Mathematics • Year 10 • Unit 3 • Lesson 16
The Cosine Rule, Skill Drill
Build fluency with the cosine rule from Lesson 16. Find the third side from SAS using c² = a² + b² − 2ab cos C, and find an angle from SSS using cos C = (a² + b² − c²) / (2ab). Identify when the rule is needed instead of the sine rule.
1. I do, fully worked example
Read every step. Each one has a short reason on the right so you can see why, not just what.
Problem. In △ABC, a = 7 cm, b = 9 cm, and ∠C = 60°. Find side c.
Step 1, Identify the setup.
Two sides (a, b) + included angle (C) → SAS → cosine rule
Reason: the unknown side c is opposite the known angle C, which sits between the two known sides.
Step 2, Write the cosine rule.
c² = a² + b² − 2ab cos C
Step 3, Substitute and calculate.
c² = 7² + 9² − 2(7)(9) cos 60°
c² = 49 + 81 − 126 × 0.5
c² = 130 − 63 = 67
Reason: cos 60° is an exact value (½), so no rounding at this step.
Step 4, Take the square root and check.
c = √67 ≈ 8.2 cm
Check: C = 60° < 90°, so c should be less than √(7² + 9²) = √130 ≈ 11.4 cm. 8.2 < 11.4 ✓
Answer: c ≈ 8.2 cm.
2. We do, fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank. 5 marks
Problem. In △ABC, a = 5 cm, b = 7 cm, c = 8 cm. Find ∠A (the angle opposite a).
Step 1, Identify the setup: three sides known (______), so use the cosine rule for an angle.
Step 2, Write the angle form (with A opposite a):
cos A = (______ + ______ − ______) / (2 × ______ × ______)
Step 3, Substitute side lengths:
cos A = (49 + 64 − 25) / (2 × 7 × 8) = ______ / ______
Step 4, Simplify the fraction and use inverse cosine:
cos A = ______ ≈ ______, so A = cos⁻¹(______) ≈ ______°
Step 5, Sanity check. Since cos A > 0, angle A must be (acute / obtuse): ______. Also, a = 5 is the shortest side, so it faces the (smallest / largest) angle: ______. ✓
3. You do, independent practice
Show your working in the space under each problem. Round to 1 decimal place unless told otherwise. The first four are foundation. The middle two are standard. The last two are extension.
Foundation, substitute and solve
3.1 In △ABC, a = 8, b = 10, ∠C = 45°. Find side c. 2 marks
3.2 In △ABC, a = 6, b = 9, ∠C = 90°. Find c using the cosine rule, then check with Pythagoras. 2 marks
3.3 In △ABC, a = 6, b = 8, c = 10. Find ∠B using cos B = (a² + c² − b²) / (2ac). 2 marks
3.4 A triangle has sides 5, 6, 7. Find the largest angle (it is opposite the longest side, 7). 2 marks
Standard, pick the right form
3.5 In △ABC, a = 9, b = 11, ∠C = 50°. Find side c. (Use cos 50° ≈ 0.643.) 2 marks
3.6 A triangle has sides 8, 10, 14. Decide whether the angle opposite 14 is acute, right or obtuse. Justify with one cosine-rule calculation. 3 marks
Extension, push your thinking
3.7 In △ABC, a = 7, b = 5, ∠C = 120°. Find c. (Use cos 120° = −½ exactly.) Why is the answer bigger than √(a² + b²)? 3 marks
3.8 In △ABC, a = 7, b = 9, c = 11. Find all three angles and verify they sum to 180°. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (5-7-8 triangle, find ∠A)
SSS → cosine rule.
cos A = (b² + c² − a²) / (2bc) = (49 + 64 − 25) / (2 × 7 × 8) = 88 / 112.
cos A ≈ 0.7857, so A = cos⁻¹(0.7857) ≈ 38.2°.
cos A > 0 → A is acute. Side a = 5 is shortest → faces the smallest angle. ✓
3.1, a = 8, b = 10, C = 45°
c² = 8² + 10² − 2(8)(10) cos 45° = 64 + 100 − 160 × 0.7071 = 164 − 113.1 = 50.9.
c = √50.9 ≈ 7.1.
3.2, a = 6, b = 9, C = 90°
c² = 36 + 81 − 2(6)(9) cos 90° = 117 − 108 × 0 = 117. c = √117 ≈ 10.8.
Pythagoras check: 6² + 9² = 36 + 81 = 117. c = √117 ≈ 10.8. ✓ (Cosine rule reduces to Pythagoras when C = 90°.)
3.3, sides 6, 8, 10; find ∠B (opposite 8)
cos B = (a² + c² − b²) / (2ac) = (36 + 100 − 64) / (2 × 6 × 10) = 72 / 120 = 0.6.
B = cos⁻¹(0.6) ≈ 53.1°. (6-8-10 is a scaled 3-4-5 right triangle; ∠C = 90°.)
3.4, sides 5, 6, 7; largest angle (opposite 7)
cos C = (5² + 6² − 7²) / (2 × 5 × 6) = (25 + 36 − 49) / 60 = 12/60 = 0.2.
C = cos⁻¹(0.2) ≈ 78.5°. Acute (cos > 0).
3.5, a = 9, b = 11, C = 50°
c² = 81 + 121 − 2(9)(11) × 0.643 = 202 − 127.3 = 74.7.
c = √74.7 ≈ 8.6.
3.6, sides 8, 10, 14; angle opposite 14
cos C = (8² + 10² − 14²) / (2 × 8 × 10) = (64 + 100 − 196) / 160 = −32/160 = −0.2.
cos C is negative → the angle is obtuse. (C ≈ 101.5°.)
3.7, a = 7, b = 5, C = 120°
c² = 49 + 25 − 2(7)(5) × (−½) = 74 − (−35) = 74 + 35 = 109.
c = √109 ≈ 10.4. This is bigger than √(49 + 25) = √74 ≈ 8.6 because cos 120° is negative, so the −2ab cos C term adds instead of subtracting. Obtuse angles always make the opposite side longer than Pythagoras predicts.
3.8, sides 7, 9, 11; all three angles
cos A = (81 + 121 − 49) / (2 × 9 × 11) = 153/198 ≈ 0.773 → A ≈ 39.4°.
cos B = (49 + 121 − 81) / (2 × 7 × 11) = 89/154 ≈ 0.578 → B ≈ 54.7°.
C = 180° − 39.4° − 54.7° = 85.9°. Sum check: 39.4 + 54.7 + 85.9 = 180.0 ✓