OxfordAQA International GCSE Mathematics: Pythagoras' theorem and right-angled trigonometry (9260) – Revision Notes

These notes condense statement G19 in section 3.3.2 of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1. For full explanations and longer worked examples, read the study guide. Test yourself with the practice questions. The course hub, the printable checklist and the free 10-minute diagnostics help you plan. Core content is 2D right-angled work. Extension only content is 3D figures. All content can be assessed on the Extension papers.

Facts to learn

The specification asks you to know these, so learn them word for word.

Result	Statement
Pythagoras	a² + b² = c², where c is the hypotenuse
Sine	sin θ = opposite / hypotenuse
Cosine	cos θ = adjacent / hypotenuse
Tangent	tan θ = opposite / adjacent

The memory aid SOH CAH TOA gives the three ratios in order.

Naming the sides

1. Mark the right angle.
2. The side opposite it is the hypotenuse.
3. Mark the angle θ you are using.
4. The side facing θ is the opposite.
5. The remaining side, touching θ, is the adjacent.

Change the angle and the opposite and adjacent swap. The hypotenuse stays put.

Pythagoras: method in steps

1. Draw the right-angled triangle and mark the hypotenuse.
2. Hypotenuse unknown: c² = a² + b² (add).
3. Short side unknown: b² = c² − a² (subtract).
4. Take the square root.
5. Check: the hypotenuse is the longest side.

Reminder. Short sides 9.6 cm and 12.8 cm: c² = 92.16 + 163.84 = 256, so c = 16 cm. Hypotenuse 16 cm and short side 9.6 cm: b² = 256 − 92.16 = 163.84, so b = 12.8 cm.

Trigonometry: choosing the ratio

Sides involved	Ratio
opposite and hypotenuse	sin
adjacent and hypotenuse	cos
opposite and adjacent	tan

Finding a length

1. Label the three sides from the given angle.
2. Pick the ratio that holds your known side and your unknown side.
3. Write the equation, for example sin 40° = x/15.
4. Unknown on top: multiply. x = 15 × sin 40°.
5. Unknown underneath: swap places. If cos 40° = 15/h then h = 15 / cos 40°.

Finding an angle

1. Write the ratio of the two known sides.
2. Apply the inverse function.
3. Round only at the end, and state the accuracy asked.

Reminder. Adjacent 7.2 cm, hypotenuse 11.5 cm: cos θ = 7.2/11.5 = 0.6261, so θ = cos⁻¹(0.6261) = 51.2° (1 d.p.).

The other acute angle is 90° − θ.

Which method? A decision table

Information given	What you want	Use
two sides	third side	Pythagoras
one side and one acute angle	another side	sin, cos or tan
two sides	an acute angle	inverse sin, cos or tan
a triangle without a right angle	any	add a perpendicular, or use another unit

2D figures

• Isosceles triangle. Split it down the middle to get two right-angled triangles. The base halves.
• Rectangle. The diagonal makes a right-angled triangle with two sides.
• Rhombus. The diagonals cross at right angles and bisect each other, so four matching right-angled triangles form.
• Trapezium with a right angle. Draw a perpendicular height to make a right-angled triangle.
• Two triangles in a row. Find a shared side first, and keep its unrounded value.

Worked reminders: mixed 2D problems

A leaning pole. A pole 5.6 m long rests against a vertical wall and makes an angle of 68° with the ground. Hypotenuse and angle are known, and the height up the wall is opposite the angle, so use sine: height = 5.6 × sin 68° = 5.19 m. The foot is adjacent to the angle, so use cosine: distance from wall = 5.6 × cos 68° = 2.10 m. Check with Pythagoras: 5.19² + 2.10² is about 31.4, and 5.6² = 31.36.

A diagonal brace. A rectangular frame is 1.8 m wide and 0.9 m high. Brace length = √(1.8² + 0.9²) = √4.05 = 2.01 m. The angle between the brace and the long side has tan θ = 0.9/1.8 = 0.5, so θ = 26.6° (1 d.p.).

Two answers from one triangle. When a question asks for a length and an angle, find the length with Pythagoras if two sides are given. Then find the angle from the original given sides, not from your rounded length.

Checking an answer

• A short side must be smaller than the hypotenuse.
• The side opposite the larger acute angle is the longer short side.
• If sin, cos or tan is used for an angle, the ratio must make sense: sin and cos of an acute angle lie between 0 and 1.
• Estimate first. A triangle with short sides about 3 and 4 has a hypotenuse near 5.

Extension only: 3D figures

1. Sketch the 3D figure.
2. Find the line or angle you need.
3. Pick out one flat right-angled triangle that contains it.
4. Redraw that triangle in 2D and write in its sides.
5. Use Pythagoras or trigonometry as in 2D.
6. For a long calculation, break it into two triangles: base first, then the vertical triangle.

3D result	How
Base diagonal of cuboid	√(l² + w²)
Space diagonal of cuboid	√(l² + w² + h²)
Angle between a line and a plane	angle between the line and its shadow on the plane
Pyramid with apex above centre	vertical height meets the base at the centre

Reminder (Extension only). Cuboid 12 by 5 by 9: base diagonal = 13, space diagonal = √(169 + 81) = √250 = 15.8 cm, and the angle with the base has tan θ = 9/13, so θ = 34.7° (1 d.p.).

Reminder (Extension only). Pyramid on a 6 by 6 square base with height 9: half the base diagonal is 3√2 = 4.243, the edge is √(18 + 81) = √99 = 9.95 cm, and the edge makes tan θ = 9/4.243, so θ = 64.8° (1 d.p.) with the base.

Must-know distinctions

• Theorem versus ratio. Pythagoras links three sides and uses no angle. Trigonometry links an angle to two sides.
• sin θ versus sin⁻¹. sin θ gives a ratio from an angle. sin⁻¹ gives an angle from a ratio.
• Hypotenuse versus longest line. The hypotenuse is the longest side of the right-angled triangle only. In a 3D figure the longest line is the space diagonal.
• Angle with a plane versus angle with an edge. In a 3D figure, use the shadow of the line on the plane, not a random edge.
• Rounding. Keep unrounded values in the calculator until the final line.

Quick self-test

Answer each question, then check the answers below.

1. A right-angled triangle has short sides 6.5 cm and 7.2 cm. Find the hypotenuse to 3 s.f.
2. The hypotenuse is 9.4 cm and one short side is 5.6 cm. Find the other short side to 3 s.f.
3. In a right-angled triangle the angle is 28° and the hypotenuse is 12 cm. Find the side opposite the angle to 3 s.f.
4. The angle is 61° and the adjacent side is 4.5 cm. Find the hypotenuse to 3 s.f.
5. Opposite 3.9 cm and adjacent 8.2 cm. Find the angle to 1 d.p.
6. Hypotenuse 13 cm and opposite 6 cm. Find the angle to 1 d.p.
7. One acute angle in a right-angled triangle is 36.5°. What is the other?
8. Which ratio links the adjacent and the hypotenuse?
9. (Extension only) A box is 20 cm by 15 cm by 12 cm. Find its space diagonal to 3 s.f.
10. (Extension only) A pyramid has a square base of side 10 cm and a vertical height of 12 cm above the centre. Find the length of a sloping edge to 3 s.f.

Answers

1. √(42.25 + 51.84) = √94.09 = 9.70 cm.
2. √(88.36 − 31.36) = √57 = 7.55 cm.
3. 12 × sin 28° = 5.63 cm.
4. 4.5 / cos 61° = 9.28 cm.
5. tan⁻¹(3.9/8.2) = 25.4°.
6. sin⁻¹(6/13) = 27.5°.
7. 90° − 36.5° = 53.5°.
8. Cosine.
9. √(400 + 225 + 144) = √769 = 27.7 cm.
10. Half base diagonal = √50, so edge = √(50 + 144) = √194 = 13.9 cm.

Where marks are usually lost

• Adding when you should subtract, or the other way round, in Pythagoras.
• Leaving the answer as c² and forgetting the square root.
• Labelling opposite and adjacent from the wrong angle.
• Using the wrong ratio and then quoting a bare answer with no working to show the intended method.
• Inverting the fraction when finding an angle.
• Rounding early so the final digit is out.
• Giving an angle with no degree symbol or a length with no unit.
• In 3D, using a sloping edge as a vertical height.
• In 3D, measuring an angle to an edge when the question asks for the angle with a plane.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.3.2 Mensuration and calculation, statement G19.