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.

Build · I Do / We Do / You Do

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.

50° a = 10 b = 12 c = ? A B C
Two sides + included angle → cosine rule for c first, then sine rule for the angles.

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.

Stuck? Revisit lesson § "Which Formula to Use?" table. SAS → cosine rule for side, sine rule for angles, area formula for area.

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 = √(______ × ______ × ______ × ______) = √______ ≈ ______

Stuck? Revisit lesson § Worked Example 2 (SSS problem). The denominator for cos A is 2bc, for cos B is 2ac.

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

Stuck on 3.8? 6-8-10 is a right-angled scaled 3-4-5: C (opposite 10) = 90°; sin A = 6/10 = 0.6, etc.

How did this worksheet feel?

What I'll revisit before next class:

Answers, Do not peek before attempting

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 ✓.

HSCScience • Mathematics • Year 10 • Unit 3 • Lesson 18 • Worksheet 1 (Build)