(i) From the diagram, we have OQ = x + OC = 20. Using trigonometry in the right triangle, we have:

\(\sin 0.6 = \frac{x}{OC} \Rightarrow OC = \frac{x}{\sin 0.6}\)

Substitute into the equation: \(x + \frac{x}{\sin 0.6} = 20\)

Solve for \(x\): \(x = 7.218\)

(ii) The area of the sector is \(\frac{1}{2} \times 20^2 \times 1.2\).

The area of the circle with centre C is \(\pi \times 7.218^2\).

Total area outside the circle: \(\frac{1}{2} \times 20^2 \times 1.2 - \pi \times 7.218^2 = 76.3 \text{ cm}^2\)

(iii) Angle PCR is \(\pi - 1.2\).

Arc PR is \(7.218 \times (\pi - 1.2) = 14.01 \text{ cm}\).

Using trigonometry, \(OP = OR = \frac{x}{\tan 0.6}\).

Perimeter of OPSR: \(14.01 + 2 \times \frac{7.218}{\tan 0.6} = 35.1 \text{ cm}\)

(i) To find \(AC\), note that \(AC = OA - OC\). Since \(OC = r \cos \theta\), we have:

\(AC = r - r \cos \theta\).

(ii) For the perimeter of the shaded region ABD:

1. Calculate arc AB: \(\text{arc } AB = \frac{4\pi}{3}\).

2. Calculate arc AD: \(\text{arc } AD = \frac{\pi}{2} \times \text{their } AC = \frac{\pi}{2} \times (4 - 4 \cos \frac{\pi}{3}) = \pi\).

3. Calculate BD: \(BD = 4 \sin \frac{\pi}{3} - \text{their } AC = 2\sqrt{3} - 2\).

4. Add these to find the perimeter: \(\text{Perimeter} = \frac{7\pi}{3} + 2\sqrt{3} - 2\).

(i) Since OBX is a right angle (90°), we use the cosine rule: \(\cos \theta = \frac{r}{2r} = \frac{1}{2}\). Therefore, \(\theta = \frac{1}{3}\pi\) radians.

(ii) The arc length AB is \(\frac{1}{3}r\pi\). The length BX is \(r\tan(\frac{1}{3}\pi) = r\sqrt{3}\). Thus, the perimeter P is \(r + \frac{1}{3}r\pi + r\sqrt{3}\).

(iii) The area of the shaded region is the area of triangle OBX minus the area of sector AOB. The area of triangle OBX is \(\frac{1}{2}r^2\sqrt{3}\) and the area of sector AOB is \(\frac{1}{6}r^2\pi\). Therefore, the area of the shaded region is \(\frac{1}{2}r^2\sqrt{3} - \frac{1}{6}r^2\pi\).

(i) To find the length of AB, use the formula for the chord length:

AB = 2 imes r imes ext{sin} rac{ heta}{2} = 2 imes 8 imes ext{sin} rac{2.4}{2} = 14.9 ext{ cm}.

(ii) The perimeter of the plate is the sum of the chord length and the arc length. The arc length is given by:

s = r heta = 8 imes (2 ext{π} - 2.4) = 31.1 ext{ cm}.

Thus, the perimeter is:

14.9 + 31.1 = 46.0 ext{ cm}.

(iii) The area of the plate is the area of the sector minus the area of the triangle. The area of the sector is:

rac{1}{2} imes r^2 imes heta = rac{1}{2} imes 8^2 imes (2 ext{π} - 2.4) = 124.3 ext{ cm}^2.

The area of the triangle is:

rac{1}{2} imes r^2 imes ext{sin} heta = rac{1}{2} imes 8^2 imes ext{sin} 2.4 = 21.6 ext{ cm}^2.

Thus, the area of the plate is:

124.3 + 21.6 = 146 ext{ cm}^2.

First, calculate the length \(AQ\) which is the radius of the arc. Since \(Q\) is the midpoint of \(BC\), and \(ABC\) is an equilateral triangle, \(AQ = \sqrt{3}\).

Next, calculate the area of triangle \(AQC\). The height from \(A\) to \(BC\) is \(\sqrt{3}\), so the area of \(\triangle AQC\) is \(\frac{1}{2} \times 2 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}\).

Now, calculate the area of the sector \(APR\). The angle \(\angle BAC\) is \(60^\circ\) or \(\frac{\pi}{3}\) radians. The area of the sector is \(\frac{1}{2} \times (\sqrt{3})^2 \times \frac{\pi}{3} = \frac{\pi}{2}\).

The area of the shaded region is the area of \(\triangle AQC\) minus the area of the sector \(APR\):

\(\text{Shaded area} = \frac{\sqrt{3}}{2} - \frac{\pi}{2} = \sqrt{3} - \frac{\pi}{2}\)