Mathematics • Year 10 • Unit 3 • Lesson 15
Sine Rule — Mixed Challenge
Pull every Lesson 15 idea together: choose between the side form and the angle form of the sine rule, complete an AAS triangle, decide whether the SSA setup is unique or ambiguous, spot the classic "wrong pairing" mistake, and design your own ambiguous-case problem with two valid triangles.
1. Mixed problems
Each problem requires a different sine-rule decision. Decide first whether you are finding a side or an angle, and whether the setup is AAS (unique) or SSA (possibly ambiguous). Give answers to 1 decimal place. 2-3 marks each
1.1 In △ABC, ∠A = 45°, ∠B = 60° and side a = 9 cm. Find side b. 2 marks
1.2 In △ABC, ∠A = 35°, ∠C = 110° and side c = 20 cm. Find side a. 2 marks
1.3 In △ABC, a = 7 cm, b = 9 cm, ∠A = 25°. Find ∠B and ∠C, and side c. (Take the acute value of ∠B.) 3 marks
1.4 Show that for the data ∠A = 40°, a = 5 cm and b = 10 cm, no triangle exists. (Hint: try to compute sin B and inspect what you get.) 2 marks
1.5 A surveyor sights a distant peak P from two points A and B on level ground, 200 m apart on a straight road. From A, ∠PAB = 70°; from B, ∠PBA = 80°. Find the distance AP from point A to the peak. 3 marks
1.6 In △ABC, ∠A = 30°, a = 6 cm, b = 9 cm. Find both possible values of ∠B and the corresponding side c in each case. 3 marks
2. Find the mistake
Another Year 10 student has tried to find side b using the sine rule. Their working is shown. Exactly one line contains a mistake. Spot it, explain why it is wrong, and re-do the working correctly. 3 marks
Problem the student tried: In △ABC, ∠A = 50°, ∠B = 60° and side a = 10 cm. Find side b. (Same as the I-do example from Worksheet 1.)
Line 1: a / sin A = b / sin B.
Line 2: 10 / sin 60° = b / sin 50°.
Line 3: b = 10 × sin 50° / sin 60°.
Line 4: b ≈ (10 × 0.7660) / 0.8660 ≈ 8.8 cm.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected line and give the correct value of b.
Stuck? Each side pairs with the sine of the angle opposite it. Side a = 10 is opposite ∠A = 50°, not ∠B = 60°.3. Open-ended challenge — engineer an ambiguous case
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design a sine-rule problem with two valid triangles. That is, choose values of one angle ∠A and two sides a and b such that:
- a, b are positive (use integer cm),
- ∠A is acute (a clean value such as 30°, 45° or 60° is recommended),
- applying the sine rule for ∠B gives sin B between 0 and 1, with both the acute and the obtuse solutions for ∠B giving a valid triangle (i.e. ∠A + acute ∠B < 180° AND ∠A + obtuse ∠B < 180°).
In your answer:
(i) Choose ∠A, a and b.
(ii) Compute sin B and both possible values of ∠B.
(iii) For each value of ∠B, find ∠C and side c.
(iv) State why both triangles are geometrically valid (the ∠A + ∠B < 180° check).
(v) Sketch both triangles to scale.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — ∠A = 45°, ∠B = 60°, a = 9
b = (9 × sin 60°) / sin 45° ≈ (9 × 0.8660) / 0.7071 ≈ 11.0 cm.
1.2 — ∠A = 35°, ∠C = 110°, c = 20
a = (20 × sin 35°) / sin 110° ≈ (20 × 0.5736) / 0.9397 ≈ 12.2 cm.
1.3 — a = 7, b = 9, ∠A = 25° (acute ∠B)
sin B = (9 × sin 25°) / 7 ≈ (9 × 0.4226) / 7 ≈ 0.5433.
∠B ≈ sin⁻¹(0.5433) ≈ 32.9°.
∠C ≈ 180° − 25° − 32.9° ≈ 122.1°.
c / sin 122.1° = 7 / sin 25°, so c ≈ (7 × sin 122.1°) / sin 25° ≈ (7 × 0.8467) / 0.4226 ≈ 14.0 cm.
1.4 — No triangle exists
sin B = (10 × sin 40°) / 5 ≈ (10 × 0.6428) / 5 ≈ 1.286.
Since sin B cannot exceed 1, there is no valid angle B, so no triangle exists with the given data. (Geometrically: side b is too long to "reach" with the given ∠A and the given a.)
1.5 — Surveyor and distant peak
∠APB = 180° − 70° − 80° = 30°.
AP is opposite ∠ABP = 80°. AB = 200 is opposite ∠APB = 30°.
AP / sin 80° = 200 / sin 30°.
AP = (200 × sin 80°) / sin 30° ≈ (200 × 0.9848) / 0.5 ≈ 393.9 m.
1.6 — Ambiguous case, ∠A = 30°, a = 6, b = 9
sin B = (9 × sin 30°) / 6 = (9 × 0.5) / 6 = 0.75.
Acute ∠B = sin⁻¹(0.75) ≈ 48.6°. Obtuse ∠B = 180° − 48.6° ≈ 131.4°.
Check both: 30° + 48.6° = 78.6° < 180° ✓ and 30° + 131.4° = 161.4° < 180° ✓.
Case 1 (acute ∠B): ∠C = 180° − 30° − 48.6° = 101.4°. c = (6 × sin 101.4°) / sin 30° ≈ (6 × 0.9803) / 0.5 ≈ 11.8 cm.
Case 2 (obtuse ∠B): ∠C = 180° − 30° − 131.4° = 18.6°. c = (6 × sin 18.6°) / sin 30° ≈ (6 × 0.3190) / 0.5 ≈ 3.8 cm.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student has paired side a (= 10) with sin 60° (which is sin B), and side b with sin 50° (sin A). The sine rule pairs each side with the sine of the angle opposite it — side a is opposite ∠A = 50°, not ∠B = 60°.
(c) Corrected Line 2: 10 / sin 50° = b / sin 60°.
b = (10 × sin 60°) / sin 50° ≈ (10 × 0.8660) / 0.7660 ≈ 11.3 cm.
3 — Open-ended (sample solution)
Choice: ∠A = 30°, a = 5 cm, b = 8 cm.
(ii) sin B = (8 × sin 30°) / 5 = (8 × 0.5) / 5 = 0.8.
Acute ∠B = sin⁻¹(0.8) ≈ 53.1°. Obtuse ∠B = 180° − 53.1° ≈ 126.9°.
(iii) Triangle 1 (acute ∠B ≈ 53.1°): ∠C = 180° − 30° − 53.1° = 96.9°. c = (5 × sin 96.9°) / sin 30° ≈ (5 × 0.9928) / 0.5 ≈ 9.9 cm.
Triangle 2 (obtuse ∠B ≈ 126.9°): ∠C = 180° − 30° − 126.9° = 23.1°. c = (5 × sin 23.1°) / sin 30° ≈ (5 × 0.3923) / 0.5 ≈ 3.9 cm.
(iv) Both are valid because 30° + 53.1° = 83.1° < 180° (Triangle 1) and 30° + 126.9° = 156.9° < 180° (Triangle 2). So both give a real, closeable triangle.
(v) Both sketches share the same side a = 5 (opposite ∠A) and ∠A = 30° at one vertex; the difference is whether ∠B is acute (a wide, thin triangle) or obtuse (a short, fat triangle).
Marking: 1 for a valid choice of ∠A, a, b that genuinely produces an ambiguous case, 1 for both ∠B values calculated correctly, 1 for both ∠C and c calculated correctly, 1 for the ∠A + ∠B < 180° validity check on both triangles. Any valid ambiguous case is acceptable.