Mathematics • Year 10 • Unit 3 • Lesson 18
Choosing the Right Rule — Skill Drill
Build fluency in matching the given information (AAS, ASA, SSA, SAS, SSS) to the right tool — sine rule, cosine rule, area formula, or Heron's. Lesson 18 is all about identification first, calculation second: pick the formula, then chain them when one is not enough.
1. I do — fully worked example
Read every step. The point of this worked example is the decision, not just the arithmetic.
Problem. In △ABC, a = 10, b = 12, ∠C = 50°. Find side c, then ∠A, then the area.
Step 1 — Identify the pattern.
Given two sides and the included angle → SAS
Reason: a, b are two sides and C is between them. SAS triggers the cosine rule (for the third side) or the area formula.
Step 2 — Find c using the cosine rule.
c² = 10² + 12² − 2(10)(12) cos 50°
c² = 100 + 144 − 240 × 0.6428 = 244 − 154.3 = 89.7
c = √89.7 ≈ 9.5
Step 3 — Find ∠A using the sine rule.
Now we have an angle-side pair (C, c), so sine rule works.
sin A / 10 = sin 50° / 9.5
sin A = 10 × 0.766 / 9.5 = 0.806
A = sin⁻¹(0.806) ≈ 53.7°
Step 4 — Find the area using A = ½ ab sin C.
Area = ½ × 10 × 12 × sin 50° = 60 × 0.766 ≈ 46.0
Reason: SAS triggers the area formula directly — no need to find c first.
Answers: c ≈ 9.5, A ≈ 53.7°, Area ≈ 46.0.
2. We do — fill in the missing steps
SSS problem. Fill in the blanks. 5 marks
Problem. In △ABC, a = 7, b = 9, c = 11. Find ∠A, ∠B, ∠C and the area.
Step 1 — Identify the pattern: three sides given → ______. Use the ______ rule for angles and ______'s formula for area.
Step 2 — Find ∠A (cosine rule, A opposite a):
cos A = (b² + c² − a²) / (2bc) = (______ + ______ − ______) / (2 × 9 × 11) = ______ / 198
cos A ≈ ______ , A ≈ ______°
Step 3 — Find ∠B (cosine rule, B opposite b):
cos B = (49 + 121 − 81) / (2 × 7 × 11) = ______ / 154 ≈ ______ , B ≈ ______°
Step 4 — Find ∠C:
C = 180° − A − B = ______°
Step 5 — Find the area (Heron's):
s = (7 + 9 + 11) / 2 = ______
A = √(______ × ______ × ______ × ______) = √______ ≈ ______
3. You do — independent practice
For each question, your first job is to name the pattern (AAS, ASA, SSA, SAS, SSS) and write which rule(s) you will use, then solve. Round to 1 dp / 0.1°.
Foundation — name the pattern then solve
3.1 Triangle with a = 8, b = 6, ∠C = 60°. Pattern = ____. Find c. 2 marks
3.2 Triangle with ∠A = 40°, ∠B = 75°, a = 8. Pattern = ____. Find side b. 2 marks
3.3 Triangle with a = 9, b = 10, c = 11. Pattern = ____. Find ∠C. 2 marks
3.4 Triangle with a = 6, b = 10, ∠C = 70°. Pattern = ____. Find the area. 2 marks
Standard — two-step problems
3.5 Triangle with a = 11, b = 14, ∠C = 65°. Find c, ∠A, and the area. (Three answers required.) 3 marks
3.6 A triangle has sides 9, 12, 15. (a) Verify it is right-angled using Pythagoras. (b) Confirm with the cosine rule: find the angle opposite 15 and check it equals 90°. 3 marks
Extension — push your thinking
3.7 A triangle has area 30 cm², sides a = 10 and b = 8. Find both possible values of the included angle C. State why two answers exist. 3 marks
3.8 Triangle with a = 6, b = 8, c = 10. Find all angles by two different methods: (i) by recognising 6-8-10 as a scaled 3-4-5; (ii) by direct cosine-rule. Then find the area by two methods: (i) ½ × leg × leg; (ii) Heron's. All four answers must agree. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (SSS on 7-9-11)
Pattern = SSS; use cosine rule for angles, Heron's for area.
cos A = (81 + 121 − 49)/198 = 153/198 ≈ 0.773 → A ≈ 39.4°.
cos B = (49 + 121 − 81)/154 = 89/154 ≈ 0.578 → B ≈ 54.7°.
C = 180° − 39.4° − 54.7° = 85.9°.
Area: s = 13.5; A = √(13.5 × 6.5 × 4.5 × 2.5) = √988.6 ≈ 31.4.
3.1 — SAS, find c
Pattern = SAS → cosine rule. c² = 64 + 36 − 96 × 0.5 = 100 − 48 = 52. c = √52 ≈ 7.2.
3.2 — AAS, find b
Pattern = AAS → sine rule. b / sin 75° = 8 / sin 40°. b = 8 × 0.9659 / 0.6428 ≈ 12.0.
3.3 — SSS, find ∠C
Pattern = SSS → cosine rule for angle. cos C = (81 + 100 − 121)/(2 × 9 × 10) = 60/180 = 1/3 ≈ 0.333. C ≈ 70.5°.
3.4 — SAS, find area
Pattern = SAS → area formula. A = ½ × 6 × 10 × sin 70° = 30 × 0.9397 ≈ 28.2 cm².
3.5 — SAS, find c, ∠A and area
c² = 121 + 196 − 308 × cos 65° = 317 − 308 × 0.4226 = 317 − 130.2 = 186.8. c ≈ 13.7.
sin A / 11 = sin 65° / 13.7. sin A = 11 × 0.9063 / 13.7 ≈ 0.728. A ≈ 46.7°.
Area = ½ × 11 × 14 × sin 65° = 77 × 0.9063 ≈ 69.8 cm².
3.6 — sides 9, 12, 15
(a) 9² + 12² = 81 + 144 = 225 = 15² ✓ Right-angled.
(b) cos C = (81 + 144 − 225)/(2 × 9 × 12) = 0/216 = 0. C = cos⁻¹(0) = 90° ✓.
3.7 — area 30, sides 10 and 8
30 = ½ × 10 × 8 × sin C = 40 sin C → sin C = 0.75.
C₁ = sin⁻¹(0.75) ≈ 48.6°; C₂ = 180° − 48.6° ≈ 131.4°.
Two answers exist because sin C = sin(180° − C), so both an acute and an obtuse angle give the same area. The two triangles are geometrically different.
3.8 — sides 6, 8, 10
(i) Scaled 3-4-5: right angle opposite 10 (C = 90°). sin A = 6/10 = 0.6 → A ≈ 36.9°. B = 180 − 90 − 36.9 = 53.1°.
(ii) Cosine rule: cos C = (36 + 64 − 100)/96 = 0 → C = 90° ✓. cos A = (64 + 100 − 36)/(2 × 8 × 10) = 128/160 = 0.8 → A = 36.9° ✓.
Area (i) ½ × 6 × 8 = 24.
Area (ii) Heron's: s = 12; A = √(12 × 6 × 4 × 2) = √576 = 24 ✓.