OCR GCSE Mathematics: Angle facts, polygons, triangle and quadrilateral properties, and symmetry (J560)

This guide is written for OCR GCSE (9-1) Mathematics, specification Version 2.1 (August 2026), sections 8.03 (angles) and 8.04 (properties of polygons). Most of the content is for both tiers. Where the specification places something in the Higher tier column only, it is labelled Higher tier only. For quick recall use the revision notes, then test yourself with the practice questions. The course hub, the checklist and the free diagnostics show where you stand.

What this unit covers

Ref	Skill	Tier
8.03a	Angles at a point sum to 360°	Both
8.03b	Angles on a line sum to 180°	Both
8.03c	Vertically opposite, alternate and corresponding angles	Both
8.03d	Derive the triangle angle sum and the exterior angle sum; find interior angle sums and the interior angle of a regular polygon	Both
8.04a	Properties of isosceles, equilateral and right-angled triangles, with reasons	Both
8.04b	Properties of the square, rectangle, parallelogram, trapezium, kite and rhombus, with reasons	Both
8.04c	Reflection and rotation symmetry of triangles, quadrilaterals and other polygons	Both
8.03 and 8.04	Use the facts to find angles and lengths in rectilinear figures and in simple proofs	Both
8.03 and 8.04	Use the facts in more formal proofs of geometrical results	Higher tier only

Angles at a point and on a line (8.03a, 8.03b)

Angles around a point add up to 360°. Angles that meet at a point on a straight line add up to 180°.

Example 1. Four angles meet at a point: 3x°, (2x + 20)°, (x + 40)° and a right angle. Find x.

3x + (2x + 20) + (x + 40) + 90 = 360    (angles at a point)
6x + 150 = 360
x = 35

The angles are 105°, 90°, 75° and 90°, which total 360°.

Example 2. Two angles on a straight line are (5y − 15)° and (3y + 35)°.

(5y − 15) + (3y + 35) = 180    (angles on a line)
8y + 20 = 180
y = 20

The angles are 85° and 95°.

Parallel lines (8.03c)

Three facts, each with a name you must quote as the reason:

• Vertically opposite angles are equal (where two straight lines cross).
• Alternate angles on parallel lines are equal (the Z shape).
• Corresponding angles on parallel lines are equal (the F shape).

A fourth fact follows from them: co-interior angles (the C or U shape) add up to 180°. The proof is one line. A corresponding angle equals its partner, and the partner plus its neighbour on the line makes 180°.

        E
        |
  A ----G---------- B     AB is parallel to CD
        |
  C ----H---------- D
        |
        F

Example 3. In the diagram, AB is parallel to CD and EF crosses both lines at G and H. Angle EGB = 112°. Find angles AGH, GHD and BGH, giving a reason each time.

Angle AGH = 112°   (vertically opposite to angle EGB)
Angle GHD = 112°   (alternate to angle AGH, or corresponding to angle EGB)
Angle BGH = 68°    (angles on a line: 180 − 112)

Always write the reason beside the angle. Never write “Z angles” as a reason: use the word alternate.

Angles in polygons (8.03d)

Triangle angle sum (derive). Draw a line through one vertex parallel to the opposite side. The two angles it makes with the sides are alternate to the two base angles. Those two angles and the angle at the top lie together on a straight line, so the three angles of the triangle add up to 180°.

Exterior angle of a triangle. Extend one side. The exterior angle and the interior angle next to it are on a line, so they add to 180°. The interior angles of the triangle add to 180° too, so the exterior angle equals the sum of the two opposite interior angles.

Exterior angles of a polygon (derive). Imagine walking round a polygon. At each corner you turn through the exterior angle. After the last corner you face the way you started, having turned through exactly one full turn. So the exterior angles of any polygon sum to 360°.

Interior angle sum. Each interior angle and its exterior angle make 180°. For a polygon with n sides:

sum of interior angles = 180n − 360 = (n − 2) × 180°

You can also reach this by splitting the polygon into (n − 2) triangles from one vertex.

Regular polygons have equal sides and equal angles, so:

exterior angle = 360° ÷ n
interior angle = 180° − exterior angle

Regular polygon	Exterior angle	Interior angle	Interior angle sum
15 sides	24°	156°	2340°
20 sides	18°	162°	3240°
30 sides	12°	168°	5040°

Example 4. Each interior angle of a regular polygon is 168°. How many sides has it?

exterior angle = 180 − 168 = 12°
n = 360 ÷ 12 = 30

Example 5. A hexagon has angles 128°, 117°, 140°, 95°, 2a° and (a + 30)°. Find a.

sum of interior angles = (6 − 2) × 180 = 720°
128 + 117 + 140 + 95 + 2a + a + 30 = 720
510 + 3a = 720
a = 70

The last two angles are 140° and 100°.

Triangle properties (8.04a)

• Isosceles: two equal sides, and the angles opposite them (the base angles) are equal. Reason: the triangle is symmetrical about the line from the apex to the midpoint of the base, and that line meets the base at right angles.
• Equilateral: three equal sides and three angles of 60°. Reason: it is isosceles from every vertex, so all three angles are equal, and they sum to 180°.
• Right-angled: one angle is 90°, so the other two add up to 90°. The longest side is the one opposite the right angle.

Example 6. In triangle PQR, PQ = PR. S lies on QR extended beyond R, and angle PRS = 124°. Find angle QPR.

Angle PRQ = 180 − 124 = 56°    (angles on a line)
Angle PQR = 56°                (base angles of an isosceles triangle)
Angle QPR = 180 − 56 − 56 = 68°  (angles in a triangle)

Example 7. ABCD is a square. Triangle ABE is equilateral and E is inside the square. Find angle AED.

Angle DAE = 90 − 60 = 30°
AD = AB = AE, so triangle ADE is isosceles with apex A
Angle AED = (180 − 30) ÷ 2 = 75°

Here a length fact (all sides of a square are equal) is what gives the angle.

Quadrilateral properties (8.04b)

The four angles of any quadrilateral sum to 360°: a diagonal splits it into two triangles, and 2 × 180° = 360°.

Shape	Sides	Angles	Diagonals
Square	4 equal; opposite sides parallel	All 90°	Equal; bisect each other at 90°; bisect the angles
Rectangle	Opposite sides equal and parallel	All 90°	Equal; bisect each other
Parallelogram	Opposite sides equal and parallel	Opposite angles equal	Bisect each other
Rhombus	4 equal; opposite sides parallel	Opposite angles equal	Bisect each other at 90°; bisect the angles
Kite	Two pairs of adjacent equal sides	One pair of opposite angles equal	Meet at 90°; one is bisected by the other
Trapezium	One pair of parallel sides	Co-interior angles between the parallel sides sum to 180°	No special property

An isosceles trapezium also has equal non-parallel sides and equal base angles.

Example 8. In parallelogram ABCD, angle A = (3x + 12)° and angle B = (2x + 18)°. Find angle C.

Angles A and B are co-interior between parallel lines, so they sum to 180°
5x + 30 = 180
x = 30, so angle A = 102° and angle B = 78°
Angle C = angle A = 102°   (opposite angles of a parallelogram)

Example 9. A rhombus ABCD has angle BAD = 74°. The diagonals meet at M. Find angles ABD and BCA.

Diagonal AC bisects angle BAD, so angle BAC = 37°
Triangle ABC is isosceles (AB = BC), so angle BCA = 37°
Angle ABC = 180 − 74 = 106° (co-interior), and BD bisects it, so angle ABD = 53°
Check: in triangle ABM, 37 + 90 + 53 = 180°

Example 10. Kite ABCD has AB = AD and CB = CD. Angle BAD = 64° and angle BCD = 118°. Find angle ABC.

Angles ABC and ADC are equal (the kite's one pair of equal opposite angles)
Sum of angles = 360°, so 2 × angle ABC = 360 − 64 − 118 = 178
Angle ABC = 89°

Symmetry (8.04c)

A shape has reflection symmetry if a mirror line maps it onto itself. It has rotation symmetry if turning it about its centre by less than 360° fits it onto itself. The order is the number of positions in a full turn, counting the start. Order 1 means no rotation symmetry. The turn between positions is 360° ÷ order.

Shape	Lines of symmetry	Order of rotation
Equilateral triangle	3	3
Isosceles triangle	1	1
Scalene triangle	0	1
Square	4	4
Rectangle	2	2
Rhombus	2	2
Parallelogram	0	2
Kite	1	1
Isosceles trapezium	1	1
Regular n-sided polygon	n	n

Example 11. A regular polygon has interior angle 150°. Exterior angle = 30°, so n = 360 ÷ 30 = 12. It has 12 lines of symmetry and rotation symmetry of order 12.

Proofs

A proof is a chain of statements, each with a reason taken from the facts above. Both tiers need to justify results in simple proofs.

Proof A (both tiers). The angles of a quadrilateral ABCD sum to 360°.

Draw diagonal AC, which splits ABCD into triangles ABC and ACD
Angles of triangle ABC sum to 180°   (angle sum of a triangle)
Angles of triangle ACD sum to 180°   (angle sum of a triangle)
The angles at A and C are split between the triangles, so the four angles of ABCD total 180 + 180 = 360°

Higher tier only: more formal proofs. The specification gives circle theorems as an example of results proved from these facts. This proof uses only isosceles triangles and the triangle angle sum.

Proof B (Higher tier only). AB is a diameter of a circle with centre O, and C is on the circle. Prove that angle ACB = 90°.

OA = OB = OC            (radii of the same circle)
Triangle OAC is isosceles, so angle OCA = angle OAC = p    (base angles)
Triangle OBC is isosceles, so angle OCB = angle OBC = q    (base angles)
Angles of triangle ABC: p + q + (p + q) = 180°             (angle sum of a triangle)
So 2(p + q) = 180° and angle ACB = p + q = 90°

Common errors

• Giving an angle with no reason, or naming the wrong fact (alternate and corresponding are not interchangeable).
• Using 360° where the angles sit on a line (180°), or the reverse.
• Treating a diagram as accurate. A diagram marked “not drawn accurately” cannot be measured.
• Dividing 360 by the interior angle instead of the exterior angle.
• Forgetting that a rhombus is not a square, and that a parallelogram has no line of symmetry.
• Writing a proof that shows an example but never states the general reason.

Next steps

Work through the practice questions, keep the revision notes open for recall, and tick off progress on the checklist.

Official syllabus

OCR, GCSE (9-1) Mathematics J560, specification Version 2.1 (August 2026), for first assessment in 2017: sections 8.03 and 8.04.