(a) Since the total angle around point O is 2π radians, and angle AOB is 2.8 radians, angle ACO is given by:

\(\angle ACO = \pi - 2.8 = 0.7 \text{ radians}\)

(b) Using the sine rule in triangle ACO, we have:

\(\frac{R}{\sin(2.8)} = \frac{r}{\sin(0.7)}\)

Solving for R gives:

\(R = \frac{r \cdot \sin(2.8)}{\sin(0.7)} \approx 1.53r\)

(c) The area of sector OAB is:

\(\text{Area of sector } OAB = \frac{1}{2} r^2 \times 2.8 = 1.4r^2\)

The area of sector CAB is:

\(\text{Area of sector } CAB = \frac{1}{2} R^2 \times 2 \times 0.7 = 1.638r^2\)

The area of triangle OAC is:

\(\text{Area of } \triangle OAC = \frac{1}{2} r^2 \sin(\pi - 2.8) = 0.4927r^2\)

The area of the shaded region is:

\(1.4r^2 - (1.638r^2 - 0.985r^2) = 0.747r^2 \text{ to } 0.748r^2\)

(a) To find the perimeter of the shaded segment, we first determine the radius of the circle. The perimeter of the sector is given by the sum of the arc length and twice the radius:

\(2r + 8 = 20\)

Solving for \(r\), we get \(r = 6\) cm.

The angle \(\angle AOB\) in radians is given by:

\(\frac{8}{6} = \frac{4}{3} \text{ radians}\)

Using the cosine rule in triangle \(OAB\), the length of \(AB\) is:

\(AB = \sqrt{6^2 + 6^2 - 2 \times 6 \times 6 \times \cos \left( \frac{4}{3} \right)}\)

Calculating this gives \(AB \approx 7.42\) cm.

Thus, the perimeter of the shaded segment is:

\(7.42 + 8 = 15.4 \text{ cm}\)

(b) To find the area of the shaded segment, we calculate the area of the sector and subtract the area of triangle \(OAB\).

The area of the sector is:

\(\frac{1}{2} \times 6^2 \times \frac{4}{3}\)

The area of triangle \(OAB\) is:

\(\frac{1}{2} \times 6 \times 6 \times \sin \left( \frac{4}{3} \right)\)

Calculating these gives the sector area as \(24\) cm² and the triangle area as \(17.49\) cm².

Thus, the area of the shaded segment is:

\(24 - 17.49 = 6.51 \text{ cm}^2\)

First, find the length of BC using the tangent of angle 60 degrees (\(\frac{\pi}{3}\) radians):

\(\tan 60 = \frac{BC}{6}\)

\(BC = 6\sqrt{3}\)

Next, calculate the area of triangle OBC:

\(\text{Area of } \triangle OBC = \frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\)

Calculate the area of sector AOC:

\(\text{Area of sector } AOC = \frac{1}{2} \times 6^2 \times \frac{\pi}{3} = 6\pi\)

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

\(\text{Area of shaded region} = 18\sqrt{3} - 6\pi\)

(i) To find the area of the shaded region, we calculate the area of the sector and subtract the area of triangle OCD.

The area of the sector is given by \(\frac{1}{2} r^2 \theta\), where \(r = 6 \text{ cm}\) and \(\theta = 0.8 \text{ radians}\). Thus, the area of the sector is \(\frac{1}{2} \times 6^2 \times 0.8 = 14.4 \text{ cm}^2\).

The area of triangle OCD is given by \(\frac{1}{2} ab \sin C\), where \(a = b = 10 \text{ cm}\) and \(C = 0.8 \text{ radians}\). Thus, the area of the triangle is \(\frac{1}{2} \times 10^2 \times \sin 0.8 = 35.9 \text{ cm}^2\).

The shaded area is \(14.4 - 35.9 = 21.5 \text{ cm}^2\).

(ii) To find the perimeter of the shaded region, we calculate the arc length and the length of CD.

The arc length is given by \(s = r \theta\), where \(r = 6 \text{ cm}\) and \(\theta = 0.8 \text{ radians}\). Thus, the arc length is \(6 \times 0.8 = 4.8 \text{ cm}\).

The length of CD can be found using the cosine rule or \(2 \times 10 \times \sin 0.4\). Thus, \(CD = 7.8 \text{ cm}\).

The perimeter of the shaded region is \(8 + 4.8 + 7.8 = 20.6 \text{ cm}\).

(i) The perimeter of the sector is given by the sum of the two radii and the arc length: \(2r + r\theta = 20\). Solving for \(\theta\), we have:

\(20 = 2r + r\theta\)

\(r\theta = 20 - 2r\)

\(\theta = \frac{20}{r} - 2\)

(ii) The area of the sector is given by \(\frac{1}{2}r^2\theta\). Substituting \(\theta = \frac{20}{r} - 2\), we get:

\(A = \frac{1}{2}r^2\left(\frac{20}{r} - 2\right)\)

\(A = \frac{1}{2}(20r - 2r^2)\)

\(A = 10r - r^2\)

(iii) Using the cosine rule in triangle \(OPQ\), where \(r = 8\) and \(\theta = 0.5\) radians:

\(PQ^2 = 8^2 + 8^2 - 2 \times 8 \times 8 \times \cos(0.5)\)

\(PQ^2 = 64 + 64 - 128 \times \cos(0.5)\)

\(PQ = \sqrt{128 - 128 \times \cos(0.5)}\)

Alternatively, using the sine rule:

\(PQ = 2 \times 8 \times \sin(0.25)\)

\(PQ \approx 3.96\) (allow 3.95)