(a) To find angle BAO, use the cosine rule in triangle AOB:

\(\cos BAO = \frac{8^2 + 12^2 - 8^2}{2 \times 8 \times 12}\)

\(\cos BAO = \frac{6}{8}\)

Thus, \(BAO = 0.723\) radians.

(b) The area of the shaded region is the area of sector ABC minus the area of triangle AOB.

Area of sector ABC:

\(\frac{1}{2} \times 12^2 \times 0.7227 = 52.0\)

Area of triangle AOB:

\(\frac{1}{2} \times 8 \times 12 \times \sin(0.7227) = 31.7\)

Shaded area:

\(52.0 - 31.7 = 20.3\)

(c) The perimeter of the shaded region is the sum of arc BC and line segments OC and AB.

Arc BC:

\(12 \times 0.7227 = 8.67\)

Perimeter:

\(8 + 4 + 8.67 = 20.7\)

(a) The area of triangle ABC can be calculated using the formula for the area of a triangle: \(\frac{1}{2} r^2 \sin(2θ)\).

The area of sector ABD is \(\frac{1}{2} r^2 θ\).

Therefore, the area of the shaded region is the area of triangle ABC minus the area of sector ABD: \(\frac{1}{2} r^2 \sin(2θ) - \frac{1}{2} r^2 θ = \frac{1}{2} r^2 (\sin 2θ - θ)\).

(b) The arc BD is given by \(rθ = 10 \times 0.6 = 6\) cm.

The length AC is \(2r \cos θ = 2 \times 10 \times \cos 0.6 = 16.506\) cm.

The length DC is \(AC - r = 16.506 - 10 = 6.506\) cm.

The perimeter of the shaded region is the sum of the arc BD, and the lengths DC and CD: \(6 + 6.506 + 10 = 22.5\) cm.

(a) Using the relationship \(\sin \theta = \frac{r}{OC} \\), we have \(OC = \frac{r}{\sin \theta} \\). Therefore, \(CD = r + \frac{r}{\sin \theta} \\).

(b) The radius of arc AB is \(4 + \frac{4}{\sin \frac{\pi}{6}} = 4 + 8 = 12 \\). The arc length \(AB = 12 \times \frac{2\pi}{6} = 4\pi \\). Thus, the perimeter is \(24 + 4\pi \\).

(c) The area of \(\triangle FOC = \frac{1}{2} \times 4 \times OC \times \sin \frac{\pi}{3} = 8\sqrt{3} \\). The area of sector \(FOE = \frac{1}{2} \times \frac{2\pi}{3} \times 4^2 = \frac{16\pi}{3} \\). Therefore, the shaded area is \(16\sqrt{3} - \frac{16\pi}{3} \\).

First, calculate the angle \(\angle POA\) using the cosine rule:

\(\cos \angle POA = \frac{5}{13}\)

\(\angle POA = \cos^{-1} \left( \frac{5}{13} \right) \approx 1.17 \text{ radians}\) (or 67.4°)

The reflex angle \(\angle AOB = 2\pi - 2 \times 1.17 \approx 3.93 \text{ radians}\).

The length of the major arc AB is:

\(5 \times 3.93 = 19.65 \text{ cm}\)

Calculate the length of AP (or BP) using the Pythagorean theorem:

\(AP = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}\)

The total length of the cord is:

\(19.65 + 12 + 12 = 43.7 \text{ cm}\)

(a) To find \(BC\), use the cosine rule in triangle \(OBC\):

\(BC^2 = r^2 + (2r)^2 - 2 \cdot r \cdot 2r \cdot \cos\left(\frac{\pi}{6}\right)\)

\(BC^2 = r^2 + 4r^2 - 4r^2 \cdot \frac{\sqrt{3}}{2}\)

\(BC^2 = 5r^2 - 2r^2\sqrt{3}\)

\(BC = r\sqrt{5 - 2\sqrt{3}}\)

(b) The perimeter of the shaded region is the arc length of \(AB\) plus \(AC\) and \(BC\):

Arc length \(AB = \frac{2\pi r}{6}\)

\(AC = r\)

Perimeter = \(\frac{2\pi r}{6} + r + r\sqrt{5 - 2\sqrt{3}}\)

(c) The area of the shaded region is the area of the sector minus the area of triangle \(OAB\):

Sector area = \(\frac{1}{2} \cdot 4r^2 \cdot \frac{\pi}{6}\)

Triangle area = \(\frac{1}{2} \cdot r \cdot 2r \cdot \sin\left(\frac{\pi}{6}\right)\)

Shaded area = \(r^2 \left( \frac{\pi}{3} - \frac{1}{2} \right)\)