Mathematics • Year 10 • Unit 3 • Lesson 18
Mixed Trig — Master Challenge
Pull the full Lesson 18 toolkit together: match each pattern (AAS, ASA, SSA, SAS, SSS) to the correct rule, chain rules in multi-step questions, deal with the SSA ambiguous case, and verify each answer using an alternative method.
1. Mixed problems — choose the right rule
For each: name the pattern, name the rule(s) you use, then solve. Round to 1 dp / 0.1°. 2-3 marks each
1.1 Triangle with ∠A = 35°, ∠B = 70°, side a = 12. Find b and c. 3 marks
1.2 Triangle with a = 9, b = 14, ∠C = 75°. Find c, ∠A and the area. 3 marks
1.3 Triangle with a = 8, b = 12, c = 15. Find ∠B and the area (use Heron's, then verify with ½ ab sin C using the angle you just found). 3 marks
1.4 SSA case: triangle with a = 8, b = 10, ∠A = 30°. Show that two different triangles satisfy these conditions, and find ∠B in both. 3 marks
1.5 Triangle has area 60 cm² with two sides 12 cm and 15 cm. Find the included angle C, then use the cosine rule to find the third side (use the acute value of C). 3 marks
1.6 A triangle has all angles given: ∠A = 50°, ∠B = 60°, ∠C = 70°. Without any side length, can you find the area? Justify your answer in one sentence (and say what extra piece of information would be sufficient). 2 marks
2. Find the mistake
A Year 10 student tries to solve a multi-step problem and reports their working below. The triangle has a = 8, b = 12, ∠C = 70°. They want side c and ∠A. Exactly one line contains a mistake. Spot it, explain why, and re-do correctly. 3 marks
Student's working:
Line 1: Pattern = SAS → cosine rule for c.
Line 2: c² = 8² + 12² − 2(8)(12) cos 70° = 64 + 144 − 192 × 0.342 = 208 − 65.7 = 142.3.
Line 3: c = √142.3 ≈ 11.9.
Line 4: Now use sine rule for ∠A: sin A / 8 = sin 70° / 12.
Line 5: sin A = 8 × 0.9397 / 12 = 0.6265. A = sin⁻¹(0.6265) ≈ 38.8°.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Re-do the working from that line and give the corrected angle A.
Stuck? When you set up the sine rule, the angle in the denominator must be opposite the side in the numerator. Check Line 4 carefully — is sin 70° paired with the right side?3. Open-ended challenge — design a chain problem
This question has many valid answers. Be creative but show every step. 4 marks
3.1 Design a real-world scenario (one or two sentences of context) that requires exactly three different formulas from Lesson 18 to fully solve: one use of the cosine rule, one use of the sine rule, and one use of the area formula or Heron's. Your scenario must give a clear set of numbers (no fluff).
For your scenario:
(i) State the context and all known numbers.
(ii) Identify the pattern (SAS, SSS, etc.) at each step.
(iii) Apply the three formulas in order and report the final answers (all three).
Bonus: Verify one of your answers using an alternative method (a second formula or a sanity check against Pythagoras).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — AAS, a = 12, ∠A = 35°, ∠B = 70°
∠C = 180° − 35° − 70° = 75°.
Sine rule: b / sin 70° = 12 / sin 35°. b = 12 × 0.9397 / 0.5736 ≈ 19.7.
c / sin 75° = 12 / sin 35°. c = 12 × 0.9659 / 0.5736 ≈ 20.2.
1.2 — SAS, a = 9, b = 14, C = 75°
c² = 81 + 196 − 252 × cos 75° = 277 − 252 × 0.2588 = 277 − 65.2 = 211.8. c ≈ 14.6.
sin A / 9 = sin 75° / 14.6. sin A = 9 × 0.9659 / 14.6 ≈ 0.595. A ≈ 36.5°.
Area = ½ × 9 × 14 × sin 75° = 63 × 0.9659 ≈ 60.8 cm².
1.3 — SSS, a = 8, b = 12, c = 15
cos B = (8² + 15² − 12²) / (2 × 8 × 15) = (64 + 225 − 144) / 240 = 145/240 ≈ 0.604. B ≈ 52.9°.
Heron's: s = (8 + 12 + 15)/2 = 17.5. A = √(17.5 × 9.5 × 5.5 × 2.5) = √2 285.9 ≈ 47.8 cm².
Verify: ½ × 8 × 15 × sin 52.9° = 60 × 0.798 ≈ 47.9 cm² ✓ (small rounding).
1.4 — SSA ambiguous, a = 8, b = 10, ∠A = 30°
Sine rule: sin B / 10 = sin 30° / 8. sin B = 10 × 0.5 / 8 = 0.625.
B₁ = sin⁻¹(0.625) ≈ 38.7° (acute case). Check: A + B₁ = 30 + 38.7 = 68.7° < 180° ✓ valid triangle.
B₂ = 180° − 38.7° = 141.3° (obtuse case). Check: A + B₂ = 30 + 141.3 = 171.3° < 180° ✓ also valid.
So two different triangles satisfy the data — this is the classic SSA ambiguous case.
1.5 — area 60, sides 12 and 15
60 = ½ × 12 × 15 × sin C = 90 sin C → sin C = 60/90 = 2/3 ≈ 0.6667. C (acute) = sin⁻¹(2/3) ≈ 41.8°.
Cosine rule for c: c² = 144 + 225 − 360 × cos 41.8° = 369 − 360 × 0.7454 = 369 − 268.3 = 100.7. c ≈ 10.0.
1.6 — AAA: can we find the area?
No. AAA fixes the shape of the triangle (it is determined up to similarity), but the actual size can be anything from microscopic to astronomical. To find the area we need at least one side length (turning AAA into AAS) — for example, "side a = 10 cm" — which lets the sine rule give all other sides, then A = ½ ab sin C gives the area.
2 — Find the mistake
(a) The mistake is on Line 4.
(b) The sine rule pairs each angle with the side opposite it. Side c (≈ 11.9) is opposite ∠C (= 70°). The student paired sin 70° with side b = 12 instead of with side c. So the right-hand side should be sin 70° / c, not sin 70° / 12.
(c) Corrected:
sin A / 8 = sin 70° / 11.9 → sin A = 8 × 0.9397 / 11.9 ≈ 0.6318.
A = sin⁻¹(0.6318) ≈ 39.2°. (Sanity: 39.2° + 70° = 109.2° < 180° ✓, leaving 70.8° for ∠B.)
3 — Open-ended (sample scenario)
Scenario: A surveyor at point A measures B at 50 m on bearing 070° and C at 80 m on bearing 130°. Find BC, ∠ABC and the area of triangle ABC.
Step 1 — Pattern at A: ∠BAC = 130° − 70° = 60°. So △ABC is SAS with AB = 50, AC = 80, ∠A = 60°.
Step 2 — Cosine rule for BC: BC² = 50² + 80² − 2(50)(80) cos 60° = 2 500 + 6 400 − 8 000 × 0.5 = 8 900 − 4 000 = 4 900. BC = 70 m.
Step 3 — Sine rule for ∠ABC: sin B / 80 = sin 60° / 70. sin B = 80 × 0.866 / 70 ≈ 0.989. B ≈ 81.8°.
Step 4 — Area formula: A = ½ × 50 × 80 × sin 60° = 2 000 × 0.866 ≈ 1 732 m².
Bonus verification: Heron's check on sides 50, 80, 70. s = 100; A = √(100 × 50 × 20 × 30) = √3 000 000 ≈ 1 732 m² ✓ matches.
Marking: 1 mark for a sensible scenario with clean numbers, 1 mark for correctly identifying SAS, 1 mark for the cosine + sine + area chain done correctly, 1 mark for a valid verification.