(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.

(i) The arc length is given by \(2r\alpha = 100\). The chord length is given by \(2r \sin \alpha = 99\). Dividing these equations gives \(\frac{99}{100} = \frac{\sin \alpha}{\alpha}\), leading to \(\frac{99}{100}x = \sin x\).

(ii) Calculate \(f(0.1) = \frac{99}{100} \times 0.1 - \sin(0.1) \approx 0.099 - 0.0998 = -0.0008\) and \(f(0.5) = \frac{99}{100} \times 0.5 - \sin(0.5) \approx 0.495 - 0.4794 = 0.0156\). Since \(f(0.1) < 0\) and \(f(0.5) > 0\), there is a root between 0.1 and 0.5.

(iii) The iterative formula \(x_{n+1} = 50 \sin x_n - 48.5 x_n\) rearranges to \(x = 50 \sin x - 48.5 x\), which is equivalent to \(\frac{99}{100}x = \sin x\).

(iv) Using the iterative formula with \(x_1 = 0.25\):

\(x_2 = 50 \sin(0.25) - 48.5 \times 0.25 \approx 0.2445\)

\(x_3 = 50 \sin(0.2445) - 48.5 \times 0.2445 \approx 0.245\)

\(x_4 = 50 \sin(0.245) - 48.5 \times 0.245 \approx 0.245\)

The value converges to \(\alpha = 0.245\).

(a) The area of the semicircle is \(\frac{1}{2} \pi r^2\). The area of the shaded segment is \(\frac{1}{2} r^2 (\theta - \sin \theta)\). Given that the shaded area is one third of the semicircle's area, we have:

\(\frac{1}{2} r^2 (\theta - \sin \theta) = \frac{1}{3} \times \frac{1}{2} \pi r^2\)

\(\theta - \sin \theta = \frac{1}{3} \pi\)

Using \(\theta = \pi - 2\theta\), we substitute:

\(\theta = \frac{1}{3}(\pi - 1.5 \sin 2\theta)\)

(b) Evaluate the expression at \(\theta = 0.5\) and \(\theta = 0.7\):

For \(\theta = 0.5\): \(\frac{1}{3}(\pi - 1.5 \sin 1) \approx 0.626\)

For \(\theta = 0.7\): \(\frac{1}{3}(\pi - 1.5 \sin 1.4) \approx 0.554\)

Since \(0.5 < 0.626\) and \(0.7 > 0.554\), \(0.5 < \theta < 0.7\).

(c) Using the iterative formula \(\theta_{n+1} = \frac{1}{3}(\pi - 1.5 \sin 2\theta_n)\):

Start with \(\theta_0 = 0.6\):

\(\theta_1 = 0.58899\)

\(\theta_2 = 0.58641\)

\(\theta_3 = 0.58606\)

\(\theta_4 = 0.58602\)

\(\theta_5 = 0.58601\)

Thus, \(\theta \approx 0.586\) to 3 decimal places.

(a) The length of CD is given by CD = 2r - 2r \cos x. The area of the trapezium is \(\frac{1}{2} \times (AB + CD) \times r \sin x = \frac{1}{2} \times (2r + 2r - 2r \cos x) \times r \sin x = r^2 (2 - \cos x) \sin x\).

The area of each sector is \(\frac{1}{2} r^2 x\), so the total area of the sectors is \(r^2 x\). Given that this is 90% of the area of the trapezium, we have \(r^2 x = 0.9 \times r^2 (2 - \cos x) \sin x\).

Canceling \(r^2\) from both sides gives \(x = 0.9 (2 - \cos x) \sin x\).

(b) Calculate \(0.9 (2 - \cos 0.5) \sin 0.5 \approx 0.484\) and \(0.9 (2 - \cos 0.7) \sin 0.7 \approx 0.716\). Since \(0.5 < 0.484 < 0.7\), x lies between 0.5 and 0.7.

(c) The iterative formula \(x_{n+1} = \cos^{-1} \left( \frac{2 - x_n}{0.9 \sin x_n} \right)\) rearranges to \(x = 0.9 \sin x (2 - \cos x)\), which is the equation from part (a). Therefore, if the sequence converges, it converges to the root of the equation.

\((d) Using the iterative formula, starting with an initial guess of x = 0.6:\)

1. \(x_1 = \cos^{-1} \left( \frac{2 - 0.6}{0.9 \sin 0.6} \right) \approx 0.6205\)

2. \(x_2 = \cos^{-1} \left( \frac{2 - 0.6205}{0.9 \sin 0.6205} \right) \approx 0.6199\)

3. \(x_3 = \cos^{-1} \left( \frac{2 - 0.6199}{0.9 \sin 0.6199} \right) \approx 0.6200\)

The value converges to 0.62 correct to 2 decimal places.

(a) The length of the tangent from a point to a circle is given by \(AT = r \tan x\) and \(BT = r \tan x\). The area of the sector \(AOB\) is \(\frac{1}{2} r^2 (2x) = r^2 x\). The area of the triangle \(AOT\) 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 circle minus the area of the sector, which is \(\pi r^2 - r^2 x\). Equating the area of the shaded region to the area of the circle, we have:

\(\pi r^2 - r^2 x = r^2 \tan x\)

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

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

(b) Calculate \(\tan 1 \approx 1.5574\) and \(\tan 1.4 \approx 5.7979\). Since \(\pi + 1 \approx 4.1416\) and \(\pi + 1.4 \approx 4.5416\), \(\tan x = \pi + x\) changes sign between 1 and 1.4, confirming a root in this interval.

(c) Using the iterative formula \(x_{n+1} = \arctan(\pi + x_n)\):

\(x_1 = \arctan(\pi) \approx 1.2626\)

\(x_2 = \arctan(\pi + 1.2626) \approx 1.3511\)

\(x_3 = \arctan(\pi + 1.3511) \approx 1.3521\)

\(x_4 = \arctan(\pi + 1.3521) \approx 1.3521\)

The root correct to 2 decimal places is 1.35.