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Mathematics · AQA 8300 §G1

Angles

Spec: AQA 8300 §G1 ⭐⭐ 🕐 40 mins AQA · Edexcel · OCR Grade 9
  • Apply angle rules at a point and on a straight line
  • Identify and use corresponding, alternate and co-interior angles
  • Find interior and exterior angles of polygons
  • Solve angle problems requiring multiple rules
  • Work with bearings and compass directions

🔑 Core Concepts

Types of Angles

An angle measures the amount of rotation between two rays sharing a common vertex. Before applying any rule, correctly classify every angle in the diagram — this tells you which range of values is reasonable for your answer.

📖
DEFINITION — Types of Angles
TypeRangeKey Fact
Acute$0° < \theta < 90°$Smaller than a right angle
Right$\theta = 90°$Marked with a square corner symbol
Obtuse$90° < \theta < 180°$Between a right angle and a straight line
Straight$\theta = 180°$Forms a flat, straight line
Reflex$180° < \theta < 360°$Larger than a straight angle; "going round the back"
🎯
EXAM TIP — Checking Reasonableness
After finding an angle, check its type matches the diagram. If the diagram shows an obtuse angle but your answer is 15°, you have made an error. Classifying angles first gives you a built-in sense-check.
✗
COMMON MISTAKE — Reflex Angles
When measuring or calculating a reflex angle, students often find the non-reflex angle (e.g. 70°) and forget to subtract from 360°. The reflex angle would be $360° - 70° = 290°$. Always read the question to see which angle is required.

Angles on a Straight Line and at a Point

These two rules are the foundation of every angle calculation. They arise from the definition of a straight line (a 180° arc) and a complete rotation (360°). They apply regardless of the number of angles involved.

📖
DEFINITION — Angles on a Straight Line
Any set of angles that together cover exactly one side of a straight line must sum to $180°$: $$a + b + c + \ldots = 180°$$ This is also called a supplementary relationship when only two angles are involved.
📖
DEFINITION — Angles at a Point
All angles arranged around a single point (a full turn) sum to $360°$: $$a + b + c + d + \ldots = 360°$$ This is a complete rotation and holds for any number of angles at that point.
Fundamental Angle Rules
$$\text{Straight line: } \sum \theta = 180°$$ $$\text{Full turn (point): } \sum \theta = 360°$$
$\theta$ = each angle at that location These rules hold for any number of angles
🎯
EXAM TIP — Writing the Rule
Always state the rule used, e.g. "angles on a straight line sum to 180°". Mark schemes frequently award 1 method mark for naming the rule, even when the arithmetic is wrong.
✗
COMMON MISTAKE — Confusing 180° and 360°
Using 360° for angles on a straight line (or 180° for a full turn) immediately produces a wrong equation and loses all marks. Ask yourself: "Do these angles share one side of a line, or do they fill all the way around?" The answer tells you which rule to use.

Vertically Opposite Angles

When two straight lines cross at a point, they form four angles. The pair of angles that are opposite each other across the intersection are called vertically opposite angles. They are equal because they are each supplementary to the same adjacent angle.

📖
DEFINITION — Vertically Opposite Angles
When two straight lines intersect at a point, the angles formed directly opposite each other are equal: $$\angle AOC = \angle BOD \quad \text{and} \quad \angle AOD = \angle BOC$$ Proof: $\angle AOC + \angle AOD = 180°$ (straight line). $\angle BOD + \angle AOD = 180°$ (straight line). Therefore $\angle AOC = \angle BOD$.
🎯
EXAM TIP — Stating the Reason
When using vertically opposite angles, write: "vertically opposite angles are equal" — not "X-angles" (which is not a recognised term in the UK curriculum). The examiner expects the precise name.
✗
COMMON MISTAKE — "Vertically" Does Not Mean Up-Down
"Vertically opposite" comes from the Latin vertex (a corner or point), not the word "vertical". The angles can be tilted at any orientation — the rule applies whenever two straight lines cross.

Angles Formed by Parallel Lines and a Transversal

A transversal is a line that crosses two or more other lines. When the crossed lines are parallel, three powerful angle relationships are created. These rules are tested at every GCSE grade, especially in algebraic form at Grade 8–9.

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DEFINITION — Corresponding Angles (F-Angles)
Corresponding angles occupy the same relative position at each intersection — both above-left, or both below-right, etc. They form an F-shape (or a reversed F).

Rule: When lines are parallel, corresponding angles are equal. $$\alpha = \beta \quad (\text{corresponding angles, } \ell_1 \parallel \ell_2)$$
📖
DEFINITION — Alternate Angles (Z-Angles)
Alternate angles lie on opposite sides of the transversal, between the two parallel lines. They form a Z-shape (or reversed Z / N-shape).

Rule: When lines are parallel, alternate angles are equal. $$\alpha = \beta \quad (\text{alternate angles, } \ell_1 \parallel \ell_2)$$
📖
DEFINITION — Co-interior Angles (C-Angles / Allied Angles)
Co-interior angles lie on the same side of the transversal, between the two parallel lines. They form a C-shape (or U-shape).

Rule: When lines are parallel, co-interior angles are supplementary (sum to 180°). $$\alpha + \beta = 180° \quad (\text{co-interior angles, } \ell_1 \parallel \ell_2)$$
Parallel Lines — Three Rules
$$\text{Corresponding (F-shape): } \alpha = \beta$$ $$\text{Alternate (Z-shape): } \alpha = \beta$$ $$\text{Co-interior (C-shape): } \alpha + \beta = 180°$$
All three rules require lines to be parallel F and Z give equal angles; C gives supplementary angles
🧠
MEMORY TRICK — F, Z, C
F = Forward (same position) → equal  |  Z = Zero difference → equal  |  C = Combined → 180°
Or simply: Only C is different — it adds to 180°. F and Z are always equal.
🎯
EXAM TIP — Choosing the Right Rule
You can always convert: if you identify a C-angle pair, you could also split the C into a Z-angle and a straight-line calculation. At Grade 9 you may need to chain two or three rules. Write each step with its rule clearly labelled.
✗
COMMON MISTAKE — Assuming Lines Are Parallel
These three rules ONLY apply when the lines are confirmed parallel. Look for the double arrow marks (▷▷) on a diagram or the statement "AB is parallel to CD" in the question. Lines that look parallel are not necessarily so.

Interior and Exterior Angles of Polygons

Any convex polygon with $n$ sides can be divided into $(n - 2)$ non-overlapping triangles by drawing diagonals from one vertex. Since each triangle contributes 180°, the total interior angle sum follows directly. The exterior angle result comes from the fact that traversing the perimeter of any polygon completes exactly one full rotation.

📖
DEFINITION — Interior Angle Sum of a Polygon
For any polygon with $n$ sides, the sum of all interior angles is: $$S_{\text{interior}} = (n - 2) \times 180°$$ For a regular polygon (all sides and angles equal), each interior angle is: $$\text{Each interior angle} = \frac{(n-2) \times 180°}{n}$$
📖
DEFINITION — Exterior Angles of a Polygon
An exterior angle is formed by extending one side of the polygon at a vertex. At each vertex, the interior and exterior angles are supplementary: $$\text{Interior angle} + \text{Exterior angle} = 180°$$ The sum of all exterior angles of any convex polygon is always: $$S_{\text{exterior}} = 360°$$ For a regular polygon: $$\text{Each exterior angle} = \frac{360°}{n}$$
Polygon Angle Formulae
$$S_{\text{interior}} = (n-2) \times 180°$$ $$\text{Each interior (regular)} = \frac{(n-2) \times 180°}{n}$$ $$\text{Each exterior (regular)} = \frac{360°}{n}$$ $$\text{Interior} + \text{Exterior} = 180°$$
$n$ = number of sides of the polygon "Regular" means all sides and angles are equal
🎯
EXAM TIP — Finding n from One Angle
If given an exterior angle of a regular polygon: $n = \dfrac{360°}{\text{exterior angle}}$.
If given an interior angle: first find exterior = $180° - \text{interior}$, then use $n = \dfrac{360°}{\text{exterior}}$.
The exterior angle route is almost always simpler.
✗
COMMON MISTAKE — Wrong Interior Angle Formula
Writing $n \times 180°$ instead of $(n-2) \times 180°$ is the single most common polygon error. A quick self-check: plug in $n = 3$ (triangle). You should get $1 \times 180° = 180°$. If your formula gives $3 \times 180° = 540°$ for a triangle, it is wrong.

Polygon Reference Table

Polygon$n$Interior SumEach Interior (regular)Each Exterior (regular)
Triangle3180°60°120°
Quadrilateral4360°90°90°
Pentagon5540°108°72°
Hexagon6720°120°60°
Heptagon7900°≈128.6°≈51.4°
Octagon81080°135°45°
Decagon101440°144°36°
Dodecagon121800°150°30°

Bearings

A bearing gives a direction as a single angle, measured clockwise from North. Bearings are used in navigation, maps, and trigonometry problems. The key property exploited at Grade 9 is that north lines at any two locations are parallel to each other — this allows you to apply Z, F, and C angle rules between them.

📖
DEFINITION — Bearings
A bearing is an angle measured clockwise from North, always written as exactly three digits:
  • North = 000°
  • East = 090°
  • South = 180°
  • West = 270°
The back bearing (the reverse direction from B back to A) is: $$\text{Back bearing} = \text{bearing} + 180° \quad (\text{if bearing} < 180°)$$ $$\text{Back bearing} = \text{bearing} - 180° \quad (\text{if bearing} \geq 180°)$$
🎯
EXAM TIP — Bearings and Parallel Lines
North lines drawn at any two points are always parallel. When a journey line connects two points, it acts as a transversal cutting those two parallel north lines. You can then use alternate angles (Z) or co-interior angles (C) to find bearings at different points. This technique is essential for multi-step bearing problems at Grade 9.
✗
COMMON MISTAKE — Bearings Without 3 Digits
Writing "47°" instead of "047°" loses the answer mark. Also, if a back bearing calculation gives a value over 360° (e.g. $250° + 180° = 430°$), subtract 360°: $430° - 360° = 070°$. Bearings must always be in the range 000° to 359°.
Draw North lines at every relevant point
→
Identify angle pairs (Z / F / C) formed between parallel North lines and journey lines
→
Apply the appropriate parallel-line angle rule
→
Write the final bearing as exactly 3 digits

🗺️ Visual Notes

Angles
Basic Rules
  • Straight line → 180°
  • Full turn → 360°
  • Vertically opposite → equal
  • Always state the rule used
Parallel Lines
  • Corresponding (F) → equal
  • Alternate (Z) → equal
  • Co-interior (C) → 180°
  • Must confirm lines are parallel
Polygons
  • Sum = $(n-2) \times 180°$
  • Regular interior = sum ÷ $n$
  • Exterior sum = 360° always
  • Interior + exterior = 180°
Bearings
  • Clockwise from North
  • Always 3 digits (e.g. 047°)
  • North lines are parallel
  • Back bearing = bearing ± 180°
Angle Types
  • Acute: 0°–90°
  • Obtuse: 90°–180°
  • Reflex: 180°–360°
  • Straight: exactly 180°
Grade 9 Skills
  • Algebraic angle equations
  • Multi-rule chain problems
  • Bearings + parallel lines
  • Proofs using angle rules

Comparison: Parallel Line Angle Rules

RuleLetter ShapeRelationshipHow to Spot It
Corresponding F (or reversed F) Equal: $\alpha = \beta$ Same side of transversal; same relative position at each parallel line
Alternate Z (or N) Equal: $\alpha = \beta$ Opposite sides of transversal; both between the parallel lines
Co-interior C (or U) Supplementary: $\alpha + \beta = 180°$ Same side of transversal; both between the parallel lines

Decision Tree: Polygon Angle Problems

Is the polygon regular (all sides and angles equal)?
→
YES: Use $\dfrac{(n-2)\times180°}{n}$ for each interior, or $\dfrac{360°}{n}$ for each exterior
→
NO: Calculate the total sum $(n-2)\times180°$, then form an equation using the given angles
→
Finding n: Convert interior → exterior ($= 180° - \text{interior}$), then $n = 360° \div \text{exterior}$

✏️ Worked Examples

Grade 4–5 · Angles on a Straight Line
Three angles on a straight line are $x$, $2x$, and $3x$. Find the value of $x$ and state the size of each angle.
1
Identify the angle rule
The three angles together form a straight line, so they sum to 180°: $$x + 2x + 3x = 180°$$
2
Simplify the left-hand side
$$6x = 180°$$
3
Solve for x
$$x = \frac{180°}{6} = 30°$$
4
State all three angles
$x = 30°$, $\quad 2x = 60°$, $\quad 3x = 90°$
5
Verify
$30° + 60° + 90° = 180°$ ✓
$x = 30°$. The three angles are $30°$, $60°$, and $90°$.
Grade 6–7 · Parallel Lines with Algebra
$AB$ and $CD$ are parallel lines cut by a transversal at points $P$ and $Q$ respectively. The angle at $P$ above $AB$ (on the left of the transversal) is $(4x - 12)°$. The alternate angle at $Q$ below $CD$ (on the right of the transversal) is $(2x + 18)°$. Find $x$ and the size of both angles.
1
Identify the relationship
The angles are on opposite sides of the transversal, between the parallel lines — these are alternate angles. Alternate angles are equal when lines are parallel: $$(4x - 12)° = (2x + 18)°$$
2
Solve the equation
$$4x - 12 = 2x + 18$$ $$4x - 2x = 18 + 12$$ $$2x = 30 \implies x = 15$$
3
Calculate both angles
Angle at $P$: $4(15) - 12 = 60 - 12 = 48°$
Angle at $Q$: $2(15) + 18 = 30 + 18 = 48°$ ✓ (equal, as expected for alternate angles)
$x = 15$; both alternate angles measure $48°$.
Grade 9 · Trapezium with Multiple Rules
$ABCD$ is a trapezium with $AB \parallel DC$. The angles are: $\angle DAB = (4x + 10)°$, $\angle ABC = (2x + 30)°$, $\angle BCD = (3x - 5)°$.
(a) Find $x$. (b) Find $\angle CDA$. (c) Show that $BC$ is not perpendicular to $AB$.
1
Use co-interior angles on transversal BC (part a)
Since $AB \parallel DC$, angles $\angle ABC$ and $\angle BCD$ are co-interior angles (both between the parallel lines, on the same side of transversal $BC$). Co-interior angles sum to 180°: $$(2x + 30) + (3x - 5) = 180$$ $$5x + 25 = 180$$ $$5x = 155 \implies x = 31$$
2
Calculate the known angles
$\angle ABC = 2(31) + 30 = 92°$
$\angle BCD = 3(31) - 5 = 88°$
$\angle DAB = 4(31) + 10 = 134°$
Check co-interior: $92° + 88° = 180°$ ✓
3
Find ∠CDA using co-interior angles on transversal AD (part b)
Angles $\angle DAB$ and $\angle CDA$ are co-interior (transversal $AD$, $AB \parallel DC$): $$\angle DAB + \angle CDA = 180°$$ $$134° + \angle CDA = 180°$$ $$\angle CDA = 46°$$
4
Verify using the quadrilateral angle sum
Sum of interior angles of a quadrilateral $= 360°$: $$134° + 92° + 88° + 46° = 360°\checkmark$$
5
Show BC is not perpendicular to AB (part c)
For $BC \perp AB$, we would require $\angle ABC = 90°$.
We found $\angle ABC = 92° \neq 90°$.
Therefore $BC$ is not perpendicular to $AB$. $\square$
(a) $x = 31$. (b) $\angle CDA = 46°$. (c) Since $\angle ABC = 92° \neq 90°$, $BC$ is not perpendicular to $AB$.

❓ Exam Questions

Q11 mark

Three angles on a straight line are in the ratio $2 : 3 : 4$. Find the size of the smallest angle.

Mark scheme:
Total ratio parts $= 2 + 3 + 4 = 9$
$9 \text{ parts} = 180°$, so $1 \text{ part} = 20°$
Smallest angle $= 2 \times 20° = \mathbf{40°}$ ✓ [1]
Q22 marks

$PQ$ and $RS$ are parallel lines cut by a transversal. Two co-interior angles are $(3k + 15)°$ and $(k + 5)°$. Find $k$ and state both angles.

Mark scheme:
Co-interior angles sum to 180° [M1 — correct equation set up]:
$(3k + 15) + (k + 5) = 180$
$4k + 20 = 180$
$k = 40$ [A1]
Angles: $3(40)+15 = 135°$ and $40+5 = 45°$. Check: $135°+45° = 180°$ ✓
Q33 marks

The exterior angle of a regular polygon is $24°$. Find (a) the number of sides, (b) each interior angle, (c) the sum of all interior angles.

Mark scheme:
(a) $n = \dfrac{360°}{24°} = \mathbf{15}$ sides [1]
(b) Interior $= 180° - 24° = \mathbf{156°}$ [1]
(c) Sum $= (15-2) \times 180° = 13 \times 180° = \mathbf{2340°}$ [1]
Alt: $15 \times 156° = 2340°$
Q44 marks

The interior angles of quadrilateral $ABCD$ are $\angle A = (3x+10)°$, $\angle B = (2x+20)°$, $\angle C = (x+30)°$, $\angle D = (4x-20)°$. Find all four angles.

Mark scheme:
Interior angle sum of a quadrilateral $= 360°$ [M1 — correct sum used]:
$(3x+10)+(2x+20)+(x+30)+(4x-20) = 360$
$10x + 40 = 360$ [M1 — correct simplification]
$x = 32$ [A1]
$\angle A = 106°$, $\angle B = 84°$, $\angle C = 62°$, $\angle D = 108°$ [A1 — all four correct]
Check: $106+84+62+108 = 360°$ ✓
Q56 marks

A ship travels from port $P$ on a bearing of $040°$ to point $Q$. It then turns and travels on a bearing of $130°$ to point $R$.
(a) Calculate the interior angle $PQR$. [2]
(b) The bearing of $P$ from $R$ is $265°$. Find the bearing of $R$ from $P$ and hence calculate $\angle QPR$. [2]
(c) Find the bearing of $Q$ from $R$. [2]

Mark scheme:
(a) Back bearing from $Q$ to $P$ = $040° + 180° = 220°$ [M1]
$\angle PQR = 220° - 130° = \mathbf{90°}$ [A1]

(b) Bearing of $R$ from $P$ = $265° - 180° = \mathbf{085°}$ [M1]
$\angle QPR = 085° - 040° = \mathbf{45°}$ [A1]

(c) $\angle QRP = 180° - 90° - 45° = 45°$
Bearing of $Q$ from $R$ = $265° + 45° = \mathbf{310°}$ [2]
Note: $PQR$ is an isoceles right triangle; the diagram should confirm $Q$ lies to the NW of $R$.

⭐ Grade 9 Model Answers

Full Annotated Solution to Q5 — Bearings (6 marks)

✏️
GRADE 9 MODEL ANSWER

Given: $P \to Q$ on bearing 040°; $Q \to R$ on bearing 130°; bearing of $P$ from $R$ = 265°.

Part (a) — Interior angle PQR [2 marks]

To find the angle inside triangle $PQR$ at vertex $Q$, I need the direction from $Q$ back to $P$ (not the direction from $P$ to $Q$):

$$\text{Direction } Q \to P = 040° + 180° = 220°$$

Examiner note: M1 is awarded here for adding 180° to find the back bearing.

The interior angle $\angle PQR$ is the angle swept from the direction $Q \to P$ (220°) round to the direction $Q \to R$ (130°), measured through the interior of the triangle:

$$\angle PQR = 220° - 130° = 90°$$

A1 for 90° with supporting calculation. Simply writing "90°" alone scores 0 — the method must be shown.

Part (b) — Angle QPR [2 marks]

The bearing of $R$ from $P$ is the reverse of the bearing of $P$ from $R$:

$$\text{Bearing } P \to R = 265° - 180° = 085°$$

M1 for subtracting 180° (since 265° ≥ 180°).

From $P$, the bearing to $Q$ is 040° and the bearing to $R$ is 085°. Both are measured clockwise from North at $P$, so:

$$\angle QPR = 085° - 040° = 45°$$

A1 for 45°.

Part (c) — Bearing of Q from R [2 marks]

Use the angle sum of triangle $PQR$:

$$\angle QRP = 180° - 90° - 45° = 45°$$

This interior angle at $R$ is 45°. I know the direction from $R$ towards $P$ is 265°. The direction from $R$ towards $Q$ is found by rotating 45° clockwise from the direction $R \to P$ (since $Q$ lies further clockwise round from $P$ when viewed from $R$):

$$\text{Bearing of } Q \text{ from } R = 265° + 45° = 310°$$

2 marks for correct method and answer. Award 1 mark if angle QRP = 45° is found but the final bearing is wrong.

🎯
WHY EACH STEP EARNS MARKS AT GRADE 9
  • Back bearing calculation shown explicitly (±180°): earns M1 even if a subsequent arithmetic slip occurs.
  • Angle rule named ("angles on a straight line", "co-interior", etc.): earns the method mark independent of the answer.
  • Verification step ($90° + 45° + 45° = 180°$): demonstrates mathematical rigour; examiners reward this at Grade 9.
  • 3-digit bearing format: the degree symbol and three digits are required. "310" scores 0 for the answer mark; "310°" scores 1.

📋 Revision Sheet

Key Definitions
Acute angle0° to 90° (not including endpoints)
Obtuse angle90° to 180°
Reflex angle180° to 360°
Straight lineAngles sum to 180°
Full turnAngles sum to 360°
Vertically oppositeEqual; formed where two lines cross
Corresponding (F)Equal; parallel lines cut by transversal
Alternate (Z)Equal; opposite sides of transversal
Co-interior (C)Sum to 180°; same side of transversal
BearingClockwise from North; 3 digits
Back bearingOriginal bearing ± 180°
Essential Formulae

$\text{Straight line: } \displaystyle\sum \theta = 180°$

$\text{Full turn: } \displaystyle\sum \theta = 360°$

$\text{Interior sum of polygon: } (n-2)\times180°$

$\text{Each interior (regular): } \dfrac{(n-2)\times180°}{n}$

$\text{Each exterior (regular): } \dfrac{360°}{n}$

$\text{Interior} + \text{Exterior} = 180°$

$\text{Back bearing: } b \pm 180°$

$\text{Find } n \text{ from exterior: } n = \dfrac{360°}{\text{ext. angle}}$

Memory Hooks
  • F = Forward = Equal (corresponding)
  • Z = Zero difference = Equal (alternate)
  • C = Combined = 180° (co-interior)
  • Only C is different — the other two are equal
  • Exterior angles tour the polygon: always 360°
  • North lines are always parallel — use Z and C between them in bearing problems
  • Pentagon = 5 letters = 5 sides = 540° sum
  • Hexagon = 6 sides = 120° each (regular)
  • Bearings: "Oh-four-seven" not "47" — three digits always
Exam Tips
  • Name the angle rule in every line of working
  • Mark parallel lines with arrows; equal angles with arc marks
  • For algebraic angles: always verify your answer sums correctly
  • For regular polygons: use exterior angle formula first — simpler
  • In bearing problems: draw North lines at every point before calculating
  • Co-interior angles ONLY sum to 180° — the other two parallel-line rules give equal angles
  • Check the final sum: interior angles of an $n$-gon must total $(n-2)\times180°$
  • At Grade 9: show every rule and every step — method marks matter

🔄 Flashcards

Click each card to reveal the answer. Test yourself before looking — active recall is the most effective revision technique.

✗ Common Mistakes

✗
MISTAKE 1 — Confusing 180° (Straight Line) and 360° (Full Turn)

What students do: Write angles at a point equal to 180°, or angles on a straight line equal to 360°.

Why marks are lost: The equation is wrong from the start; all subsequent arithmetic is incorrect, typically costing the full marks on a multi-mark question.

How to avoid: Ask: "Do these angles fan out on one side of a straight line (half-turn) or go all the way around a point (full turn)?" If they straddle the line on both sides, use 360°. If they are all on one side only, use 180°.

✗
MISTAKE 2 — Using Parallel Line Rules Without Confirming the Lines Are Parallel

What students do: See a Z-shape or F-shape in a diagram and immediately apply the equal-angle rule without checking whether the lines are actually parallel.

Why marks are lost: The stated reason is invalid, so the method mark is lost. If the lines are not parallel, none of the three angle rules apply.

How to avoid: Only apply corresponding, alternate, or co-interior rules when the question states the lines are parallel or the diagram shows parallel arrow marks (▷▷). Never assume from appearance alone.

✗
MISTAKE 3 — Omitting the Angle Rule from Working

What students do: Arrive at the correct numerical answer but write nothing to explain why, e.g. "angle = 48°" with no justification.

Why marks are lost: Many mark schemes award a dedicated method mark for naming the angle rule. Writing only an answer without reasoning loses that mark.

How to avoid: Develop a habit of writing the rule in brackets after every angle you find, for example: "∠PQR = 48° (alternate angles, AB ∥ CD)". This also helps you spot your own errors.

✗
MISTAKE 4 — Wrong Polygon Interior Angle Formula

What students do: Use $n \times 180°$ instead of $(n-2) \times 180°$, or correctly remember the form but substitute the wrong value of $n$.

Why marks are lost: An incorrect total means all angle calculations are wrong, potentially losing 2–3 marks on a single question.

How to avoid: Memorise $(n-2) \times 180°$ and verify with a known case: $n = 3$ (triangle) gives $(3-2) \times 180° = 180°$ ✓. If your formula gives a different answer for a triangle, it is wrong.

✗
MISTAKE 5 — Bearing Not Written as Three Digits

What students do: Write a bearing as "47°" instead of "047°", or calculate a back bearing that exceeds 360° without subtracting 360°.

Why marks are lost: The answer mark specifically requires the 3-digit convention. A value such as 380° is also an invalid bearing.

How to avoid: Always check: "Does my bearing have exactly 3 digits?" For back bearings: if bearing < 180°, add 180°; if bearing ≥ 180°, subtract 180°. If the result is ≥ 360°, subtract 360°. Example: $250° + 180° = 430° \to 430° - 360° = 070°$.

✗
MISTAKE 6 — Mixing Up Interior and Exterior Angle Calculations for Regular Polygons

What students do: Divide 180° (not 360°) by $n$ to find each exterior angle, or divide 360° by $n$ and then claim this is the interior angle.

Why marks are lost: The resulting value is plausible-looking (a number less than 180°) but incorrect, and the method mark for the correct formula is also lost.

How to avoid: Exterior angles sum to 360°: each exterior $= 360° \div n$. Interior angles sum to $(n-2) \times 180°$: each interior $= (n-2)\times180° \div n$. Cross-check: interior + exterior must always equal 180°.

✅ Final Checklist

Click each item when you are confident with it. Your progress is saved automatically in your browser.

  • I can identify and name acute, obtuse, right, straight, and reflex angles
  • I know that angles on a straight line sum to 180° and can write the correct equation
  • I know that angles at a point sum to 360° and can write the correct equation
  • I can identify vertically opposite angles and explain why they are equal
  • I can identify corresponding angles (F-shape) and state they are equal in parallel lines
  • I can identify alternate angles (Z-shape) and state they are equal in parallel lines
  • I can identify co-interior angles (C-shape) and state they sum to 180° in parallel lines
  • I can apply the formula $(n-2) \times 180°$ to find the interior angle sum of any polygon
  • I can find each interior angle of a regular polygon using $\frac{(n-2)\times180°}{n}$
  • I know the exterior angle sum of any polygon is always 360°
  • I can find the number of sides of a regular polygon given one interior or exterior angle
  • I can state and calculate bearings correctly as three-digit numbers
  • I can calculate back bearings by adding or subtracting 180°
  • I can set up and solve algebraic equations using multiple angle rules
  • I always name the angle rule used in every line of my working
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