Mathematics • Year 10 • Unit 3 • Lesson 16
Cosine Rule — Mixed Challenge
Pull together every idea from Lesson 16: choose between SAS and SSS forms, classify the triangle (acute / right / obtuse) from the sign of cos, chain the cosine rule with the sine rule, and explain why the cosine rule generalises Pythagoras.
1. Mixed problems — choose the right idea
Each question pulls on a different idea from Lesson 16. Decide whether you need the SAS form, the SSS form, both rules in sequence, or a special-case shortcut before you start writing. Round to 1 dp / 0.1°. 2-3 marks each
1.1 In △ABC, a = 12, b = 15, ∠C = 80°. Find side c. 2 marks
1.2 A triangle has sides 11, 13, 20. Find the largest angle. Is the triangle acute, right or obtuse? Justify using the sign of cos. 3 marks
1.3 In △ABC, a = 9, b = 11, ∠C = 50°. Find c (cosine rule), then find ∠A using the sine rule. 3 marks
1.4 Show, by direct substitution, that the cosine rule c² = a² + b² − 2ab cos C reduces to Pythagoras when C = 90°. 2 marks
1.5 An isosceles triangle has two equal sides of 10 cm and an included apex angle of 100°. Find the base length. 2 marks
1.6 A triangle has sides 7, 24, 25. Use the cosine rule to find the angle opposite side 25 (do not just quote Pythagoras). What does your answer confirm about the triangle? 3 marks
2. Find the mistake
Another Year 10 student has tried to find side c for a triangle with a = 6, b = 8, ∠C = 120°. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it is wrong, then re-do the working correctly. 3 marks
Student's working:
Line 1: c² = a² + b² − 2ab cos C
Line 2: c² = 6² + 8² − 2(6)(8) cos 120°
Line 3: c² = 36 + 64 − 96 × 0.5
Line 4: c² = 100 − 48 = 52
Line 5: c = √52 ≈ 7.2
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working from that line onward, and give the final value of c.
Stuck? cos 120° is not 0.5 — it is negative. Sanity check: with a 120° angle, c must be longer than √(a² + b²), not shorter.3. Open-ended challenge — design your own triangle
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design two different non-right triangles, one acute and one obtuse, where the longest side is exactly 10 cm in both. The two triangles must not be similar (their other two sides must differ).
For each triangle you design:
(i) State all three side lengths.
(ii) Use the cosine rule to find the largest angle.
(iii) State whether the triangle is acute or obtuse and justify from the sign of cos.
Bonus: Explain in one sentence what condition on a² + b² versus c² (where c is the longest side) tells you the type of triangle, without doing any arithmetic.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — a = 12, b = 15, C = 80°
c² = 144 + 225 − 2(12)(15) cos 80° = 369 − 360 × 0.1736 = 369 − 62.5 = 306.5.
c = √306.5 ≈ 17.5.
1.2 — sides 11, 13, 20; largest angle
Largest angle is opposite 20. cos C = (11² + 13² − 20²) / (2 × 11 × 13) = (121 + 169 − 400) / 286 = −110/286 ≈ −0.3846.
C ≈ 112.6°. Since cos C < 0, C is obtuse, so the triangle is obtuse.
1.3 — a = 9, b = 11, C = 50° (cosine then sine)
c² = 81 + 121 − 198 × 0.6428 = 202 − 127.3 = 74.7. c = √74.7 ≈ 8.6.
Sine rule: sin A / 9 = sin 50° / 8.6, so sin A = 9 × 0.766 / 8.6 ≈ 0.802. A = sin⁻¹(0.802) ≈ 53.3°.
1.4 — Cosine rule reduces to Pythagoras at C = 90°
cos 90° = 0, so c² = a² + b² − 2ab × 0 = a² + b². That is exactly Pythagoras. So Pythagoras is the special case of the cosine rule when the included angle is a right angle.
1.5 — Isosceles, two sides 10, apex 100°
c² = 10² + 10² − 2(10)(10) cos 100° = 200 − 200 × (−0.1736) = 200 + 34.7 = 234.7.
c = √234.7 ≈ 15.3 cm. The obtuse apex makes the base longer than √(100 + 100) ≈ 14.1 cm.
1.6 — sides 7, 24, 25
cos C = (7² + 24² − 25²) / (2 × 7 × 24) = (49 + 576 − 625) / 336 = 0 / 336 = 0.
So C = cos⁻¹(0) = 90°. This confirms 7-24-25 is a Pythagorean triple — the triangle is right-angled. (The cosine rule gives the same answer as Pythagoras, as it must.)
2 — Find the mistake
(a) The mistake is on Line 3.
(b) cos 120° = −0.5, not 0.5. The student dropped the negative sign. For obtuse angles, cosine is negative, so the −2ab cos C term becomes positive (subtracting a negative adds).
(c) Corrected:
c² = 36 + 64 − 96 × (−0.5) = 100 + 48 = 148.
c = √148 ≈ 12.2.
Sanity check: an obtuse-angle case must give c > √(a² + b²) = √100 = 10. 12.2 > 10 ✓ (and 7.2 < 10, which would have been impossible).
3 — Open-ended (sample triangles, longest side = 10)
Triangle A — acute (sides 7, 8, 10): cos C = (49 + 64 − 100)/(2 × 7 × 8) = 13/112 ≈ 0.116. C = cos⁻¹(0.116) ≈ 83.3°. cos C > 0 → C is acute, so the whole triangle is acute. ✓
Triangle B — obtuse (sides 5, 6, 10): cos C = (25 + 36 − 100)/(2 × 5 × 6) = −39/60 = −0.65. C = cos⁻¹(−0.65) ≈ 130.5°. cos C < 0 → C is obtuse, so the triangle is obtuse. ✓ (Quick existence check: 5 + 6 = 11 > 10, so the triangle is valid.)
Bonus: Comparing a² + b² with c² (longest side):
· a² + b² > c² → acute triangle.
· a² + b² = c² → right triangle (Pythagoras).
· a² + b² < c² → obtuse triangle.
This is just the sign of the cosine-rule numerator (a² + b² − c²): positive ⇒ cos C > 0 ⇒ acute; zero ⇒ right; negative ⇒ obtuse.
Marking: 1 mark for each correct triangle with sign-of-cos justification (× 2 = 2 marks), 1 mark for one valid cosine-rule calculation, 1 mark for the a² + b² vs c² rule in the bonus.