(i) To find angle BAC, use the cosine rule in triangle ABC:

\(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)

where a = 16, b = 10, c = 10. Substituting the values:

\(\cos A = \frac{10^2 + 10^2 - 16^2}{2 \times 10 \times 10} = \frac{100 + 100 - 256}{200} = \frac{-56}{200} = -0.28\)

Thus,

\(A = \cos^{-1}(-0.28) \approx 0.6435 \text{ radians}\)

(ii) To find the area of the shaded region, calculate the area of triangle ABC and subtract it from the sum of the areas of the sectors:

Area of triangle ABC:

\(\text{Area} = \frac{1}{2} \times 16 \times 10 \times \sin(0.6435) = 48\)

Area of one sector (e.g., sector ABE):

\(\text{Area} = \frac{1}{2} \times 10^2 \times 0.6435 = 32.175\)

Total area of two sectors:

\(2 \times 32.175 = 64.35\)

Shaded area:

\(\text{Shaded area} = 64.35 - 48 = 16.35 \approx 16.4\)

(i) To find angle ABC, use the sine function in the right triangle ABC:

\(\sin \angle ABC = \frac{AC}{BC} = \frac{8}{10}\)

\(\angle ABC = \arcsin\left(\frac{8}{10}\right) \approx 0.927 \text{ radians}\)

(ii) First, find AB using the Pythagorean theorem:

\(AB = \sqrt{BC^2 - AC^2} = \sqrt{10^2 - 8^2} = 6 \text{ cm}\)

Calculate the area of triangle BCD:

\(\text{Area of } \triangle BCD = \frac{1}{2} \times 8 \times 6 = 48 \text{ cm}^2\)

Calculate the area of sector BCD:

\(\text{Area of sector } BCD = \frac{1}{2} \times 10^2 \times 2 \times 0.927 = 92.73 \text{ cm}^2\)

Calculate the area of segment BCD:

\(\text{Area of segment } BCD = 92.73 - 48 = 44.73 \text{ cm}^2\)

Calculate the area of the semicircle with radius 8 cm:

\(\text{Area of semicircle} = \frac{1}{2} \times \pi \times 8^2 = 100.53 \text{ cm}^2\)

Finally, find the shaded area:

\(\text{Shaded area} = 100.53 - 44.73 = 55.8 \text{ cm}^2\)

(i) To find the perimeter of the shaded region, we need to consider the lengths of AB, the arc AB, and the sides AD and DC of the rectangle.

The length of AB can be found using the formula: \(AB = 2r\sin\theta\).

The length of the arc AB is given by: \(2r\theta\).

The perimeter P of the shaded region is: \(P = AD + DC + AB + \text{arc } AB = 2r + 2r\theta + 2r\sin\theta\).

(ii) For r = 5 and θ = \(\frac{1}{6} \pi\), we calculate the area of the shaded region.

The area of sector AOB is \(\frac{1}{2} r^2 (2\theta) = \frac{25\pi}{6}\) or 13.1.

The area of triangle AOB is \(\frac{1}{2} \times 2r\sin\theta \times r\cos\theta = \frac{25\sqrt{3}}{4}\) or 10.8.

The area of rectangle ABCD is \(r \times 2r\sin\theta = 25\).

The area of the shaded region is: \(25 - (\frac{25\pi}{6} - \frac{25\sqrt{3}}{4})\) or 22.7.

(i) Letting \(M\) be the midpoint of \(AB\), we have \(OM = 8\) cm using the Pythagorean theorem. Therefore, \(XM = 2\) cm. Using trigonometry, \(\tan AXM = \frac{6}{2}\), so \(AXB = 2 \times \tan^{-1}(3) = 2.498\) radians.

(ii) The length \(AX = \sqrt{6^2 + 2^2} = \sqrt{40}\). The arc \(AYB = r\theta = \sqrt{40} \times 2.498\). The perimeter is \(12 + \text{arc} = 27.8\) cm.

(iii) The area of sector \(AXBY = \frac{1}{2} \times (\sqrt{40})^2 \times 2.498\). The area of triangle \(AXB = \frac{1}{2} \times 12 \times 2\). Subtract these to find the shaded area: 38.0 cm\(^2\).

(i) To find \(\angle CBD\), we first find \(\angle ABC\). Since \(\angle CAB = \frac{2}{7} \pi\), and \(\triangle ABC\) is isosceles with \(AB = AC\), the remaining angle \(\angle ABC\) is:

\(\angle ABC = \pi - \angle CAB = \pi - \frac{2}{7} \pi = \frac{5}{7} \pi\)

Since \(ABD\) is a straight line, \(\angle CBD = \pi - \angle ABC = \pi - \frac{5}{7} \pi = \frac{9}{14} \pi\).

(ii) To find the perimeter of the shaded region, we need to calculate the lengths of \(BC\), arc \(CD\), and arc \(CB\).

Using the sine rule in \(\triangle ABC\):

\(\frac{\sin \frac{\pi}{7}}{8} = \frac{\sin \frac{2\pi}{7}}{BC}\)

Solving for \(BC\), we find \(BC \approx 6.94 \text{ cm}\).

The length of arc \(CD\) is given by:

\(\text{arc } CD = BC \times \frac{9\pi}{14} \approx 6.94 \times \frac{9\pi}{14} \approx 14.02 \text{ cm}\)

The length of arc \(CB\) is given by:

\(\text{arc } CB = 8 \times \frac{2\pi}{7} \approx 7.18 \text{ cm}\)

Thus, the perimeter of the shaded region is:

\(6.94 + 14.02 + 7.18 = 28.1 \text{ cm}\)