(i) The area of the sector CMB is \(\frac{1}{2} a^2 \theta\).

The area of the triangle ABC is \(\frac{1}{2} a^2\).

Given that the area of the sector is one third of the area of the triangle:

\(\frac{1}{2} a^2 \theta = \frac{1}{3} \times \frac{1}{2} a^2\)

\(a^2 \theta = \frac{1}{3} a^2\)

\(\theta = \frac{1}{3}\)

Using the formula for the tangent of angle \(\theta\):

\(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{a} = 1\)

Thus, \(\tan \theta = 3\theta\).

(ii) Using the iterative formula \(\theta_{n+1} = \arctan(3\theta_n)\):

Start with an initial guess, say \(\theta_0 = 1\).

\(\theta_1 = \arctan(3 \times 1) = \arctan(3) \approx 1.2490\)

\(\theta_2 = \arctan(3 \times 1.2490) = \arctan(3.7470) \approx 1.3084\)

\(\theta_3 = \arctan(3 \times 1.3084) = \arctan(3.9252) \approx 1.3235\)

\(\theta_4 = \arctan(3 \times 1.3235) = \arctan(3.9705) \approx 1.3259\)

\(\theta_5 = \arctan(3 \times 1.3259) = \arctan(3.9777) \approx 1.3265\)

Thus, the root correct to 2 decimal places is \(\theta = 1.32\).

(i) The length of the tangent CT can be expressed as \(CT = r \tan x\). The area of the semicircle is \(\frac{1}{2} \pi r^2\). The area of the sector BOC is \(\frac{1}{2} r^2 x\). The area of the triangle OCT is \(\frac{1}{2} r \cdot r \tan x = \frac{1}{2} r^2 \tan x\). The area of the shaded region is the area of the triangle minus the area of the sector: \(\frac{1}{2} r^2 \tan x - \frac{1}{2} r^2 x\). Setting this equal to the area of the semicircle gives:

\(\frac{1}{2} r^2 \tan x - \frac{1}{2} r^2 x = \frac{1}{2} \pi r^2\)

Dividing through by \(\frac{1}{2} r^2\) gives:

\(\tan x - x = \pi\)

Thus, \(\tan x = x + \pi\).

(ii) Using the iterative formula \(x_{n+1} = \arctan(x_n + \pi)\), start with an initial guess, say \(x_0 = 1\).

1. \(x_1 = \arctan(1 + \pi) = 1.3521\)

2. \(x_2 = \arctan(1.3521 + \pi) = 1.3490\)

3. \(x_3 = \arctan(1.3490 + \pi) = 1.3492\)

4. \(x_4 = \arctan(1.3492 + \pi) = 1.3492\)

The value converges to 1.35 when rounded to 2 decimal places.

(i) The area of the segment is given by \(\frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin \theta\). With \(r = 10\), this becomes \(50\theta - 50\sin \theta\).

The area of the circle is \(\pi r^2 = 100\pi\). The area of the segment is \(\frac{1}{5}\) of the circle, so:

\(50\theta - 50\sin \theta = \frac{1}{5} \times 100\pi\)

\(50\theta - 50\sin \theta = 20\pi\)

\(\theta - \sin \theta = \frac{2}{5}\pi\)

Thus, \(\theta = \frac{2}{5}\pi + \sin \theta\).

(ii) Using the iterative formula \(\theta_{n+1} = \frac{2}{5}\pi + \sin \theta_n\) with an initial value of \(\theta_0 = 2.1\):

\(\theta_1 = \frac{2}{5}\pi + \sin 2.1 \approx 2.1198\)

\(\theta_2 = \frac{2}{5}\pi + \sin 2.1198 \approx 2.1097\)

\(\theta_3 = \frac{2}{5}\pi + \sin 2.1097 \approx 2.1149\)

\(\theta_4 = \frac{2}{5}\pi + \sin 2.1149 \approx 2.1122\)

\(\theta_5 = \frac{2}{5}\pi + \sin 2.1122 \approx 2.1111\)

Thus, \(\theta \approx 2.11\) radians.

To find the length of AB, use the formula:

\(AB = \sqrt{200 - 200\cos \theta}\)

\(AB = \sqrt{200 - 200\cos 2.11} \approx 17.4 \text{ cm}\)

(i) The area of the semicircle is \(\frac{1}{2} \pi r^2\). The area of the sector BOC is \(\frac{1}{2} r^2 x\). The area of the triangle BOC is \(\frac{1}{2} r^2 \sin x\). The area of the segment is \(\frac{1}{2} r^2 (x - \sin x)\). Given that this is a quarter of the semicircle's area, we have:

\(\frac{1}{2} r^2 (x - \sin x) = \frac{1}{4} \times \frac{1}{2} \pi r^2\)

\(x - \sin x = \frac{1}{4} \pi\)

\(x = \frac{3}{4} \pi - \sin x\)

(ii) Evaluate \(f(x) = x - (\frac{3}{4} \pi - \sin x)\) at \(x = 1.3\) and \(x = 1.5\):

\(f(1.3) = 1.3 - (\frac{3}{4} \pi - \sin 1.3) \approx -0.047\)

\(f(1.5) = 1.5 - (\frac{3}{4} \pi - \sin 1.5) \approx 0.155\)

Since \(f(1.3) < 0\) and \(f(1.5) > 0\), there is a root between 1.3 and 1.5.

(iii) Using the iterative formula:

\(x_1 = \frac{3}{4} \pi - \sin 1.4 \approx 1.3805\)

\(x_2 = \frac{3}{4} \pi - \sin 1.3805 \approx 1.3793\)

\(x_3 = \frac{3}{4} \pi - \sin 1.3793 \approx 1.3796\)

\(x_4 = \frac{3}{4} \pi - \sin 1.3796 \approx 1.3795\)

The root correct to 2 decimal places is 1.38.

(i) The perimeter of the rectangle is \(2(3a + a) = 8a\). Half of this is \(4a\).

The perimeter of the sector AMN is \(r \cdot x + 2r\).

Equating the two perimeters: \(r \cdot x + 2r = 4a\).

Factor out \(r\): \(r(x + 2) = 4a\).

Since \(r = a \csc x\), substitute to get \(a \csc x (x + 2) = 4a\).

Cancel \(a\): \(\csc x (x + 2) = 4\).

\(\csc x = \frac{1}{\sin x}\), so \(\frac{1}{\sin x} (x + 2) = 4\).

Rearrange to get \(\sin x = \frac{1}{4}(2 + x)\).

(ii) Using the iterative formula \(x_{n+1} = \sin^{-1}\left(\frac{2 + x_n}{4}\right)\):

Start with \(x_1 = 0.8\).

\(x_2 = \sin^{-1}\left(\frac{2 + 0.8}{4}\right) = \sin^{-1}(0.7) \approx 0.7754\).

\(x_3 = \sin^{-1}\left(\frac{2 + 0.7754}{4}\right) = \sin^{-1}(0.69385) \approx 0.7672\).

\(x_4 = \sin^{-1}\left(\frac{2 + 0.7672}{4}\right) = \sin^{-1}(0.6918) \approx 0.7650\).

\(x_5 = \sin^{-1}\left(\frac{2 + 0.7650}{4}\right) = \sin^{-1}(0.69125) \approx 0.7643\).

The root correct to 2 decimal places is \(0.76\).