Angle Rules
On a line · In a triangle · Parallel lines · Polygons
Angles on a Straight Line
\(\text{Angles on a line} = 180°\)
Angles at a point sum to 360°. Vertically opposite angles are equal.
Angles in a Triangle
\(A + B + C = 180°\)
Exterior angle = sum of the two non-adjacent interior angles.
Angles in a Polygon
\(\text{Sum of interior angles} = (n-2) \times 180°\)
\(\text{Each interior} = \dfrac{(n-2) \times 180°}{n}\)
\(\text{Each exterior} = \dfrac{360°}{n}\)
Parallel Line Angle Rules
ALTERNATE ANGLES
Z-angles. Equal. On opposite sides of the transversal.
CO-INTERIOR ANGLES
C-angles. Add to 180°. Same side of the transversal.
CORRESPONDING ANGLES
F-angles. Equal. Same position at each intersection.
Exam language: Always state the reason. "Alternate angles, AB ∥ CD" scores the mark — just "Z-angles" does not.
Polygon Summary
| Shape | Sides (n) | Sum of interior | Each interior (regular) | Each exterior (regular) |
|---|---|---|---|---|
| Triangle | 3 | 180° | 60° | 120° |
| Quadrilateral | 4 | 360° | 90° | 90° |
| Pentagon | 5 | 540° | 108° | 72° |
| Hexagon | 6 | 720° | 120° | 60° |
| Octagon | 8 | 1080° | 135° | 45° |
Properties of Shapes
Triangles · Quadrilaterals · Circles · Similarity & Congruence
Types of Triangles
Equilateral: all sides equal, all angles 60°
Isosceles: 2 equal sides, 2 equal base angles
Scalene: no equal sides or angles
Right-angled: one angle = 90°
Quadrilateral Properties
Square: 4 equal sides, 4 right angles, diagonals bisect perpendicularly
Rectangle: opposite sides equal, 4 right angles, equal diagonals
Rhombus: 4 equal sides, opposite angles equal, diagonals bisect at 90°
Parallelogram: opposite sides equal and parallel, opposite angles equal
Trapezium: exactly one pair of parallel sides
Kite: 2 pairs of adjacent equal sides, one pair of equal opposite angles
Circle Theorems
THEOREM 1
Angle at centre = 2 × angle at circumference (same arc)
THEOREM 2
Angles in the same segment are equal
THEOREM 3
Angle in semicircle = 90° (angle in a semicircle)
THEOREM 4
Cyclic quadrilateral: opposite angles add to 180°
THEOREM 5
Tangent ⊥ radius at the point of contact
THEOREM 6
Tangent from external point: two tangents are equal in length
Similarity & Congruence
CONGRUENT TRIANGLES (same shape AND size)
SSS — 3 sides equal
SAS — 2 sides + included angle equal
ASA / AAS — 2 angles + 1 side equal
RHS — right angle, hypotenuse, side equal
SAS — 2 sides + included angle equal
ASA / AAS — 2 angles + 1 side equal
RHS — right angle, hypotenuse, side equal
SIMILAR TRIANGLES (same shape, different size)
All angles equal (AA sufficient). Sides in proportion.
If linear scale factor = \(k\):
Area scale factor = \(k^2\)
Volume scale factor = \(k^3\)
If linear scale factor = \(k\):
Area scale factor = \(k^2\)
Volume scale factor = \(k^3\)
Common trap: Don't forget to square for area and cube for volume.
Pythagoras & Trigonometry
Pythagoras' theorem · SOH-CAH-TOA · Sine & cosine rules · Bearings
Pythagoras' Theorem
\(a^2 + b^2 = c^2\)
Where \(c\) is the hypotenuse (longest side, opposite the right angle).
Find hyp
\(c = \sqrt{a^2 + b^2}\)
Find side
\(a = \sqrt{c^2 - b^2}\)
3D Pythagoras: Apply in two steps. Find a diagonal on the base first, then use it as the base for the 3D calculation.
SOH-CAH-TOA (Right-angled triangles)
SOH
\(\sin\theta = \dfrac{opp}{hyp}\)CAH
\(\cos\theta = \dfrac{adj}{hyp}\)TOA
\(\tan\theta = \dfrac{opp}{adj}\)Finding the angle: Use the inverse function — \(\theta = \sin^{-1}\left(\frac{opp}{hyp}\right)\). Always check your calculator mode (deg/rad)!
Sine Rule · Cosine Rule · Area Formula
SINE RULE
\(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)
Use when given a side and its opposite angle. Watch for the ambiguous case (SSA).
COSINE RULE
\(a^2 = b^2 + c^2 - 2bc\cos A\)
\(\cos A = \dfrac{b^2+c^2-a^2}{2bc}\)
Use with 3 sides or 2 sides + included angle.
AREA
\(\text{Area} = \dfrac{1}{2}ab\sin C\)
Use when given 2 sides and the included angle \(C\) between them.
Bearings
Rule 1
Bearings are always measured clockwise from North.
Rule 2
Always written as 3 digits (e.g. 045°, not 45°).
Rule 3
To find the reverse bearing: add or subtract 180°.
Back bearing: If A is on a bearing of 070° from B, then B is on bearing \(070° + 180° = 250°\) from A.
Worked Example
A ship sails 12 km due East, then 9 km due North. Find its distance and bearing from start.
Distance
\(d = \sqrt{12^2 + 9^2} = \sqrt{225} = 15\text{ km}\)
Angle
\(\tan\theta = \frac{12}{9} \Rightarrow \theta = \tan^{-1}(1.\overline{3}) \approx 53.1°\)
Bearing
Bearing from North = \(90° - 53.1° = 036.9°\) ≈ 037°
Practice Questions
3 Easy · 3 Medium · 3 Hard — attempt first, then reveal full solution
Easy 1
[2]
Find the size of each interior angle of a regular octagon.
STEP-BY-STEP SOLUTION
Formula
Sum of interior angles = \((n-2)\times180° = (8-2)\times180° = 1080°\)
Each angle
\(\dfrac{1080°}{8} = \boldsymbol{135°}\)
Alternatively: each exterior = 360°÷8 = 45°, so interior = 180°−45° = 135° ✓
Easy 2
[2]
A right-angled triangle has legs of length 8 cm and 15 cm. Find the hypotenuse.
STEP-BY-STEP — PYTHAGORAS
Apply
\(c^2 = 8^2 + 15^2 = 64 + 225 = 289\)
Answer
\(c = \sqrt{289} = \boldsymbol{17 \text{ cm}}\)
8-15-17 is a Pythagorean triple. Other common ones: 3-4-5, 5-12-13.
Easy 3
[2]
In triangle ABC, \(\angle A = 90°\), AB = 7 cm, BC = 10 cm. Find angle B.
STEP-BY-STEP — SOH-CAH-TOA
Identify
From angle B: AB = adjacent = 7, BC = hypotenuse = 10.
Use CAH
\(\cos B = \dfrac{7}{10} = 0.7\)
Answer
\(B = \cos^{-1}(0.7) = \boldsymbol{45.6°}\) (1 d.p.)
Medium 1
[3]
AB and CD are parallel. Angle ABE = 55° and angle CDE = 38°. Find angle BED, giving a reason for each step.
STEP-BY-STEP — PARALLEL LINE ANGLES
Draw alt. angle
Draw a line through E parallel to AB and CD.
Angle 1
Alternate angle to ABE = 55° (AB ∥ through-E-line, alternate angles equal)
Angle 2
Alternate angle to CDE = 38° (CD ∥ through-E-line, alternate angles equal)
BED
\(\angle BED = 55° + 38° = \boldsymbol{93°}\)
Medium 2
[4]
Triangle PQR has PQ = 8 cm, QR = 11 cm, and angle PQR = 65°. Find the area of triangle PQR and the length PR.
STEP-BY-STEP — AREA FORMULA + COSINE RULE
Area
\(\text{Area} = \frac{1}{2} \times 8 \times 11 \times \sin 65°\)
\(= 44\sin 65° \approx \boldsymbol{39.9 \text{ cm}^2}\)
\(= 44\sin 65° \approx \boldsymbol{39.9 \text{ cm}^2}\)
Cosine rule
\(PR^2 = 8^2 + 11^2 - 2(8)(11)\cos 65°\)
\(= 64 + 121 - 176\cos 65°\)
\(= 185 - 176(0.4226) \approx 185 - 74.4 = 110.6\)
\(= 64 + 121 - 176\cos 65°\)
\(= 185 - 176(0.4226) \approx 185 - 74.4 = 110.6\)
Answer
\(PR = \sqrt{110.6} \approx \boldsymbol{10.5 \text{ cm}}\)
Medium 3
[3]
Two similar triangles have areas of 16 cm² and 36 cm². If the perimeter of the smaller triangle is 20 cm, find the perimeter of the larger triangle.
STEP-BY-STEP — SIMILARITY SCALE FACTOR
Area ratio
Area ratio = \(\dfrac{36}{16} = \dfrac{9}{4}\)
Linear ratio
Linear scale factor \(k = \sqrt{\dfrac{9}{4}} = \dfrac{3}{2}\)
Perimeter
Larger perimeter \(= 20 \times \dfrac{3}{2} = \boldsymbol{30 \text{ cm}}\)
⚠ Areas scale by \(k^2\) — always take the square root to get the linear scale factor.
Hard 1
[5]
ABCD is a cyclic quadrilateral. Angle ABС = 3x + 10° and angle ADC = 5x − 30°. Find angle ABC and show that AC is a diameter of the circle.
STEP-BY-STEP — CIRCLE THEOREMS
Theorem
Opposite angles of a cyclic quadrilateral add to 180°.
\((3x+10) + (5x-30) = 180\)
\((3x+10) + (5x-30) = 180\)
Solve
\(8x - 20 = 180 \Rightarrow 8x = 200 \Rightarrow x = 25\)
ABC
\(\angle ABC = 3(25)+10 = \boldsymbol{85°}\)
ADC
\(\angle ADC = 5(25)-30 = 95°\)
Diameter?
For AC to be a diameter, angle ADC or ABC must be 90° (angle in semicircle). But 85° ≠ 90° and 95° ≠ 90°, so AC is not a diameter. (Exam note: check the question wording — you may need to compute a different angle.)
Hard 2
[5]
A cuboid has dimensions 6 cm × 4 cm × 3 cm. Find (a) the length of the space diagonal, (b) the angle this diagonal makes with the base.
STEP-BY-STEP — 3D PYTHAGORAS
(a) Base diagonal
\(d_{\text{base}} = \sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13}\)
(a) Space diagonal
\(d = \sqrt{52 + 3^2} = \sqrt{61} \approx \boldsymbol{7.81 \text{ cm}}\)
(b) Angle with base
\(\tan\theta = \dfrac{3}{\sqrt{52}} \Rightarrow \theta = \tan^{-1}\!\left(\dfrac{3}{\sqrt{52}}\right) \approx \boldsymbol{22.6°}\)
Strategy: First find the base diagonal — this becomes adjacent. Height (3) is opposite. Then use tan.
Hard 3
[5]
In triangle XYZ, XY = 9 cm, YZ = 7 cm, and XZ = 5 cm. Find angle XYZ and hence find the area of the triangle.
STEP-BY-STEP — COSINE RULE THEN AREA
Cosine rule
Find angle Y (opposite XZ = 5):
\(\cos Y = \dfrac{9^2+7^2-5^2}{2(9)(7)} = \dfrac{81+49-25}{126} = \dfrac{105}{126} = \dfrac{5}{6}\)
\(\cos Y = \dfrac{9^2+7^2-5^2}{2(9)(7)} = \dfrac{81+49-25}{126} = \dfrac{105}{126} = \dfrac{5}{6}\)
Angle Y
\(Y = \cos^{-1}\!\left(\tfrac{5}{6}\right) \approx \boldsymbol{33.6°}\)
Area
\(\text{Area} = \frac{1}{2}(9)(7)\sin(33.6°) \approx \frac{63}{2}\times0.5528 \approx \boldsymbol{17.4 \text{ cm}^2}\)
Past Year Paper Style Questions
Cambridge IGCSE 0580 style — multi-part, show all working
0580 Style
Polygons & Parallel Lines
[9]
(a) The interior angle of a regular polygon is 156°. How many sides does it have? [2]
(b) In the diagram, AB is parallel to CD. Angle ABE = 48° and angle ECD = 75°. Find angle BEC, giving reasons. [4]
(c) The exterior angle of a regular polygon is \((x+15)°\) and the interior angle is \((5x-3)°\). Find \(x\) and the number of sides. [3]
(b) In the diagram, AB is parallel to CD. Angle ABE = 48° and angle ECD = 75°. Find angle BEC, giving reasons. [4]
(c) The exterior angle of a regular polygon is \((x+15)°\) and the interior angle is \((5x-3)°\). Find \(x\) and the number of sides. [3]
FULL WORKED SOLUTION
(a)
Each exterior = \(180°-156°=24°\). Number of sides = \(360°\div24°=\boldsymbol{15}\)
(b) Method
Draw line through E parallel to AB and CD.
Alt. angle to ABE from E = 48° (alternate angles, AB ∥ parallel-line)
Alt. angle to ECD from E = 75° (alternate angles, CD ∥ parallel-line)
\(\angle BEC = 48°+75° = \boldsymbol{123°}\)
Alt. angle to ABE from E = 48° (alternate angles, AB ∥ parallel-line)
Alt. angle to ECD from E = 75° (alternate angles, CD ∥ parallel-line)
\(\angle BEC = 48°+75° = \boldsymbol{123°}\)
(c) Equation
Interior + exterior = 180°:
\((5x-3)+(x+15)=180 \Rightarrow 6x+12=180 \Rightarrow x=28\)
\((5x-3)+(x+15)=180 \Rightarrow 6x+12=180 \Rightarrow x=28\)
(c) Sides
Exterior = \(28+15=43°\). Sides = \(360\div43\) — not a whole number, so re-check: actually exterior = \(x+15 = 43°\). \(360/43\) is not integer → likely the exam intends exterior = \((x+14)°\) or similar. Using the values: \(\boldsymbol{n = 360°/(x+15)° = 360°/43° \approx 8.4}\). The answer expected may vary — set up the equations as shown.
0580 Style
Trigonometry & Bearings
[10]
(a) A ship sails from port P on a bearing of 040° for 15 km to reach port Q. It then sails on a bearing of 130° for 20 km to reach port R. Find the distance PR and the bearing of R from P. [5]
(b) Three towns A, B and C are such that B is 12 km from A on a bearing of 065°, and C is 18 km from A on a bearing of 155°. Find BC and angle ABC. [5]
(b) Three towns A, B and C are such that B is 12 km from A on a bearing of 065°, and C is 18 km from A on a bearing of 155°. Find BC and angle ABC. [5]
FULL WORKED SOLUTION
(a) Angle PQR
Bearing 040° then 130°. Angle at Q between directions: south of PQ is bearing 220°. Angle PQR = 130°−40°+90° ... use co-interior: angle between north at Q and north at P with 040° bearing means angle PQR = 180°−(130°−040°) = 180°−90° = 90°. So PQR = 90°!
(a) PR
\(PR = \sqrt{15^2+20^2} = \sqrt{625} = \boldsymbol{25\text{ km}}\)
(a) Bearing
\(\tan(\angle RPQ) = \frac{20}{15} \Rightarrow \angle RPQ \approx 53.1°\). Bearing of R from P = \(40°+53.1° \approx \boldsymbol{093°}\)
(b) Angle BAC
Angle between AB (065°) and AC (155°) = 155°−65° = 90°
(b) BC
\(BC = \sqrt{12^2+18^2} = \sqrt{468} \approx \boldsymbol{21.6\text{ km}}\)
(b) Angle ABC
\(\tan(\angle ABC) = \frac{18}{12} = 1.5 \Rightarrow \angle ABC \approx \boldsymbol{56.3°}\)
0580 Style
Circle Theorems & Similarity
[9]
(a) O is the centre of a circle. Points A, B, C lie on the circumference. Angle BOC = 112°. Find angle BAC and state the theorem used. [2]
(b) A tangent from external point P touches the circle at T. PT = 12 cm and the radius of the circle is 5 cm. Find the distance from P to the centre O. [3]
(c) Two similar cones have heights in the ratio 2:5. The volume of the smaller cone is 32 cm³. Find the volume of the larger cone. [4]
(b) A tangent from external point P touches the circle at T. PT = 12 cm and the radius of the circle is 5 cm. Find the distance from P to the centre O. [3]
(c) Two similar cones have heights in the ratio 2:5. The volume of the smaller cone is 32 cm³. Find the volume of the larger cone. [4]
FULL WORKED SOLUTION
(a)
Angle at centre = 2 × angle at circumference.
\(\angle BAC = \frac{112°}{2} = \boldsymbol{56°}\)
Theorem: Angle at the centre is twice the angle at the circumference, subtended by the same arc.
\(\angle BAC = \frac{112°}{2} = \boldsymbol{56°}\)
Theorem: Angle at the centre is twice the angle at the circumference, subtended by the same arc.
(b)
Tangent ⊥ radius, so triangle OTP has a right angle at T.
\(OP = \sqrt{OT^2+PT^2} = \sqrt{5^2+12^2} = \sqrt{169} = \boldsymbol{13\text{ cm}}\)
\(OP = \sqrt{OT^2+PT^2} = \sqrt{5^2+12^2} = \sqrt{169} = \boldsymbol{13\text{ cm}}\)
(c)
Linear scale factor = \(\frac{5}{2}\). Volume scale factor = \(\left(\frac{5}{2}\right)^3 = \frac{125}{8}\)
Larger volume = \(32 \times \frac{125}{8} = \boldsymbol{500\text{ cm}^3}\)
Larger volume = \(32 \times \frac{125}{8} = \boldsymbol{500\text{ cm}^3}\)