First, find the angle \\ \angle AOB \\ using the arc length formula: \\ \text{Arc length} = r \theta \\ where \\ r = 6 \\ and \\ \text{Arc length} = 15. \\ Thus, \\ \theta = \frac{15}{6} = 2.5 \\text{ radians}.

The angle \\ \angle BOC = \pi - 2.5 \\text{ radians}.

Using the cosine rule in triangle \\ \triangle BOC, \\ BC = 6(\pi - 2.5) = 3.850 \\text{ cm}.

Using the sine rule, \\ \sin(\pi - 2.5) = \frac{BX}{6}, \\ so \\ BX = 6 \sin(\pi - 2.5) = 3.59 \\text{ cm}.

To find \\ OX, \\ use either \\ OX = 6 \cos(\pi - 2.5) \\ or Pythagoras' theorem: \\ OX = 4.807 \\text{ cm}.

Finally, \\ XC = 6 - OX = 1.193 \\text{ cm}.

The perimeter \\ P = BX + XC + BC = 3.59 + 1.193 + 3.850 = 8.63 \\text{ cm}.

First, calculate the length of OC using the cosine of angle AOB:

\(OC = 6 \cos 0.8 = 4.18 \text{ cm}\).

Next, find the area of sector OCD:

\(\text{Area of sector } OCD = \frac{1}{2} \times (4.18)^2 \times 0.8\).

Then, calculate the area of triangle OCA:

\(\Delta OCA = \frac{1}{2} \times 6 \times 4.18 \times \sin 0.8\).

Finally, find the required area of the shaded region by subtracting the area of sector OCD from the area of triangle OCA:

\(\text{Required area} = \Delta OCA - \text{sector } OCD\).

The calculated area is approximately 2.01 cm².

(i) Since arc OC is part of a circle with centre A, the angle CAO is \(\frac{\pi}{3}\) radians.

(ii) The area of sector AOC is given by \(\frac{1}{2} r^2 \times \frac{\pi}{3}\).

The area of triangle ABC is \(\frac{1}{2} (r)(2r) \sin\left(\frac{\pi}{3}\right)\).

Using \(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\), the area of triangle ABC becomes \(\frac{1}{2} (2r^2) \frac{\sqrt{3}}{2} = \frac{r^2 \sqrt{3}}{2}\).

The area of the shaded region is the area of triangle ABC minus the area of sector AOC, which is \(\frac{r^2 \sqrt{3}}{2} - \frac{1}{2} r^2 \frac{\pi}{3}\).

(a) The angle \(ACB\) is given as \(k\pi\) radians. From the mark scheme, \(k = \frac{2}{3}\).

(b) The perimeter of the shaded motif is the circumference of the circle with radius \(r\), which is \(2\pi r\).

(c) To find the area of the shaded motif:

- The area of the major sector \(OAB\) is \(\frac{1}{2} r^2 \times \frac{4\pi}{3} = \frac{2\pi r^2}{3}\).

- The area of the minor sector \(CAOB\) is \(\frac{1}{2} r^2 \times \frac{\pi}{3} = \frac{\pi r^2}{6}\).

- The area of triangle \(ACB\) is \(\frac{1}{2} r^2 \sin \frac{\pi}{3} = \frac{r^2 \sqrt{3}}{4}\).

- The area of the shaded motif is the area of the major sector minus the area of the minor sector and the triangle: \(\frac{2\pi r^2}{3} - \left( \frac{\pi r^2}{6} + \frac{r^2 \sqrt{3}}{4} \right) = \frac{\pi r^2}{3} + \frac{r^2 \sqrt{3}}{2}\).

(i) The arc length AB is given by \(2r\theta\). The tangent segments AT and BT are equal, and each is \(r \tan \theta\) because \(\tan \theta = \frac{AT}{r}\). Therefore, the perimeter \(P\) of the shaded region is:

\(P = 2r\theta + 2r \tan \theta\)

(ii) For r = 5 and \theta = 1.2, the area of triangle AOT is \(\frac{1}{2} \times 5 \times 5 \times \tan 1.2\). The area of sector AOB is \(\frac{1}{2} \times 5^2 \times 2.4\). The shaded area is twice the area of triangle AOT minus the area of sector AOB:

\(\text{Area} = 2 \times \left( \frac{1}{2} \times 5 \times 5 \times \tan 1.2 \right) - \frac{1}{2} \times 25 \times 2.4\)

Calculating gives:

\(\text{Area} = 34.3 \text{ cm}^2\)