first 4 decimal places don't change anymore.
$θ_{1}$ = 2.3
$θ_{2}$ = 2.3 - $\frac{f(2.3)}{f'(2.3)}$ $\approx$ 2.2634
$θ_{3}$ = 2.2634 - $\frac{f(2.263480)}{f'(2.263480)}$ $\approx$ 2.2622
$θ_{4}$ = 2.2622 - $\frac{f(2.262207)}{f'(2.262207)}$ $\approx$ 2.2622
θ $\approx$ 2.2622 radians
