G20: Pythagoras & Trigonometry
Know and use Pythagoras' theorem; trigonometric ratios in right-angled triangles; use these in 3D (Higher)
📋 Key Concepts
Pythagoras' theorem applies to right-angled triangles and relates the three sides.
Trigonometry relates angles and sides in right-angled triangles using sin, cos, and tan.
📝 Pythagoras' Theorem
a² + b² = c²
Where c is the hypotenuse (longest side, opposite the right angle).
Finding the hypotenuse: c = √(a² + b²)
Finding a shorter side: a = √(c² - b²)
Example 1
Find the hypotenuse of a right-angled triangle with sides 5 cm and 12 cm.
Solution:
c² = 5² + 12² = 25 + 144 = 169
c = √169 = 13 cm
Example 2
Find the missing side in a right-angled triangle with hypotenuse 15 cm and one side 9 cm.
Solution:
a² = 15² - 9² = 225 - 81 = 144
a = √144 = 12 cm
📝 Trigonometric Ratios
Label the triangle:
- Opposite (opp) - side opposite the angle
- Adjacent (adj) - side next to the angle
- Hypotenuse (hyp) - longest side
sin θ = opp/hyp
cos θ = adj/hyp
tan θ = opp/adj
cos θ = adj/hyp
tan θ = opp/adj
Memory aid: SOHCAHTOA
- Sin = Opposite/Hypotenuse
- Cos = Adjacent/Hypotenuse
- Tan = Opposite/Adjacent
📝 Finding a Side
Example 3
Find x in a right-angled triangle with angle 30° and hypotenuse 10 cm, where x is the opposite side.
Solution:
sin 30° = x/10
x = 10 × sin 30°
x = 10 × 0.5 = 5 cm
Example 4
Find the adjacent side when the angle is 40° and the hypotenuse is 8 cm.
Solution:
cos 40° = adj/8
adj = 8 × cos 40°
adj = 8 × 0.766 = 6.13 cm (to 2 d.p.)
📝 Finding an Angle
Use inverse functions: sin⁻¹, cos⁻¹, tan⁻¹
Example 5
Find angle θ when opposite = 7 cm and hypotenuse = 10 cm.
Solution:
sin θ = 7/10 = 0.7
θ = sin⁻¹(0.7)
θ = 44.4° (to 1 d.p.)
Example 6
Find angle θ when opposite = 5 cm and adjacent = 12 cm.
Solution:
tan θ = 5/12
θ = tan⁻¹(5/12)
θ = 22.6° (to 1 d.p.)
📝 Higher: 3D Pythagoras
3D Pythagoras: Find the diagonal of a cuboid.
Diagonal = √(a² + b² + c²)
Example 7
Find the length of the space diagonal of a cuboid 3 cm × 4 cm × 5 cm.
Solution:
Diagonal = √(3² + 4² + 5²)
= √(9 + 16 + 25)
= √50 = 5√2 cm ≈ 7.07 cm
📝 Higher: 3D Trigonometry
Example 8
Find the angle between the space diagonal of a 3 cm × 4 cm × 5 cm cuboid and the base.
Solution:
Height = 5 cm, diagonal of base = √(3² + 4²) = 5 cm
tan θ = 5/5 = 1
θ = tan⁻¹(1) = 45°
❓ Practice Questions
Q1: Find the hypotenuse of a triangle with sides 8 cm and 15 cm.
Q2: Find the missing side when hypotenuse = 20 cm and one side = 16 cm.
Q3: Find the opposite side when angle = 35° and hypotenuse = 12 cm.
Q4: Find angle θ when opposite = 9 cm and adjacent = 12 cm.
Q5: (Higher) Find the space diagonal of a 2 cm × 6 cm × 9 cm cuboid.
✅ Answers
- 17 cm
- 12 cm
- 12 × sin 35° = 6.88 cm (to 2 d.p.)
- tan⁻¹(9/12) = 36.9° (to 1 d.p.)
- √(4 + 36 + 81) = √121 = 11 cm
🎯 Exam Tips
- Always identify the hypotenuse first (longest side, opposite right angle)
- Use SOHCAHTOA to choose the correct ratio
- For finding angles, use inverse functions
- Check your calculator is in degrees mode (DEG)
- Round answers appropriately
- For 3D, find the diagonal of the base first, then use height
🧠 Problem-Solving Strategies
Problem-Solving
Pythagoras: a squared + b squared = c squared (c is the hypotenuse). SOH CAH TOA for right-angled triangles. Identify hypotenuse (longest side, opposite right angle), opposite (across from the angle) and adjacent (next to the angle, not the hypotenuse). For 3D problems, find right-angled triangles within the 3D shape.
Multi-Step Problem
A ladder of length 10 m leans against a wall making an angle of 65 degrees with the ground. How high up the wall does it reach? How far is the foot of the ladder from the wall?
Solution: Height = 10 x sin(65) = 9.06 m. Distance from wall = 10 x cos(65) = 4.23 m. Check: 9.06 squared + 4.23 squared = 82.1 + 17.9 = 100 = 10 squared.
⚠️ Common Errors
Watch Out!
1. Wrong: Using the hypotenuse as one of the shorter sides in Pythagoras: c squared + a squared = b squared Correct: The hypotenuse (longest side) is always c. a squared + b squared = c squared. Never add the hypotenuse squared to another side squared.
2. Wrong: Mixing up opposite and adjacent sides when using SOH CAH TOA Correct: Always label from the given angle: opposite is across from it, adjacent is next to it (not the hypotenuse). The hypotenuse is always the longest side.
3. Wrong: Using sin, cos or tan with the right angle itself Correct: SOH CAH TOA uses one of the OTHER two angles, not the 90 degree angle. The right angle is never the reference angle for trig ratios.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A ship at point S is 12 km from lighthouse A on a bearing of 040 degrees and 9 km from lighthouse B on a bearing of 300 degrees. (a) Find the distance between the two lighthouses. (b) Find the bearing of lighthouse B from lighthouse A.
(a) The angle at S between the bearings: from 300 degrees to 360 degrees = 60 degrees, plus 40 degrees = 100 degrees. Using cosine rule: AB squared = 12 squared + 9 squared - 2(12)(9)cos(100) = 144 + 81 + 216(0.1736) = 225 + 37.5 = 262.5. AB = 16.2 km.
(b) Using sine rule: sin(angle A)/9 = sin(100)/16.2. sin(angle A) = 9 x 0.9848/16.2 = 0.5471. Angle A = 33.2 degrees. Bearing of B from A = 040 + 180 + 33.2 = 253 degrees (approx, depends on diagram orientation).
Mark scheme: M1 finding angle at S, A1 100 degrees, M1 cosine rule, A1 16.2 km, M1 sine rule, A1 bearing
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A ladder 5 m long leans against a vertical wall. The foot of the ladder is 3 m from the wall.
(a) How high up the wall does the ladder reach?
(b) The foot of the ladder slips 0.5 m further from the wall. Does the top drop by more or less than 0.5 m? Explain without full calculation.
(c) A builder says "Moving the base 1 m out will always drop the top by less than 1 m for any starting position." Is this true?
Answers: (a) h = root(25 - 9) = 4 m. (b) It drops by MORE than 0.5 m. At the top of the wall the height is decreasing faster because the ladder is nearly flat (the relationship between base distance and height is non-linear — Pythagoras is quadratic, not linear). (c) Yes — by Pythagoras, if the base moves out by 1, the height drops by less than 1 for any initial position. This is because the height change involves square roots which decrease slower than linear for large heights but the total change is always less than the base change. Formal proof: new height squared = L squared - (b+1) squared = L squared - b squared - 2b - 1 = h squared - 2b - 1, so the drop = h - root(h squared - 2b - 1) which is less than 1.
📝 Exam Questions by Topic
🎬 Video Resources
Ready to ace your GCSE Mathematics exams?
Get the best revision books and guides to boost your grades.