(i) The area of triangle ONB is \(\frac{1}{2} r^2 \cos \alpha \sin \alpha\).

The area of sector OAB is \(\frac{1}{2} r^2 \alpha\).

Given that the area of triangle ONB is half the area of sector OAB, we have:

\(\frac{1}{2} r^2 \cos \alpha \sin \alpha = \frac{1}{2} \left( \frac{1}{2} r^2 \alpha \right)\)

Simplifying gives \(\sin 2\alpha = \alpha\).

(ii) Sketch the graphs of \(y = \sin 2x\) and \(y = x\) over the interval \(0 < x < \frac{1}{2}\pi\). The graphs intersect exactly once in this interval, indicating one root.

(iii) Using the iterative formula \(x_{n+1} = \sin(2x_n)\) with \(x_1 = 1\):

\(x_2 = \sin(2 \times 1) = \sin(2) \approx 0.909\)

\(x_3 = \sin(2 \times 0.909) \approx 0.973\)

\(x_4 = \sin(2 \times 0.973) \approx 0.929\)

\(x_5 = \sin(2 \times 0.929) \approx 0.964\)

\(x_6 = \sin(2 \times 0.964) \approx 0.945\)

\(x_7 = \sin(2 \times 0.945) \approx 0.955\)

\(x_8 = \sin(2 \times 0.955) \approx 0.949\)

\(x_9 = \sin(2 \times 0.949) \approx 0.953\)

The iterations converge to \(\alpha = 0.95\) correct to 2 decimal places.