Answer
$\theta\approx2.9916$, $r\approx1.0028$
Work Step by Step
Step 1. Using the figure given in the exercise, we have $r\theta=3$ and $r^2+r^2-2r^2 cos\theta = 2^2$ (the Laws of Cosines) or $r^2(1-cos\theta)=2$
Step 2. The first relation gives $r=\frac{3}{\theta}$. Use it in the second equation to get $(\frac{3}{\theta})^2(1-cos\theta)=2$ or $9(1-cos\theta)=2\theta^2$ where $0\lt\theta\lt\pi$
Step 3. Use $x$ in place of $\theta$ and let $f(x)=2x^2-9(1-cos(x))=2x^2+9cos(x)-9$; we have $f'(x)=4x-9sin(x)$.
Step 4. Test signs of $f(x)$, $f(\frac{\pi}{2})\approx-4\lt0$ and $f(3)\approx0.09\gt0$; thus there is a zero in the interval of $(\frac{\pi}{2},3)$ based on the Intermediate Value Theorem of a continuous function.
Step 5. Using Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with a starting point $x_0=3$, we can obtain the zero shown in the table as $\theta\approx2.991563136\approx2.9916$ radians, which in turn gives $r\approx1.002820219\approx1.0028$
