Angle \(AOB\) is at the centre of the circle and angle \(ACB\) is at the circumference. Both stand on the same arc \(AB\).

The angle at the centre is twice the angle at the circumference, so the angle at the circumference is half the angle at the centre.

\(x = 130 \div 2 = 65^\circ\)

The mark is for naming the rule. "The angle at the centre is twice the angle at the circumference" earns it.

Saying the two angles "stand on the same arc" describes the picture without naming a rule, so on its own it earns nothing.

Angle \(ACB\) and angle \(ADB\) both stand on the chord \(AB\), and \(C\) and \(D\) are on the same side of it — the same segment of the circle.

Angles in the same segment are equal, so \(x\) is the same as angle \(ACB\).

\(x = 35^\circ\)

The mark is for naming the rule: "angles in the same segment are equal".

The wording matters more here than in most of these questions, because the rule only holds while \(C\) and \(D\) are on the same side of \(AB\). On opposite sides the two angles add up to \(180^\circ\) instead.

\(AB\) is a diameter, so \(C\) sits on a semicircle standing on \(AB\).

\(\text{angle } ACB = 90^\circ\)

The mark is for naming the rule: "the angle in a semicircle is a right angle".

"Because \(AB\) goes through the centre" is the reason the semicircle is there, not the rule itself, so it does not earn the mark on its own.

The three angles of triangle \(ABC\) add up to \(180^\circ\).

\(x = 180 - 90 - 59\)

\(x = 31^\circ\)

\(ABCD\) is a cyclic quadrilateral, because all four vertices lie on the circle.

Angle \(DAB\) and angle \(BCD\) are opposite angles, so they add up to \(180^\circ\).

\(x = 180 - 78 = 102^\circ\)

The mark is for naming the rule: "opposite angles of a cyclic quadrilateral add up to \(180^\circ\)".

"Angles in a quadrilateral add up to \(360^\circ\)" is true of every quadrilateral and does not get you to a single angle here, so it is not the rule being used.

Angle \(ADC\) is opposite angle \(ABC\), which is marked \(101^\circ\).

Opposite angles of a cyclic quadrilateral add up to \(180^\circ\).

\(\text{angle } ADC = 180 - 101 = 79^\circ\)

\(OA\) is a radius and \(PA\) is a tangent touching the circle at \(A\).

\(\text{angle } OAP = 90^\circ\)

The mark is for naming the rule: "a tangent is perpendicular to the radius at the point of contact".

It is the radius drawn to the point where the tangent touches that is at right angles — not any other radius — so the wording is worth getting whole.

\(OAPB\) is a quadrilateral, so its four angles add up to \(360^\circ\).

Angle \(OAP\) and angle \(OBP\) are both \(90^\circ\), and angle \(APB\) is \(60^\circ\).

\(x = 360 - 90 - 90 - 60\)

\(x = 120^\circ\)

\(OA\) and \(OB\) are radii, so triangle \(OAB\) is isosceles and angle \(OBA = \text{angle } OAB = 59^\circ\).

\(OB\) and \(OC\) are radii, so triangle \(OBC\) is isosceles and angle \(OBC = \text{angle } OCB = 28^\circ\).

\(\text{angle } ABC = 59^\circ + 28^\circ = 87^\circ\)

If \(AC\) were a diameter, the angle in a semicircle, angle \(ABC\), would be \(90^\circ\). Here it is \(87^\circ\), so \(AC\) is not a diameter.

No centre is marked, and the angle between the tangent and \(AB\) is at \(A\), a point on the circumference, so it is not an angle at the centre. Line 2 is the misapplied theorem; lines 1 and 3 are correct. The angle between the tangent and a chord equals the angle in the alternate segment, so angle \(ACB\) is equal to it.

Working correctly gives \(m = 64^\circ\).

\(PA\) and \(PB\) are both tangents drawn from \(P\), so \(PA = PB\) and triangle \(ABP\) is isosceles.

The two base angles \(PAB\) and \(PBA\) are therefore equal, and the three angles of the triangle add up to \(180^\circ\).

\(x = (180 - 58) \div 2\)

\(x = 61^\circ\)

The mark is for the rule about the lengths: "the two tangents from a point to a circle are equal", giving \(PA = PB\).

A triangle with two equal sides is isosceles, and that is what lets you split the remaining \((180 - 58)^\circ\) equally between the two base angles.

\(x\) is the angle between the tangent and the chord \(TA\).

The segment on the other side of \(TA\) is the alternate segment, and \(B\) is in it, so angle \(TBA\) is the angle in the alternate segment.

\(x = 70^\circ\)

The mark is for naming the rule: "the angle between a tangent and a chord equals the angle in the alternate segment".

The word that does the work is alternate: the matching angle is in the segment on the far side of the chord \(TA\), which is the one containing \(B\), not the segment \(x\) itself sits against.

\(OP\) and \(OQ\) are both radii, so they are equal and triangle \(POQ\) is isosceles.

The base angles of an isosceles triangle are equal, and the angles of a triangle add up to \(180^\circ\), so

\(\text{angle } PQO = \frac{ 180 - x }{2} = 90 - \frac{ x }{2}\)

\(OQ\) is a radius and the line through \(Q\) and \(R\) is a tangent at \(Q\), so the angle between them is \(90^\circ\) (a tangent is perpendicular to the radius at the point of contact).

\(\text{angle } OQR = 90^\circ\)

\(P\) and \(R\) are on opposite sides of the line \(OQ\), so the two angles are combined with a +:

\(\text{angle } PQR = 90 + \left(90 - \frac{ x }{2}\right)\)

\(= 90 + 90 - \frac{ x }{2}\)

\(= 180° - \frac{ x }{2}°\)