<p(a)(i)(a) Diameter
(a)(i)(b) Tangent
(a)(ii) Angle between tangent and radius = 90
Angles in a triangle add to 180
(a)(iii) Opposite angles are equal
(a)(iv)(a) 90
(a)(iv)(b) 59
(a)(iv)(c) 149
(b) Equal sides
Equal angles
(c) 144

Detailed Solution 

(a) (i)
(a) Line BOD: This is a diameter because it passes through the center O and touches two points on the circumference (B and D).
(b) Line ABC: This is a tangent because it touches the circle at only one point (B) and does not enter the circle.

(ii)

1. The angle between a tangent and the radius at the point of contact is always 90°
The line AC is a tangent to the circle at B.
The radius OB meets the tangent at B, 
$
\angle OBA = 90^\circ
$

2. The sum of the angles in a triangle is 180°
 In triangle AOB, 
$
\angle OAB = 28^\circ, \quad \angle OBA = 90^\circ
$
 the sum of angles in a triangle
$
\angle AOB = 180^\circ – 90^\circ – 28^\circ = 62^\circ
$

(iii)

Opposite Angles
$
\angle DOE = \angle AOB = 62^\circ
$

(a)(iv)(a) If an angle subtended by a chord on the circumference of a circle is 90 degrees, then the chord is the circle’s diameter

(a)(iv)(b)

Triangle ODE is isosceles because OD and OE are radii of the circle,
$
\angle ODE = \angle OED
$
 The sum of angles in triangle ODE is 180°:
$
\angle DOE + \angle ODE + \angle OED = 180^\circ
$
$
62^\circ + \angle ODE + \angle ODE = 180^\circ
$
$
2 \times \angle ODE = 180^\circ – 62^\circ
$
$
2 \times \angle ODE = 118^\circ
$
$
\angle ODE = 59^\circ
$

(a)(iv)(c)

Angles on a straight line sum to 180°. Since ∠OEB and ∠BEF lie on line DF,
$
\angle OEB + \angle BEF = 180^\circ
$
∠OEB = 31°
$
31^\circ + \angle BEF = 180^\circ
$
$
\angle BEF = 180^\circ – 31^\circ = 149^\circ
$

(b) Two geometrical properties of a regular polygon
All sides are equal
All interior angles are equal

(c)
$
\text{Interior Angle} = \frac{(n-2) \times 180^\circ}{n}
$
For (\( n = 10 \))
$
\text{Interior Angle} = \frac{(10-2) \times 180}{10}
$
$
= \frac{8 \times 180}{10}
$
$
= \frac{1440}{10} = 144^\circ
$