Mathematics • Year 10 • Unit 3 • Lesson 15
Sine Rule — Skill Drill
Build fluency with the sine rule from Lesson 15: a / sin A = b / sin B = c / sin C. Practise pairing each angle with the side opposite it, finding sides (AAS), finding angles (SSA), and watching out for the ambiguous case.
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 = 50°, ∠B = 60°, and side a = 10 cm (side opposite ∠A). Find side b to 1 decimal place.
Step 1 — Write the sine rule.
a / sin A = b / sin B
Reason: each side is paired with the sine of the angle opposite it.
Step 2 — Substitute the known values.
10 / sin 50° = b / sin 60°
Reason: a = 10, ∠A = 50°, ∠B = 60°.
Step 3 — Rearrange for b.
b = 10 × sin 60° / sin 50°
b ≈ 10 × 0.8660 / 0.7660 ≈ 11.3 cm
Reason: cross-multiply then divide.
Step 4 — Sanity-check.
∠B (60°) > ∠A (50°), so b should be greater than a = 10. ✓
Answer: b ≈ 11.3 cm.
2. We do — fill in the missing steps (finding an angle)
Same structure as Section 1, but with the working faded. Fill in each blank. 5 marks
Problem. In △ABC, a = 8 cm, b = 12 cm and ∠A = 30°. Find ∠B to 1 decimal place. Then check whether the obtuse solution is also valid.
Step 1 — Write the sine rule with angle on top:
sin B / b = sin A / a, so sin B = (b × sin A) / a = (______ × sin ______°) / ______
Step 2 — Calculate sin B:
sin B = (______ × 0.5) / ______ = ______ / ______ = ______
Step 3 — Take inverse sine (acute solution):
∠B = sin⁻¹(______) ≈ ______°
Step 4 — Obtuse alternative:
Obtuse ∠B = 180° − ______° = ______°
Step 5 — Check the obtuse case is geometrically possible:
∠A + obtuse ∠B = 30° + ______° = ______° (must be < 180° for the triangle to exist). Valid? ______
3. You do — independent practice
Show your working. Foundation (3.1-3.4), standard (3.5-3.6), extension (3.7-3.8). Give angle answers to 1 decimal place and side answers to 1 decimal place.
Foundation — sine rule mechanics
3.1 Write the sine rule formula for a general triangle ABC. 1 mark
3.2 In △ABC, ∠A = 40°, ∠B = 70° and a = 12 cm. Find side b. 2 marks
3.3 In △ABC, ∠A = 35°, ∠C = 85° and a = 7 cm. Find side c. 2 marks
3.4 In △ABC, ∠A = 50°, ∠B = 60° and a = 8 cm. Find ∠C and side c. 2 marks
Standard — finding angles
3.5 In △ABC, a = 10 cm, b = 15 cm and ∠A = 35°. Find ∠B (acute solution only). 2 marks
3.6 In △ABC, ∠A = 40°, ∠B = 70° and side c = 12 cm. Find ∠C, then sides a and b. 3 marks
Extension — the ambiguous case
3.7 In △ABC, a = 5 cm, b = 8 cm and ∠A = 30°. How many possible triangles exist? Explain. (Find both possible values of ∠B and check whether each gives a valid triangle.) 3 marks
3.8 Explain in one sentence why the sine rule cannot determine an angle uniquely when the given angle is acute and opposite the shorter of the two known sides. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (a = 8, b = 12, ∠A = 30°)
Step 1: sin B = (12 × sin 30°) / 8.
Step 2: sin B = (12 × 0.5) / 8 = 6/8 = 0.75.
Step 3: ∠B = sin⁻¹(0.75) ≈ 48.6°.
Step 4: Obtuse ∠B = 180° − 48.6° = 131.4°.
Step 5: 30° + 131.4° = 161.4° < 180°. Valid — both solutions give a real triangle.
3.1 — Sine rule
a / sin A = b / sin B = c / sin C. Each side over the sine of the angle opposite it. Equivalently sin A / a = sin B / b = sin C / c.
3.2 — ∠A = 40°, ∠B = 70°, a = 12
b = (12 × sin 70°) / sin 40° ≈ (12 × 0.9397) / 0.6428 ≈ 17.5 cm.
3.3 — ∠A = 35°, ∠C = 85°, a = 7
c = (7 × sin 85°) / sin 35° ≈ (7 × 0.9962) / 0.5736 ≈ 12.2 cm.
3.4 — Find ∠C and c
∠C = 180° − 50° − 60° = 70°.
c = (8 × sin 70°) / sin 50° ≈ (8 × 0.9397) / 0.7660 ≈ 9.8 cm.
3.5 — a = 10, b = 15, ∠A = 35°
sin B = (15 × sin 35°) / 10 ≈ (15 × 0.5736) / 10 ≈ 0.8604.
∠B = sin⁻¹(0.8604) ≈ 59.4° (acute).
3.6 — ∠A = 40°, ∠B = 70°, c = 12
∠C = 180° − 40° − 70° = 70°.
a = (12 × sin 40°) / sin 70° ≈ (12 × 0.6428) / 0.9397 ≈ 8.2 cm.
b = (12 × sin 70°) / sin 70° = 12 cm (because ∠B = ∠C, so b = c).
3.7 — Ambiguous case
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°.
Check acute: 30° + 53.1° = 83.1° < 180° → valid; ∠C ≈ 96.9°.
Check obtuse: 30° + 126.9° = 156.9° < 180° → valid; ∠C ≈ 23.1°.
Therefore there are two possible triangles (ambiguous case).
3.8 — Why ambiguity arises
Because sin x = sin (180° − x): for any value of sin B between 0 and 1, the equation has two solutions in (0°, 180°) — one acute and one obtuse — and when the given angle is acute and opposite a side shorter than the other known side, both solutions can still give a valid triangle (∠A + ∠B < 180°), so the sine rule alone cannot pick between them.