Angles: Parallel Lines, Triangles and Polygons

This lesson covers the fundamental angle facts and properties you need for AQA GCSE Mathematics. You will learn about angles on straight lines, around a point, and vertically opposite angles, then move on to angles formed by parallel lines and a transversal. Finally, you will explore the interior and exterior angles of polygons. These topics appear frequently on both Foundation and Higher tier papers.

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Basic Angle Facts

Before tackling more advanced problems, you must be confident with the core angle rules.

Rule	Statement	Diagram Hint
Angles on a straight line	Angles on a straight line add up to 180 degrees	A straight line with angles above it
Angles around a point	Angles around a point add up to 360 degrees	Angles radiating from a single point
Vertically opposite angles	Vertically opposite angles are equal	Two intersecting lines forming an X
Angles in a triangle	Angles in a triangle add up to 180 degrees	Any triangle
Angles in a quadrilateral	Angles in a quadrilateral add up to 360 degrees	Any four-sided shape

Worked Example 1

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

Solution: Angles on a straight line sum to 180 degrees. x + 130 = 180 x = 180 - 130 x = 50 degrees

Worked Example 2

Three angles meet at a point: 90 degrees, 145 degrees and y. Find y.

Solution: Angles around a point sum to 360 degrees. 90 + 145 + y = 360 235 + y = 360 y = 360 - 235 y = 125 degrees

Exam Tip: Always state the angle rule you are using when showing your working. Simply writing the calculation without naming the rule will cost you marks on "give a reason" questions.

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Angles in Parallel Lines

When a straight line (called a transversal) crosses two parallel lines, several angle relationships are formed.

graph TD
    A[Transversal crosses parallel lines] --> B[Alternate angles]
    A --> C[Corresponding angles]
    A --> D[Co-interior angles]
    B --> B1[Equal — Z shape]
    C --> C1[Equal — F shape]
    D --> D1[Sum to 180 degrees — C or U shape]

Angle Type	Relationship	Memory Aid
Alternate angles	Equal	Look for a Z shape (or reversed Z)
Corresponding angles	Equal	Look for an F shape (or rotated F)
Co-interior angles (also called allied or same-side interior)	Add up to 180 degrees	Look for a C or U shape

Worked Example 3

A transversal crosses two parallel lines. One angle is 72 degrees. Find its alternate angle and the co-interior angle on the same side.

Solution:

• Alternate angle = 72 degrees (alternate angles are equal)
• Co-interior angle = 180 - 72 = 108 degrees (co-interior angles sum to 180 degrees)

Worked Example 4

In a diagram with parallel lines, angle a = 3x + 10 and its corresponding angle b = 5x - 30. Find x.

Solution: Corresponding angles are equal: 3x + 10 = 5x - 30 10 + 30 = 5x - 3x 40 = 2x x = 20

Exam Tip: Questions often combine parallel-line rules with other angle facts. For example, you might need to use alternate angles AND angles in a triangle in the same question. Look for the chain of reasoning.

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Angles in Triangles

Types of Triangle by Angle

Type	Properties
Acute-angled	All angles less than 90 degrees
Right-angled	One angle is exactly 90 degrees
Obtuse-angled	One angle is greater than 90 degrees

Types of Triangle by Side

Type	Properties
Equilateral	All three sides equal; all angles are 60 degrees
Isosceles	Two sides equal; two base angles equal
Scalene	All three sides different; all three angles different

Worked Example 5

An isosceles triangle has one angle of 40 degrees at the apex. Find the base angles.

Solution: Let each base angle be b. 40 + b + b = 180 40 + 2b = 180 2b = 140 b = 70 degrees

Exam Tip: In isosceles triangle questions, always identify which angle is the "odd one out." If the given angle is a base angle, then the other base angle is the same. If the given angle is the apex, subtract from 180 and halve the remainder.

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Interior Angles of Polygons

The sum of the interior angles of any polygon depends on the number of sides.

Formula: Sum of interior angles = (n - 2) x 180 degrees, where n is the number of sides.

Polygon	Sides (n)	Sum of Interior Angles
Triangle	3	(3-2) x 180 = 180 degrees
Quadrilateral	4	(4-2) x 180 = 360 degrees
Pentagon	5	(5-2) x 180 = 540 degrees
Hexagon	6	(6-2) x 180 = 720 degrees
Heptagon	7	(7-2) x 180 = 900 degrees
Octagon	8	(8-2) x 180 = 1080 degrees
Decagon	10	(10-2) x 180 = 1440 degrees

For a regular polygon (all sides and angles equal), each interior angle = (n - 2) x 180 / n.

Worked Example 6

Find the size of each interior angle of a regular nonagon (9 sides).

Solution: Sum of interior angles = (9 - 2) x 180 = 7 x 180 = 1260 degrees Each interior angle = 1260 / 9 = 140 degrees

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Exterior Angles of Polygons

The exterior angles of any polygon always sum to 360 degrees.

For a regular polygon, each exterior angle = 360 / n.

Also note: interior angle + exterior angle = 180 degrees (they form a straight line).

Worked Example 7

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

Solution: n = 360 / exterior angle = 360 / 40 = 9 sides (a nonagon)

Worked Example 8

A regular polygon has interior angles of 156 degrees. Find the number of sides.

Solution: Exterior angle = 180 - 156 = 24 degrees n = 360 / 24 = 15 sides

Exam Tip: If you are asked to find the number of sides from an interior angle, always convert to the exterior angle first, then divide 360 by it. This is quicker and less error-prone than using the interior angle formula directly.

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Summary

• Angles on a straight line sum to 180 degrees; around a point sum to 360 degrees
• Vertically opposite angles are equal
• Alternate angles (Z-shape) and corresponding angles (F-shape) are equal when lines are parallel
• Co-interior angles (C/U-shape) sum to 180 degrees
• Angles in a triangle sum to 180 degrees; in a quadrilateral to 360 degrees
• Sum of interior angles of an n-sided polygon = (n - 2) x 180
• Exterior angles of any polygon sum to 360 degrees
• For a regular polygon: each exterior angle = 360 / n

Exam Tip: When answering multi-step angle problems in the exam, present your working as a chain of reasoning. State the angle rule at each step — for example, "Angle ABD = 72 degrees (alternate angles are equal)." This is essential for full marks on proof-style questions.

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Extended Worked Examples

Worked Example 9 (Foundation): Chain of angle reasoning

In the diagram, lines $AB$AB and $CD$CD are parallel. A transversal cuts $AB$AB at $P$P and $CD$CD at $Q$Q. The angle $APQ = 68^{\circ}$APQ=68∘. The line $QR$QR is drawn such that triangle $PQR$PQR is isosceles with $QP = QR$QP=QR and $R$R lies below $CD$CD. The angle $PQR = 68^{\circ}$PQR=68∘. Find the angle $QRP$QRP.

Step 1 — work with the parallel lines. Because $AB \parallel CD$AB∥CD and $PQ$PQ is the transversal, alternate angles are equal:

$\text{angle } PQD = \text{angle } APQ = 68^{\circ}.$angle PQD=angle APQ=68∘.

Step 2 — use the isosceles triangle. In triangle $PQR$PQR, the sides $QP$QP and $QR$QR are equal, so the base angles $QPR$QPR and $QRP$QRP are equal. Let each base angle equal $x$x.

Step 3 — use the triangle angle sum. Angles in a triangle sum to $180^{\circ}$180∘:

$68 + x + x = 180$68+x+x=180 $2x = 112$2x=112 $x = 56^{\circ}.$x=56∘.

So angle $QRP = 56^{\circ}$QRP=56∘ (base angles of an isosceles triangle are equal; angles in a triangle sum to $180^{\circ}$180∘).

Worked Example 10 (Foundation/Higher): Regular polygon with a twist

Three interior angles of a pentagon measure $100^{\circ}$100∘, $120^{\circ}$120∘ and $95^{\circ}$95∘. The remaining two interior angles are equal. Find their size.

Step 1 — pentagon angle sum. Using $(n-2)\times 180^{\circ}$(n−2)×180∘ with $n=5$n=5:

$(5-2)\times 180 = 3 \times 180 = 540^{\circ}.$(5−2)×180=3×180=540∘.

Step 2 — set up an equation. Let each unknown angle be $y$y. Then

$100 + 120 + 95 + y + y = 540$100+120+95+y+y=540 $315 + 2y = 540$315+2y=540 $2y = 225$2y=225 $y = 112.5^{\circ}.$y=112.5∘.

Each of the two equal angles is $\mathbf{112.5^{\circ}}$112.5∘.

Worked Example 11 (Higher): Interior/exterior angle mixed problem

The interior angle of a regular polygon is seven times its exterior angle. Find the number of sides.

Step 1 — set up. Let the exterior angle be $e$e. Then the interior angle is $7e$7e, and interior $+$+ exterior $= 180^{\circ}$=180∘:

$7e + e = 180$7e+e=180 $8e = 180$8e=180 $e = 22.5^{\circ}.$e=22.5∘.

Step 2 — apply the exterior-angle sum. For a regular polygon, $n = 360 / e$n=360/e:

$n = \frac{360}{22.5} = 16.$n=22.5360​=16.

The polygon has 16 sides (a hexadecagon).

Common mistake: Students try to use the interior-angle formula directly and get tangled in algebra. Always convert to exterior angles first — the calculation is much cleaner.

Worked Example 12 (Higher): Two regular polygons meeting at a vertex

Two regular polygons meet at a vertex, along with a square, so that their interior angles fit together exactly around a point with no gap. One polygon is a regular hexagon. Find the number of sides of the other polygon.

Step 1 — interior angles. Hexagon interior angle $= 120^{\circ}$=120∘; square interior angle $= 90^{\circ}$=90∘.

Step 2 — angles around a point. The three interior angles meeting at the point sum to $360^{\circ}$360∘:

$120 + 90 + x = 360$120+90+x=360 $x = 150^{\circ}.$x=150∘.

Step 3 — find $n$n. The unknown polygon has interior angle $150^{\circ}$150∘, so its exterior angle is $180 - 150 = 30^{\circ}$180−150=30∘. Hence

$n = \frac{360}{30} = 12.$n=30360​=12.

The other polygon is a regular dodecagon (12 sides).

Worked Example 13 (Higher): Parallel lines with algebraic angles

In a diagram, $AB \parallel CD$AB∥CD and the co-interior angles formed with transversal $EF$EF are $(4x + 25)^{\circ}$(4x+25)∘ and $(6x - 15)^{\circ}$(6x−15)∘. Find the value of $x$x and state both angles.

Step 1 — use the co-interior rule. Co-interior angles sum to $180^{\circ}$180∘:

$(4x + 25) + (6x - 15) = 180$(4x+25)+(6x−15)=180 $10x + 10 = 180$10x+10=180 $10x = 170$10x=170 $x = 17.$x=17.

Step 2 — substitute back.

• First angle: $4(17) + 25 = 68 + 25 = 93^{\circ}$4(17)+25=68+25=93∘.
• Second angle: $6(17) - 15 = 102 - 15 = 87^{\circ}$6(17)−15=102−15=87∘.

Check: $93 + 87 = 180^{\circ}$93+87=180∘ as expected.

Answering at different grade levels

Question: The diagram shows two parallel lines cut by a transversal. One of the angles formed is $112^{\circ}$112∘. A triangle is then drawn with one side along the transversal, one side along one parallel line, and a third angle of $35^{\circ}$35∘. Find the remaining angle inside the triangle.

Grades 3–4 answer. The angle on the other parallel line opposite the $112^{\circ}$112∘ is also $112^{\circ}$112∘ because the lines are parallel. The angle inside the triangle next to it is $180 - 112 = 68^{\circ}$180−112=68∘ (angles on a straight line). The three angles in the triangle sum to $180^{\circ}$180∘, so the missing angle is $180 - 68 - 35 = 77^{\circ}$180−68−35=77∘.

Grades 5–6 answer. Using corresponding angles (equal when lines are parallel), the angle on the lower parallel line is also $112^{\circ}$112∘. This angle and the triangle's interior angle at that vertex lie on a straight line, so the interior angle is $180 - 112 = 68^{\circ}$180−112=68∘. By the angle sum of a triangle ($180^{\circ}$180∘), the remaining angle is $180 - 68 - 35 = \mathbf{77^{\circ}}$180−68−35=77∘.

Grades 7–9 answer. Let the interior angle of the triangle at the transversal be $\alpha$α. By co-interior angles with the given $112^{\circ}$112∘, $\alpha = 180 - 112 = 68^{\circ}$α=180−112=68∘. The triangle angle sum gives the remaining angle as $180 - \alpha - 35 = 180 - 68 - 35 = \mathbf{77^{\circ}}$180−α−35=180−68−35=77∘. Justification at each step: co-interior angles between parallel lines sum to $180^{\circ}$180∘; angles in a triangle sum to $180^{\circ}$180∘. A complete answer names each angle rule used so the examiner can award the reasoning marks.

AQA alignment: This content is aligned with AQA GCSE Mathematics (8300) specification — specifically Topic G G3 Angles / parallel lines, G6 Angles in polygons, and the angle-reasoning requirements that underpin G14 Circle theorems [H]. Assessed on Papers 1, 2, and 3 (calculator / non-calculator).