(i) The area of the semicircle with diameter AB is \(\frac{\pi a^2}{8}\) and similarly for AC. The total area of the semicircles is \(\frac{\pi a^2}{4}\).

The area of the circular segment is \(\frac{1}{2}a^2\theta - \frac{1}{2}a^2\sin \theta\).

Equating the areas: \(\frac{1}{2}a^2\theta - \frac{1}{2}a^2\sin \theta = \frac{\pi a^2}{4}\).

Dividing through by \(\frac{1}{2}a^2\): \(\theta - \sin \theta = \frac{\pi}{2}\).

Rearranging gives \(\theta = \frac{1}{2}\pi + \sin \theta\).

(ii) Calculate \(f(\theta) = \frac{\pi}{2} + \sin \theta\) at \(\theta = 2.2\) and \(\theta = 2.4\):

\(f(2.2) = 2.3793\) and \(f(2.4) = 2.2463\).

Since \(2.2 < f(2.2)\) and \(f(2.4) < 2.4\), \(\theta\) lies between 2.2 and 2.4.

(iii) Using the iterative formula \(\theta_{n+1} = \frac{1}{2}\pi + \sin \theta_n\):

Start with \(\theta_0 = 2.2\):

\(\theta_1 = 2.3793\)

\(\theta_2 = 2.2614\)

\(\theta_3 = 2.3417\)

\(\theta_4 = 2.2881\)

\(\theta_5 = 2.3244\)

\(\theta_6 = 2.3000\)

\(\theta_7 = 2.3165\)

\(\theta_8 = 2.3054\)

\(\theta_9 = 2.3129\)

\(\theta_{10} = 2.3072\)

The value converges to \(\theta = 2.31\) to 2 decimal places.