This is somewhat of a trick question. Sorry! It seemed like a good idea at the time.

Draw a perpendicular from \(O\) to \(A B\), meeting \(A B\) at \(M\). Let \(\theta=\) \(\angle A O M\). Standard trigonometry shows that the length of \(A B\) is \(2 \sin \theta\). The shorter arc joining \(A\) and \(B\) has length \(2 \theta\), so the longer arc has length \(2 \pi-2 \theta\). The longer arc is twice the chord, and therefore

\[

2 \pi-2 \theta=4 \sin \theta .

\]

We can use the Newton Method to solve this equation as it stands. Let \(f(\theta)=2 \sin \theta+\theta-\pi\). Then \(f^{\prime}(\theta)=2 \cos \theta+1\), and the Newton Method recurrence is

\[

\theta_{n+1}=\theta_{n}-\frac{2 \sin \theta_{n}+\theta_{n}-\pi}{2 \cos \theta_{n}+1} .

\]

An ordinary calculator will only handle this to 8 or 9 places. The scientific calculator that comes bundled with Microsoft Windows can handle about 30 .

But we can find the answer without doing any work by looking back at a calculation done in the notes. We are solving \(\pi-\theta=2 \sin \theta\). Note that by the symmetry of the sine function, we have \(\sin \theta=\sin (\pi-\theta)\). Let \(x=\pi-\theta\). Then our equation is equivalent to \(x=2 \sin x\).

It so happens that in the notes this equation is solved to high accuracy. The positive root of this equation, to 19 places, is given there as \(x=\) \(1.8954942670339809471\). But the length of the chord is \(2 \sin \theta\), that is, \(2 \sin x\), and that is equal to \(x\).