Chapter 8-17: Trigonometry Exam Technique

Higher Tier ONLY Grades 6-9 Calculator Paper 2 Paper 3

⚠️ Higher Tier Essential Content

This chapter covers advanced trigonometry that is Higher tier ONLY. Sine rule, cosine rule, and area of triangles (½ab sin C) are worth 4-6 marks per question in Paper 3. These topics are the difference between Grade 6 and Grade 8-9. Foundation students should focus on basic Pythagoras and SOH CAH TOA only.

📝 Trigonometry Exam Technique (Papers 2 & 3)

  • Sine/Cosine rules appear in every Higher Paper 3 - worth 5-6 marks each time
  • Draw clear diagrams - label all given information immediately
  • Decide which rule to use - check what you have and what you need
  • Sine rule: Use when you have 2 sides and 2 angles (opposite pairs)
  • Cosine rule: Use when you have 3 sides, or 2 sides and the included angle
  • Area formula (½ab sin C): Often missed - easy 2-3 marks!
  • Bearings with trig: Common Paper 3 multi-step question (6 marks)
  • Calculator mode: MUST be in DEGREES not radians!

📐 Formula Sheet Reference (2025)

These formulas ARE on the GCSE formula sheet, but you must know when to use them!

Sine Rule:
a/sin A = b/sin B = c/sin C
Cosine Rule (finding side):
a² = b² + c² - 2bc cos A
Cosine Rule (finding angle):
cos A = (b² + c² - a²) / 2bc
Area of Triangle:
Area = ½ab sin C

Real Past Paper Questions (Grade 7-9 Focus)

Question 1 (AQA 2023 Paper 3) 4 marks 6-8 mins
ABC is a triangle.
AB = 8 cm
AC = 12 cm
Angle BAC = 35°

Calculate the area of triangle ABC.
Give your answer to 3 significant figures.
[Triangle ABC diagram: AB = 8 cm, AC = 12 cm, angle at A = 35°]

✓ Mark Scheme

Key insight: Two sides and included angle → use Area = ½ab sin C (1 mark)

Step 1: Identify a = 8, b = 12, C = 35° (angle between them)

Step 2: Area = ½ × 8 × 12 × sin(35°) (1 mark for formula)

Step 3: Area = 48 × sin(35°) (1 mark)

Step 4: Area = 48 × 0.5736... = 27.533... (1 mark)

Step 5: Area = 27.5 cm² (to 3 s.f.)

Answer: 27.5 cm²

⚠️ Common Mistakes (Examiner Reports)

  • Using Area = ½ × base × height (need to find height first - longer method)
  • Calculator in radians mode: sin(35) in radians = -0.428... (WRONG!)
  • Rounding too early: using sin(35°) = 0.57 instead of 0.5736...
  • Forgetting to give answer to 3 s.f. as requested
  • No units (cm²) - can lose a mark
Question 2 (Edexcel 2023 Paper 3) 5 marks 8-10 mins
PQR is a triangle.
PQ = 9 cm
QR = 7 cm
Angle PQR = 42°

Calculate the length of PR.
Give your answer to 3 significant figures.
[Triangle PQR diagram: PQ = 9 cm, QR = 7 cm, angle at Q = 42°]

✓ Mark Scheme

Key insight: Two sides and included angle → use Cosine rule (1 mark)

Step 1: a² = b² + c² - 2bc cos A

Let PR = a, PQ = 9, QR = 7, angle Q = 42°

Step 2: PR² = 9² + 7² - 2(9)(7)cos(42°) (1 mark for formula)

Step 3: PR² = 81 + 49 - 126 × 0.7431... (1 mark)

Step 4: PR² = 130 - 93.636... = 36.364 (1 mark)

Step 5: PR = √36.364 = 6.030... cm (1 mark)

Answer: PR = 6.03 cm (to 3 s.f.)

⚠️ Common Mistakes (Examiner Reports)

  • Using Sine rule instead (doesn't work - need opposite pairs)
  • Sign error: adding 2bc cos A instead of subtracting
  • Forgetting to square root at the end (giving 36.4 instead of 6.03)
  • Calculator errors with cos(42°) - must be in degrees!
  • Rounding 6.030 to 6.0 instead of 6.03 (need 3 s.f.)
Question 3 (OCR 2023 Paper 3) 5 marks 8-10 mins
XYZ is a triangle.
XY = 11 cm
YZ = 8 cm
Angle XYZ = 50°

Calculate angle XZY.
Give your answer to 1 decimal place.
[Triangle XYZ diagram: XY = 11 cm, YZ = 8 cm, angle at Y = 50°]

✓ Mark Scheme

Key insight: Two sides and included angle, need opposite angle → Sine rule (1 mark)

Step 1: First find XZ using cosine rule (1 mark)

XZ² = 11² + 8² - 2(11)(8)cos(50°)

XZ² = 121 + 64 - 176 × 0.6428 = 185 - 113.13 = 71.87

XZ = 8.478 cm

Step 2: Now use sine rule: a/sin A = b/sin B (1 mark)

11/sin(angle XZY) = 8.478/sin(50°)

Step 3: sin(angle XZY) = 11 × sin(50°) / 8.478 (1 mark)

sin(angle XZY) = 11 × 0.7660 / 8.478 = 0.9939

Step 4: angle XZY = sin⁻¹(0.9939) = 83.58...° (1 mark)

Answer: angle XZY = 83.6° (to 1 d.p.)

⚠️ Common Mistakes (Examiner Reports)

  • Trying to use sine rule without finding third side first
  • Using wrong sides in sine rule (must use opposite pairs)
  • Calculator error: not using sin⁻¹ (inverse sine) to find angle
  • Getting confused between the three angles - label diagram clearly!
  • Rounding intermediate values too much (use full calculator value)
Question 4 (AQA 2022 Paper 3 - Grade 8-9) 6 marks 10-12 mins
A ship sails from port A on a bearing of 065° for 40 km to point B.
It then sails on a bearing of 140° for 55 km to point C.

Calculate:
(a) The distance AC
(b) The bearing of C from A

Give your answers to 3 significant figures.
[Bearing diagram: North arrows at A and B, bearing 065° from A to B (40 km), bearing 140° from B to C (55 km)]

✓ Mark Scheme

(a) Finding distance AC:

Step 1: Find angle ABC using bearings (1 mark)

Angle at B = 140° - 65° = 75° (angle between AB and BC)

Step 2: Use cosine rule with AB = 40, BC = 55, angle = 75° (1 mark)

AC² = 40² + 55² - 2(40)(55)cos(75°)

AC² = 1600 + 3025 - 4400 × 0.2588 = 4625 - 1138.88 = 3486.12

Step 3: AC = √3486.12 = 59.04... km (1 mark)

Answer (a): AC = 59.0 km (to 3 s.f.)

(b) Finding bearing of C from A:

Step 4: Use sine rule to find angle CAB (1 mark)

55/sin(angle CAB) = 59.04/sin(75°)

sin(angle CAB) = 55 × sin(75°) / 59.04 = 0.8993

angle CAB = sin⁻¹(0.8993) = 64.1°

Step 5: Bearing = 65° + 64.1° = 129.1° (1 mark)

Answer (b): Bearing of C from A = 129° (to 3 s.f.) (1 mark)

⚠️ Common Mistakes (Examiner Reports)

  • Not drawing a clear diagram - leads to angle confusion
  • Calculating angle ABC incorrectly from bearings
  • Using 140° as the angle in triangle (should be 75°)
  • Forgetting that bearing must be measured from North clockwise
  • Getting confused between bearing "from A to C" vs "from C to A"
  • Not showing multi-step working - loses method marks
Question 5 (Edexcel 2023 Paper 3 - Grade 8-9) 6 marks 10-12 mins
Triangle DEF has:
DE = 15 cm
EF = 18 cm
DF = 20 cm

(a) Calculate the size of the largest angle in the triangle.
(b) Hence, calculate the area of triangle DEF.
Give your answers to 3 significant figures.
[Triangle DEF diagram: DE = 15 cm, EF = 18 cm, DF = 20 cm]

✓ Mark Scheme

(a) Finding largest angle:

Key insight: Largest angle is opposite longest side (EF = 18 cm) (1 mark)

So we need angle D (opposite to EF)

Step 1: Use cosine rule for finding angle (1 mark)

cos D = (DE² + DF² - EF²) / (2 × DE × DF)

Step 2: cos D = (15² + 20² - 18²) / (2 × 15 × 20) (1 mark)

cos D = (225 + 400 - 324) / 600 = 301 / 600 = 0.5017

Step 3: angle D = cos⁻¹(0.5017) = 59.88...° (1 mark)

Answer (a): Largest angle = 59.9° (to 3 s.f.)

(b) Finding area (using "hence"):

Step 4: Use Area = ½ab sin C with the angle just found (1 mark)

Area = ½ × 15 × 20 × sin(59.88°)

Area = 150 × 0.8658 = 129.87 cm²

Answer (b): Area = 130 cm² (to 3 s.f.) (1 mark)

⚠️ Common Mistakes (Examiner Reports)

  • Not identifying that largest angle is opposite longest side
  • Using sine rule instead of cosine rule (need 3 sides)
  • Sign errors in cosine rule numerator
  • In part (b), not using the "hence" - calculating area differently
  • "Hence" means use your previous answer - examiners penalize ignoring this
  • Using wrong formula: ½ × base × height requires finding height first

📊 What Different Grades Look Like (Higher Tier)

Grade 5-6:

• Can use basic SOH CAH TOA in right-angled triangles

• Understands Pythagoras' theorem

• Struggles with sine and cosine rules

• May not recognize when to use which formula

Grade 7:

• Can use sine rule with two sides and two angles (Question 3)

• Can use cosine rule to find a side (Question 2)

• Recognizes when to use area formula ½ab sin C (Question 1)

• Makes occasional errors with bearings

Grade 8-9:

• Solves complex bearing problems combining multiple rules (Question 4)

• Uses "hence" correctly in multi-part questions (Question 5)

• Can find angles using both sine and cosine rules confidently

• Recognizes largest angle is opposite longest side

• Shows clear, logical working through 6-mark questions

• Checks answers make sense (e.g., angles sum to 180°)

🎯 Grade 8-9 Trigonometry Tips

  • Decision tree: 2 sides + included angle = Cosine rule OR Area formula
  • Decision tree: 2 sides + 2 angles (opposite pairs) = Sine rule
  • Decision tree: 3 sides only = Cosine rule to find angle
  • Bearings: Always draw North arrow at EACH point, measure angles carefully
  • Calculator check: Press Mode button, ensure "Deg" is displayed
  • "Hence" questions: Must use your previous answer - it's testing multi-step thinking
  • Check angles sum to 180°: Quick way to verify you haven't made mistakes

⚡ Quick Revision Summary

  • Sine rule: a/sin A = b/sin B = c/sin C (opposite pairs)
  • Cosine rule (side): a² = b² + c² - 2bc cos A
  • Cosine rule (angle): cos A = (b² + c² - a²) / 2bc
  • Area formula: Area = ½ab sin C (two sides and included angle)
  • When to use sine rule: Two sides and two angles (opposite pairs)
  • When to use cosine rule: Three sides, OR two sides and included angle
  • Bearings: Measured clockwise from North (000° to 360°)
  • Largest angle: Always opposite the longest side
  • Calculator mode: MUST be in degrees (Deg) not radians (Rad)
  • Rounding: Only round final answer, keep full values in working