Chapter 1: Angles and Shapes
Angles are the foundation of all geometry. In this chapter we build up from the basic types of angle to the powerful rules about parallel lines, then study the angle properties of triangles, quadrilaterals, and polygons. We finish with bearings, an important applied topic that appears on almost every exam paper.
Exam Board Coverage
All content in this chapter is assessed at both Foundation and Extended tier by Cambridge (0580) and Edexcel (4MA1). Bearings and multi-step angle problems are more common at Extended level.
1.1 Types of Angles
An angle measures the amount of turn between two straight lines that meet at a point, called the vertex. Angles are measured in degrees (°).
The Six Types of Angle
| Type | Range | Example |
|---|---|---|
| Zero angle | $0°$ | Two rays pointing in the same direction |
| Acute | $0° < \theta < 90°$ | $37°$, $62°$, $89°$ |
| Right angle | $\theta = 90°$ | Corner of a square; marked with a small square symbol |
| Obtuse | $90° < \theta < 180°$ | $91°$, $120°$, $179°$ |
| Straight angle | $\theta = 180°$ | A straight line |
| Reflex | $180° < \theta < 360°$ | $200°$, $315°$ |
| Complete turn | $\theta = 360°$ | Full rotation back to start |
Naming Angles
An angle at vertex $B$ formed by points $A$, $B$, $C$ is written $\angle ABC$ or $\widehat{ABC}$, with $B$ always in the middle. A single letter $\angle B$ may be used when only one angle is at that vertex.
1.2 Angle Relationships
Several important rules follow directly from the definition of angle. These are used constantly when solving multi-step geometry problems.
Four Fundamental Rules
- Angles on a straight line sum to $180°$. If $a$ and $b$ are on a straight line: $a + b = 180°$.
- Angles at a point sum to $360°$. If angles $a, b, c$ share a vertex: $a + b + c = 360°$.
- Vertically opposite angles are equal. When two straight lines cross, the angles diagonally opposite are equal: $a = c$ and $b = d$.
- Complementary angles sum to $90°$. Supplementary angles sum to $180°$.
Worked Example 1.1 — Angles at a Point
Three angles meet at a point: $x$, $2x$, and $3x + 10°$. Find the value of $x$.
Step 1 Angles at a point sum to $360°$:
$$x + 2x + (3x + 10°) = 360°$$
Step 2 Collect like terms: $6x + 10° = 360°$
Step 3 Solve: $6x = 350°$, so $x = 58.3°$ (1 d.p.).
Worked Example 1.2 — Vertically Opposite and Straight Line
Two straight lines cross. One of the four angles is $74°$. Find the other three angles.
Step 1 The angle vertically opposite to $74°$ is also $74°$.
Step 2 The two remaining angles are on a straight line with $74°$:
$$\text{adjacent angle} = 180° - 74° = 106°$$
Result The four angles are $74°$, $106°$, $74°$, $106°$.
1.3 Parallel Lines and Transversals
When a transversal (a line crossing two others) cuts a pair of parallel lines, it creates eight angles. These angles have special relationships: three pairs of equal or supplementary angles that every IGCSE student must recognise on sight.
The Three Parallel Line Rules
- Alternate angles (Z-angles) are equal. They lie on opposite sides of the transversal, between the parallel lines, forming a Z or N shape.
- Corresponding angles (F-angles) are equal. They lie on the same side of the transversal, in the same position at each intersection, forming an F shape.
- Co-interior angles (C-angles, also called same-side interior or allied angles) are supplementary: they sum to $180°$. They lie on the same side of the transversal, between the parallel lines, forming a C or U shape.
GEOM·1
Figure 1: Parallel lines $d_1 \parallel d_2$ cut by transversal $t$. Alternate angles $\alpha = \alpha$ (blue); corresponding angles $\beta = \beta$ (green). Co-interior angles are supplementary: $\alpha + \beta = 180°$.
Worked Example 1.3 — Parallel Lines
Lines $AB$ and $CD$ are parallel. A transversal crosses them, making an angle of $65°$ with line $AB$. Find all the other angles formed at both intersections.
Step 1 Label the given angle $\theta_1 = 65°$ at the first intersection. Its vertically opposite angle is also $65°$.
Step 2 The two angles on the straight line at the first intersection: $180° - 65° = 115°$ (two of them, vertically opposite).
Step 3 At the second intersection on line $CD$:
- Corresponding angle (F) to $65°$ = $65°$
- Alternate angle (Z) to $65°$ = $65°$
- Co-interior angle (C) to $65°$ = $180° - 65° = 115°$
Result Eight angles are formed: four of $65°$ and four of $115°$ (alternating around each intersection).
Exam Technique
Always state the reason when finding angles using parallel line rules. Write "alternate angles" or "corresponding angles" in your working — not just the calculation. Marks are allocated for the reason.
1.4 Triangles
A triangle has three sides and three interior angles. No matter its shape, the interior angles always add up to the same total.
Triangle Angle Sum
The sum of the interior angles of any triangle is $180°$:
$$\angle A + \angle B + \angle C = 180°$$
Proof sketch: Draw a line through the apex parallel to the base. The two alternate angles account for two angles of the triangle; the remaining angle is the apex angle. These three together form a straight line, giving $180°$.
Exterior Angle Theorem
An exterior angle of a triangle equals the sum of the two non-adjacent interior angles:
$$\text{exterior angle} = \angle A + \angle B$$
Types of Triangle
Classification by Angles
- Acute-angled triangle: all three angles less than $90°$.
- Right-angled triangle: one angle exactly $90°$. The side opposite the right angle is the hypotenuse.
- Obtuse-angled triangle: one angle greater than $90°$.
Classification by Sides
- Scalene: all three sides (and angles) different.
- Isosceles: two sides equal; the angles opposite the equal sides are also equal (base angles).
- Equilateral: all three sides equal; all angles $= 60°$.
Worked Example 1.4 — Triangle Angles
(a) An isosceles triangle has a vertex angle of $40°$. Find the base angles.
(b) The exterior angle of a triangle is $135°$. One interior angle is $72°$. Find the other two interior angles.
(a) Step 1 The two base angles are equal. Let each base angle $= x$.
$40° + x + x = 180°$, so $2x = 140°$, giving $x = 70°$.
Result The base angles are each $70°$.
(b) Step 1 By the exterior angle theorem, the exterior angle equals the sum of the two non-adjacent interior angles:
$135° = 72° + \text{(other interior angle)}$, so the other interior angle $= 63°$.
Step 2 Third angle: $180° - 72° - 63° = 45°$.
1.5 Quadrilaterals
A quadrilateral has four sides and four interior angles that always sum to $360°$. Memorising the properties of the special quadrilaterals is essential for IGCSE.
Angle Sum of a Quadrilateral
The sum of interior angles of any quadrilateral $= 360°$.
Reason: Any quadrilateral can be split into two triangles, each contributing $180°$, giving $2 \times 180° = 360°$.
Properties of Special Quadrilaterals
| Shape | Sides | Angles | Diagonals |
|---|---|---|---|
| Square | All 4 equal; opposite sides parallel | All $90°$ | Equal; bisect at $90°$; bisect angles |
| Rectangle | Opposite sides equal and parallel | All $90°$ | Equal; bisect each other |
| Rhombus | All 4 equal; opposite sides parallel | Opposite angles equal | Unequal; bisect at $90°$; bisect angles |
| Parallelogram | Opposite sides equal and parallel | Opposite angles equal; co-interior $= 180°$ | Bisect each other (not equal, not $90°$) |
| Trapezium | One pair of parallel sides | Co-interior angles on same side sum to $180°$ | No special property |
| Kite | Two pairs of adjacent equal sides | One pair of opposite angles equal | One diagonal bisects the other at $90°$ |
Worked Example 1.5 — Quadrilateral Angles
In parallelogram $ABCD$, angle $A = 3x + 10°$ and angle $B = 5x - 20°$. Find $x$ and all four angles.
Step 1 In a parallelogram, co-interior angles (between the parallel sides) sum to $180°$. Angles $A$ and $B$ are co-interior:
$$(3x + 10°) + (5x - 20°) = 180°$$
Step 2 $8x - 10° = 180°$, so $8x = 190°$, giving $x = 23.75°$.
Step 3 $\angle A = 3(23.75°) + 10° = 81.25°$. $\angle B = 5(23.75°) - 20° = 98.75°$.
Step 4 Opposite angles are equal: $\angle C = \angle A = 81.25°$ and $\angle D = \angle B = 98.75°$.
Check $81.25° + 98.75° + 81.25° + 98.75° = 360°$ ✓
1.6 Regular Polygons
A regular polygon has all sides equal and all interior angles equal. The formulae below follow from dividing the polygon into triangles from the centre.
Polygon Angle Formulae
For a regular polygon with $n$ sides:
$$\text{Sum of interior angles} = (n-2) \times 180°$$
$$\text{Each interior angle} = \frac{(n-2) \times 180°}{n}$$
$$\text{Each exterior angle} = \frac{360°}{n}$$
Key fact: Interior angle $+$ exterior angle $= 180°$ (angles on a straight line).
Polygon Names
| $n$ | Name | Interior angle | Exterior angle |
|---|---|---|---|
| 3 | Triangle | $60°$ | $120°$ |
| 4 | Quadrilateral (Square) | $90°$ | $90°$ |
| 5 | Pentagon | $108°$ | $72°$ |
| 6 | Hexagon | $120°$ | $60°$ |
| 7 | Heptagon | $128.6°$ | $51.4°$ |
| 8 | Octagon | $135°$ | $45°$ |
| 9 | Nonagon | $140°$ | $40°$ |
| 10 | Decagon | $144°$ | $36°$ |
| 12 | Dodecagon | $150°$ | $30°$ |
Worked Example 1.6 — Finding the Number of Sides
(a) A regular polygon has an interior angle of $156°$. How many sides does it have?
(b) The sum of the interior angles of a polygon is $2340°$. How many sides does it have?
(a) Step 1 Exterior angle $= 180° - 156° = 24°$.
Step 2 $n = \dfrac{360°}{24°} = 15$ sides.
(b) Step 1 Use $(n-2) \times 180° = 2340°$:
$n - 2 = \dfrac{2340°}{180°} = 13$, so $n = 15$ sides.
1.7 Bearings
A bearing is a direction measured as a clockwise angle from north. Bearings are always written as three-figure numbers (e.g., $045°$, not $45°$).
Bearing Rules
- Always measure from North, going clockwise.
- Always give bearings as three figures: write $045°$ not $45°$.
- The range is $000°$ to $360°$ (or just below $360°$).
Back Bearing
If the bearing from $A$ to $B$ is $\theta$, then the bearing from $B$ to $A$ (the back bearing or reverse bearing) is:
$$\text{back bearing} = \begin{cases} \theta + 180° & \text{if } \theta < 180° \\ \theta - 180° & \text{if } \theta \geq 180° \end{cases}$$
Worked Example 1.7 — Bearings and Back Bearings
(a) The bearing from town $A$ to town $B$ is $072°$. What is the bearing from $B$ to $A$?
(b) Town $C$ is due East of town $D$. Town $E$ is on a bearing of $310°$ from $D$. Find the angle $\angle CDE$.
(a) $072° < 180°$, so back bearing $= 072° + 180° = 252°$.
(b) Step 1 Due East from $D$ corresponds to a bearing of $090°$. Bearing of $E$ from $D$ is $310°$.
Step 2 $\angle CDE = 360° - 310° + 90° = 140°$. Alternatively, measure clockwise from $DE$ to $DC$: the angle from North to $E$ is $310°$, and from North to $C$ is $090°$. Going anti-clockwise from $C$ to $E$ via North: $360° - 310° + 90° = 140°$.
Worked Example 1.8 — Bearings with Parallel Lines
Ship $S$ travels on a bearing of $130°$. Lines on a chart represent North at two observation points $A$ and $B$ on the same North–South line. If the ship's track makes an angle of $130°$ with north at $A$, find the bearing of the ship as seen from $B$, explaining your reasoning.
Step 1 The two North lines (at $A$ and $B$) are parallel (both pointing North).
Step 2 The ship's track is a transversal across those parallel North lines.
Step 3 By the alternate angles (Z-angles) rule, the angle the track makes with the North line at $B$ on the opposite side = $130°$. Since bearings are measured clockwise from North, the bearing from $B$ is also $130°$.
Reason Alternate angles between parallel lines — applicable because both North lines are parallel.
Exercises
Exercise 1.1
Angles on a straight line are $2x + 15°$, $x + 5°$, and $4x - 10°$. Find $x$ and state the size of each angle.
Show Solution
$(2x+15°) + (x+5°) + (4x-10°) = 180°$
$7x + 10° = 180°$, so $7x = 170°$, giving $x = \dfrac{170}{7} \approx 24.3°$.
Angles: $2(24.3°)+15°=63.6°$; $24.3°+5°=29.3°$; $4(24.3°)-10°=87.1°$. Check: $63.6+29.3+87.1=180°$ ✓
Exercise 1.2
Two parallel lines are cut by a transversal. One of the co-interior angles is $3y - 12°$. The other is $5y + 8°$. Find $y$ and both angles.
Show Solution
Co-interior angles sum to $180°$: $(3y-12°)+(5y+8°)=180°$
$8y - 4° = 180°$, so $8y = 184°$, giving $y = 23°$.
Angles: $3(23°)-12°=57°$ and $5(23°)+8°=123°$. Check: $57+123=180°$ ✓
Exercise 1.3
An isosceles triangle has one angle of $100°$. Find the other two angles and state the type of isosceles triangle (acute or obtuse).
Show Solution
The $100°$ angle is the vertex angle (since if $100°$ were a base angle, both base angles would be $100°$ and the sum would exceed $180°$).
Base angles: $\dfrac{180°-100°}{2} = 40°$ each.
This is an obtuse-angled isosceles triangle (one angle $> 90°$).
Exercise 1.4
The exterior angle of a regular polygon is $24°$. (a) How many sides does the polygon have? (b) What is the sum of the interior angles?
Show Solution
(a) $n = \dfrac{360°}{24°} = 15$ sides.
(b) Sum $= (15-2) \times 180° = 13 \times 180° = 2340°$.
Exercise 1.5
In parallelogram $PQRS$, angle $P = 2a + 30°$ and angle $Q = 3a + 10°$. Find $a$ and all four angles.
Show Solution
Adjacent angles in a parallelogram are supplementary: $(2a+30°)+(3a+10°)=180°$
$5a + 40° = 180°$, so $a = 28°$.
$\angle P = 86°$, $\angle Q = 94°$, $\angle R = 86°$, $\angle S = 94°$.
Exercise 1.6
The bearing of lighthouse $L$ from harbour $H$ is $055°$. A ship travels from $H$ directly to $L$, then continues in the same direction to port $P$. What bearing must the ship's captain set to return directly from $P$ to $H$?
Show Solution
Bearing from $H$ to $L$ (and $P$) is $055°$. Since $055° < 180°$, back bearing $= 055° + 180° = 235°$.
The captain must set a bearing of $\mathbf{235°}$.
Exercise 1.7 Extended
In the diagram, $AB \parallel CD$. Angle $ABE = 47°$ and angle $DCE = 85°$. Find angle $BEC$, giving reasons for each step.
Show Solution
Draw a line $EF$ through $E$ parallel to both $AB$ and $CD$.
$\angle BEF = 47°$ (alternate angles, $EF \parallel AB$).
$\angle CEF = 85°$ (alternate angles, $EF \parallel CD$).
$\angle BEC = \angle BEF + \angle CEF = 47° + 85° = 132°$.
Exercise 1.8 Extended
A regular polygon has an interior angle that is five times its exterior angle. Find the number of sides.
Show Solution
Let exterior angle $= e$. Interior angle $= 5e$. They sum to $180°$: $e + 5e = 180°$, so $e = 30°$.
Number of sides $= \dfrac{360°}{30°} = 12$. It is a regular dodecagon.
Exercise 1.9
Three of the interior angles of a hexagon are $130°$, $95°$, and $115°$. The other three angles are equal. Find the equal angle.
Show Solution
Sum of interior angles of a hexagon $= (6-2)\times 180° = 720°$.
Sum of known angles $= 130°+95°+115° = 340°$.
Remaining sum $= 720°-340° = 380°$. Each equal angle $= \dfrac{380°}{3} = 126.\overline{6}° \approx 126.7°$.
Exercise 1.10 Extended
$ABC$ is a triangle with $\angle BAC = 65°$. $D$ is a point on $BC$ such that $AD$ bisects angle $BAC$. $E$ is a point on $AB$ such that $CE$ is perpendicular to $AB$. Find angle $AEC$ and angle $DAC$.
Show Solution
$\angle AEC = 90°$ (given: $CE \perp AB$).
$\angle DAC = \dfrac{1}{2}\angle BAC = \dfrac{1}{2}(65°) = 32.5°$ (AD bisects $\angle BAC$).
Exam Tips
Top Tips for Angles and Shapes
- Always give a reason when finding angles. Write "angles on a straight line", "vertically opposite", "alternate angles (AB ∥ CD)", etc. Reasons earn marks.
- Mark equal angles with the same letter on your diagram. This prevents confusion in multi-step problems.
- Parallel line problems: look for Z, F, and C shapes to identify which rule to use. Co-interior (C) angles sum to 180° — not equal.
- Bearings: always measure clockwise from North; always write three figures. Double-check with a quick sketch.
- Polygon check: if the question asks whether a polygon is regular, check if the exterior angle divides exactly into 360°.
- Unknown angles in polygons: first find the total interior angle sum, then subtract the known angles.