Here is the solution for question 14: a) State the special name of the triangle AOB. Since AB is a diameter and O is the center, points A, O, and B are collinear. OA and OB are both radii, so OA = OB. A triangle with two equal sides is isosceles. Since the points are collinear, it is a degenerate isosceles triangle. b) Find the value of angle AOB. Since AB is a diameter and O is the center, A, O, and B lie on a straight line. The angle on a straight line is $180^\circ$. Therefore, $\angle AOB = \boxed{180^\circ}$. c) Find the value of angle AOC. The angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the circumference. Arc AC subtends $\angle AOC$ at the center and $\angle ABC$ at the circumference. Given $\angle ABC = 30^\circ$. $$ \angle AOC = 2 \times \angle ABC $$ $$ \angle AOC = 2 \times 30^\circ $$ $$ \angle AOC = \boxed{60^\circ} $$ d) If the radius of the circle is 7cm, calculate the length of the arc AC. Step 1: Identify the given values. Radius $r = 7 \text{ cm}$. Angle subtended by arc AC at the center, $\angle AOC = 60^\circ$. Step 2: Use the formula for arc length. The length of an arc $L$ is given by the formula: $$ L = \frac{\theta}{360^\circ} \times 2\pi r $$ where $\theta$ is the central angle in degrees. Step 3: Substitute the values into the formula. $$ L_{AC} = \frac{60^\circ}{360^\circ} \times 2\pi (7 \text{ cm}) $$ Step 4: Simplify the expression. $$ L_{AC} = \frac{1}{6} \times 14\pi \text{ cm} $$ $$ L_{AC} = \frac{14\pi}{6} \text{ cm} $$ $$ L_{AC} = \boxed{\frac{7\pi}{3} \text{ cm}} $$ That's 2 down. 3 left today — send the next one.