(i) The area of the sector is \(\frac{1}{2} r^2 \alpha\) and the area of triangle \(AOB\) is \(\frac{1}{2} r^2 \sin \alpha\). Setting the area of the triangle to be half the area of the sector gives:

\(\frac{1}{2} r^2 \sin \alpha = \frac{1}{4} r^2 \alpha\)

\(\sin \alpha = \frac{1}{2} \alpha\)

Let \(x = \alpha\), then \(x = 2 \sin x\).

(ii) Evaluate \(x - 2 \sin x\) at \(x = \frac{1}{2} \pi\) and \(x = \frac{2}{3} \pi\):

For \(x = \frac{1}{2} \pi\), \(x - 2 \sin x = \frac{1}{2} \pi - 2 \cdot 1 = \frac{1}{2} \pi - 2\).

For \(x = \frac{2}{3} \pi\), \(x - 2 \sin x = \frac{2}{3} \pi - 2 \cdot \frac{\sqrt{3}}{2} = \frac{2}{3} \pi - \sqrt{3}\).

Both values are negative, indicating \(\alpha\) lies between these values.

(iii) The iterative formula is \(x_{n+1} = \frac{1}{3}(x_n + 4 \sin x_n)\). Rearranging gives:

\(x = \frac{1}{3}(x + 4 \sin x)\)

\(3x = x + 4 \sin x\)

\(2x = 4 \sin x\)

\(x = 2 \sin x\)

Thus, the sequence converges to a root of \(x = 2 \sin x\).

(iv) Using the iterative formula with \(x_1 = 1.8\):

\(x_2 = \frac{1}{3}(1.8 + 4 \sin 1.8) \approx 1.8950\)

\(x_3 = \frac{1}{3}(1.8950 + 4 \sin 1.8950) \approx 1.9025\)

\(x_4 = \frac{1}{3}(1.9025 + 4 \sin 1.9025) \approx 1.9048\)

\(x_5 = \frac{1}{3}(1.9048 + 4 \sin 1.9048) \approx 1.9054\)

The value converges to 1.90, correct to 2 decimal places.