Angles and Polygons

Angles are one of the most fundamental ideas in geometry. For Edexcel GCSE Mathematics (1MA1), you need to be confident with angle facts on lines and at points, angles in triangles and quadrilaterals, and interior and exterior angles of regular and irregular polygons. This lesson covers all of these topics with worked examples and Edexcel-style exam practice.

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Key Vocabulary

Term	Meaning	Example
Acute angle	An angle less than 90°	35°, 60°
Right angle	An angle of exactly 90°	Corner of a square
Obtuse angle	An angle between 90° and 180°	120°, 150°
Reflex angle	An angle between 180° and 360°	210°, 300°
Polygon	A closed 2D shape with straight sides	Triangle, hexagon
Regular polygon	A polygon where all sides and all angles are equal	Equilateral triangle, square
Interior angle	An angle inside a polygon at a vertex	—
Exterior angle	The angle between one side and the extension of the adjacent side	—

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Basic Angle Facts

These are the building blocks you must memorise — they are not given on the Edexcel formula sheet.

Angles on a Straight Line

Angles on a straight line add up to 180°.

Worked Example 1

Two angles on a straight line are x° and 130°. Find x.

x + 130 = 180 x = 180 - 130 = 50°

Angles at a Point

Angles around a point add up to 360°.

Worked Example 2

Three angles around a point are 140°, 85° and y°. Find y.

140 + 85 + y = 360 225 + y = 360 y = 135°

Vertically Opposite Angles

When two straight lines cross, the opposite angles are equal.

Worked Example 3

Two straight lines cross. One of the angles is 72°. State the sizes of the other three angles.

The vertically opposite angle = 72°. The other pair of vertically opposite angles = 180 - 72 = 108° each.

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Angles in Triangles

The angles in any triangle add up to 180°.

Special Triangles

Triangle	Properties
Equilateral	All sides equal, all angles 60°
Isosceles	Two sides equal, two base angles equal
Scalene	No sides or angles equal
Right-angled	One angle of 90°

Worked Example 4

An isosceles triangle has an angle of 40° between the two equal sides. Find the base angles.

Let each base angle = b. 40 + b + b = 180 40 + 2b = 180 2b = 140 b = 70°

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Angles in Quadrilaterals

The angles in any quadrilateral add up to 360°.

Worked Example 5

Three angles of a quadrilateral are 90°, 85° and 110°. Find the fourth angle.

90 + 85 + 110 + d = 360 285 + d = 360 d = 75°

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Exterior Angles of a Triangle

An exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Worked Example 6

In triangle PQR, angle P = 55° and angle Q = 70°. Side QR is extended to point S. Find the exterior angle PRS.

Exterior angle PRS = P + Q = 55 + 70 = 125°

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Interior and Exterior Angles of Polygons

Sum of Interior Angles

For a polygon with n sides:

Sum of interior angles = $(n - 2) \times 180°$(n−2)×180°

Polygon	Sides (n)	Sum of interior angles
Triangle	3	$(3-2) \times 180 = 180°$(3−2)×180=180°
Quadrilateral	4	$(4-2) \times 180 = 360°$(4−2)×180=360°
Pentagon	5	$(5-2) \times 180 = 540°$(5−2)×180=540°
Hexagon	6	$(6-2) \times 180 = 720°$(6−2)×180=720°
Octagon	8	$(8-2) \times 180 = 1080°$(8−2)×180=1080°
Decagon	10	$(10-2) \times 180 = 1440°$(10−2)×180=1440°

Each Interior Angle of a Regular Polygon

Each interior angle = $(n - 2) \times 180°$(n−2)×180° / n

Worked Example 7

Find the size of each interior angle of a regular octagon.

Sum = $(8-2) \times 180 = 1080°$(8−2)×180=1080° Each angle = 1080 / 8 = 135°

Exterior Angles

The exterior angles of any convex polygon always add up to 360°.

For a regular polygon: Each exterior angle = $\frac{360°}{n}$n360°​

Important: Interior angle + Exterior angle = 180° (they form a straight line).

Worked Example 8

Each exterior angle of a regular polygon is 40°. How many sides does it have?

n = 360 / 40 = 9 sides (nonagon)

Worked Example 9

Each interior angle of a regular polygon is 156°. Find the number of sides.

Exterior angle = 180 - 156 = 24° n = 360 / 24 = 15 sides

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Angles in Parallel Lines

When a transversal crosses two parallel lines, the following angle relationships hold:

Type	Rule	How to spot
Alternate angles	Equal	Z-shape (or reversed Z)
Corresponding angles	Equal	F-shape (or reversed F)
Co-interior (allied) angles	Add up to 180°	C-shape or U-shape

Worked Example 10

A transversal crosses two parallel lines. One of the alternate angles is 64°. Find the co-interior angle on the same side.

Alternate angle = 64°, so the other alternate angle = 64°. Co-interior angle = 180 - 64 = 116°

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Common Mistakes and Misconceptions

• Forgetting that angle sums work for irregular polygons too. The formula $(n - 2) \times 180°$(n−2)×180° applies to all polygons, not just regular ones.
• Confusing interior and exterior angles. Remember: they add up to 180° at each vertex.
• Using the wrong angle fact on parallel lines. Check the shape formed: Z = alternate, F = corresponding, C/U = co-interior.
• Writing "angles on a straight line = 360°." No — angles on a straight line = 180°; angles at a point = 360°.
• Not giving reasons in exam answers. Edexcel requires you to state the angle fact used (e.g. "co-interior angles sum to 180°").

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Edexcel Exam Tips

• Always state which angle fact you are using — marks are awarded for the reasoning, not just the numerical answer.
• "Give reasons for your answer" means you must name the geometric property (e.g. "vertically opposite angles are equal").
• Polygon interior angle sums are NOT on the Edexcel formula sheet — you must memorise $(n - 2) \times 180°$(n−2)×180°.
• If a question involves parallel lines, look for arrows on the diagram indicating which lines are parallel.

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Practice Problems

1. Two angles on a straight line are 3x° and (x + 20)°. Find x. Answer: 3x + x + 20 = 180, 4x = 160, x = 40.
2. The angles of a triangle are (2y + 10)°, (3y)° and (y + 20)°. Find y and each angle. Answer: 6y + 30 = 180, y = 25; angles = 60°, 75°, 45°.
3. Find the sum of the interior angles of a heptagon (7 sides). Answer: (7 - 2) x 180 = 900°.
4. Each interior angle of a regular polygon is 144°. How many sides? Answer: Exterior = 36°, n = 360/36 = 10 sides.
5. In the diagram, lines AB and CD are parallel. A transversal makes an angle of 52° with AB. Find the co-interior angle at CD. Answer: 180 - 52 = 128°.

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Summary

• Angles on a straight line = 180°; at a point = 360°; vertically opposite angles are equal.
• Angles in a triangle = 180°; in a quadrilateral = 360°.
• Sum of interior angles of an n-sided polygon = $(n - 2) \times 180°$(n−2)×180°.
• Exterior angles of any convex polygon = 360°.
• Interior + exterior angle = 180° at each vertex.
• Parallel line angles: alternate (equal), corresponding (equal), co-interior (sum to 180°).
• Always give geometric reasons in your exam answers.

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Extended Worked Examples

Example A — mixing polygon rules with parallel-line reasoning

A regular pentagon ABCDE has a line drawn from vertex A to vertex C. A separate line through C is parallel to AB. Find the angle between AC and the line through C that is parallel to AB, giving reasons.

Step 1 — interior angle of a regular pentagon. Sum of interior angles = $(5 - 2) \times 180° = 540°$(5−2)×180°=540°. Since the pentagon is regular, each interior angle $= 540° \div 5 = 108°$=540°÷5=108°.

Step 2 — identify the isosceles triangle. In triangle ABC the sides AB and BC are equal (sides of a regular polygon), so triangle ABC is isosceles. Angle ABC $= 108°$=108°. The two base angles are equal and their sum is $180° - 108° = 72°$180°−108°=72°, so angle BAC $=$= angle BCA $= 36°$=36°.

Step 3 — apply the parallel-line rule. The angle between AC and the line through C parallel to AB is the alternate angle to angle BAC (the Z-shape formed by transversal AC across the two parallel lines AB and the new line at C). Alternate angles are equal, so the required angle $= 36°$=36°.

Reasoning to write in the exam: "Interior angle of regular pentagon $= (5-2)\times 180° \div 5 = 108°$=(5−2)×180°÷5=108°. Triangle ABC is isosceles (sides of regular polygon equal), so base angles $= (180° - 108°) \div 2 = 36°$=(180°−108°)÷2=36°. Alternate angles between parallel lines are equal, so the required angle $= 36°$=36°."

Example B — exterior angle of a regular polygon

The exterior angle of a regular polygon is $15°$15°. How many sides does it have, and what is the sum of its interior angles?

Exterior angles of any convex polygon sum to $360°$360°, and for a regular polygon each exterior angle $= 360° \div n$=360°÷n. So $n = 360° \div 15° = 24$n=360°÷15°=24 sides. Sum of interior angles $= (24 - 2) \times 180° = 22 \times 180° = 3960°$=(24−2)×180°=22×180°=3960°.

Example C — finding an unknown on parallel lines

Two parallel lines are cut by a transversal. On one line an angle of $(3x + 20)°$(3x+20)° is formed on the right of the transversal; the co-interior angle on the other line is $(5x - 40)°$(5x−40)°. Find $x$x.

Co-interior angles between parallel lines sum to $180°$180°: $(3x + 20) + (5x - 40) = 180$(3x+20)+(5x−40)=180, so $8x - 20 = 180$8x−20=180, giving $8x = 200$8x=200 and $x = 25$x=25.

Example D — angles in an irregular polygon

A pentagon has four interior angles of $100°$100°, $115°$115°, $130°$130° and $95°$95°. Find the fifth interior angle.

Sum of interior angles of a pentagon $= (5 - 2) \times 180° = 540°$=(5−2)×180°=540°. Fifth angle $= 540° - (100 + 115 + 130 + 95) = 540° - 440° = 100°$=540°−(100+115+130+95)=540°−440°=100°.

Example E — bearings and parallel north lines

North lines at two points A and B are parallel. The bearing of B from A is $070°$070°. Find the bearing of A from B.

The back-bearing lies $180°$180° from the original. $070° + 180° = 250°$070°+180°=250°. Because both north lines are parallel, alternate angles confirm that the angle to the south-west at B measures the same $70°$70° as at A, and adding the $180°$180° half-turn to reverse direction gives $250°$250°.

Answering at different grade levels

Exam-style question: ABCD is a quadrilateral with AB parallel to DC. Angle DAB $= 78°$=78° and angle ABC $= 112°$=112°. (a) Find angle ADC, giving a reason. (b) Is ABCD a trapezium or a parallelogram? Justify your answer.

• Grades 3–4 answer. "Angle DAB and angle ADC are co-interior angles. They add to $180°$180°. So angle ADC $= 180 - 78 = 102°$=180−78=102°. The shape is a trapezium because only one pair of sides is parallel."
• Grades 5–6 answer. "AB is parallel to DC, so DA is a transversal cutting the two parallel lines. Angles DAB and ADC are co-interior angles, which sum to $180°$180°. Therefore angle ADC $= 180° - 78° = 102°$=180°−78°=102°. For ABCD to be a parallelogram we would need AD parallel to BC as well. Checking the other pair: angles ABC and BCD are also co-interior and sum to $180°$180°, so angle BCD $= 180° - 112° = 68°$=180°−112°=68°. Because angle DAB $\neq$= angle BCD the shape cannot be a parallelogram (opposite angles of a parallelogram are equal), so ABCD is a trapezium."
• Grades 7–9 answer. "Let AB $\parallel$∥ DC. Then DA is a transversal and angles DAB and ADC are co-interior, so angle ADC $= 180° - 78° = 102°$=180°−78°=102° (co-interior angles sum to $180°$180°). Similarly, BC is a transversal so angle BCD $= 180° - 112° = 68°$=180°−112°=68°. A parallelogram requires both pairs of opposite angles equal; here angle DAB $= 78° \neq 68° =$=78°=68°= angle BCD, so ABCD is not a parallelogram. Having exactly one pair of parallel sides, ABCD is a trapezium by definition. (Check: sum of interior angles $= 78 + 112 + 68 + 102 = 360°$=78+112+68+102=360° as required.)"

Edexcel alignment: This content is aligned with Edexcel GCSE Mathematics (1MA1) specification — specifically Topic G1 Angles, lines and parallel line geometry, G3 Angle facts and properties of triangles and quadrilaterals, and G6 Properties of polygons including interior and exterior angles of regular polygons. Assessed on Papers 1, 2, 3.