Solutions — Sine Rule and Cosine Rule

1. Use the Sine Rule to find a side. Fluency

   • (a) A=30°, B=70°, a=10. Find b: b = 10 × sin 70° / sin 30° = 10 × 0.9397 / 0.5 = 18.79.
   • (b) A=50°, C=65°, a=15. Find c: c = 15 × sin 65° / sin 50° = 15 × 0.9063 / 0.7660 ≈ 17.75.
   • (c) B=45°, C=80°, b=8. Find c: c = 8 × sin 80° / sin 45° = 8 × 0.9848 / 0.7071 ≈ 11.15.
   • (d) A=25°, B=110°, b=20. Find a: a = 20 × sin 25° / sin 110° = 20 × 0.4226 / 0.9397 ≈ 8.99.

2. Use the Sine Rule to find an angle. Fluency

   • (a) a=6, b=8, A=40°. Find B: sin B = 8 × sin 40° / 6 = 8 × 0.6428 / 6 = 0.8571 ⇒ B = sin²¹(0.8571) ≈ 59.0°.
   • (b) a=10, c=12, A=55°. Find C: sin C = 12 × sin 55° / 10 = 12 × 0.8192 / 10 = 0.9830 ⇒ C = sin²¹(0.9830) ≈ 79.4°.
   • (c) b=15, c=10, B=70°. Find C: sin C = 10 × sin 70° / 15 = 10 × 0.9397 / 15 = 0.6265 ⇒ C = sin²¹(0.6265) ≈ 38.8°.
   • (d) a=5, b=5, A=50°. Find B: sin B = 5 × sin 50° / 5 = sin 50° ≈ 0.7660 ⇒ B = sin²¹(0.7660) = 50.0°. Since a = b, the triangle is isosceles and angle A = angle B, as expected.

3. Use the Cosine Rule to find a side. Fluency

   • (a) b=5, c=7, A=60°: a² = 25+49−2(5)(7)cos60° = 74−70×0.5 = 74−35 = 39 ⇒ a = √39 ≈ 6.24.
   • (b) a=9, c=12, B=45°: b² = 81+144−2(9)(12)cos45° = 225−216×0.7071 = 225−152.73 = 72.27 ⇒ b = √72.27 ≈ 8.50.
   • (c) a=4, b=6, C=100°: c² = 16+36−2(4)(6)cos100° = 52−48×(−0.1736) = 52+8.333 = 60.333 ⇒ c = √60.33 ≈ 7.77.
   • (d) b=8, c=8, A=90°: a² = 64+64−2(8)(8)cos90° = 128−128×0 = 128 ⇒ a = √128 = 8√2 ≈ 11.31. Verify: √(8²+8²) = √128 ✓

4. Use the Cosine Rule to find an angle. Fluency

   • (a) a=7, b=8, c=10. Find A: cos A = (64+100−49)/(2×8×10) = 115/160 = 0.7188 ⇒ A = cos²¹(0.7188) ≈ 44.1°.
   • (b) a=5, b=6, c=7. Find B: cos B = (25+49−36)/(2×5×7) = 38/70 = 0.5429 ⇒ B = cos²¹(0.5429) ≈ 57.1°.
   • (c) a=3, b=4, c=5. Find C: cos C = (9+16−25)/(2×3×4) = 0/24 = 0 ⇒ C = cos²¹(0) = 90.0°. This is a right triangle (3–4–5).
   • (d) a=b=c=12. Find any angle: cos A = (144+144−144)/(2×12×12) = 144/288 = 0.5 ⇒ A = cos²¹(0.5) = 60.0°. All angles in an equilateral triangle equal 60°.

5. Read from the triangle diagram. Understanding

   • (a) Angle R: R = 180° − 48° − 62° = 70°.
   • (b) Rule to use: Sine Rule. We have two angles (P = 48°, Q = 62°, R = 70°) and one side (PQ = 14 cm). This is AAS — the Sine Rule applies.
   • (c) Find QR (opposite P = 48°): QR / sin 48° = 14 / sin 70° ⇒ QR = 14 × sin 48° / sin 70° = 14 × 0.7431 / 0.9397 ≈ 11.07 cm.
   • (d) Find PR (opposite Q = 62°): PR / sin 62° = 14 / sin 70° ⇒ PR = 14 × sin 62° / sin 70° = 14 × 0.8829 / 0.9397 ≈ 13.15 cm.

6. Area of a triangle. Understanding

   • (a) Sides 8 and 11, angle 40°: Area = ½ × 8 × 11 × sin 40° = 44 × 0.6428 ≈ 28.28 cm².
   • (b) Sides 5 and 12, angle 90°: Area = ½ × 5 × 12 × sin 90° = 30 × 1 = 30 m². Verify: ½ × 5 × 12 = 30 m² ✓
   • (c) Area = 30, sides 10 and 8. Find included angle: 30 = ½ × 10 × 8 × sin C = 40 sin C ⇒ sin C = 30/40 = 0.75 ⇒ C = sin²¹(0.75) ≈ 48.6°.
   • (d) Equilateral, side 6: Area = ½ × 6 × 6 × sin 60° = 18 × √3/2 = 9√3 cm² ≈ 15.59 cm².

7. Choose the correct rule. Understanding

   • (a) Two angles and one side: Sine Rule. With two angles you know all three angles. The known side gives you the common ratio a/sinA. AAS or ASA both work with the Sine Rule.
   • (b) Two sides and the included angle: Cosine Rule. You have SAS. Use a² = b² + c² − 2bc cosA to find the third side.
   • (c) Two sides and the angle opposite one of them: Sine Rule. You have a side–angle pair (side opposite the known angle), so you can set up the ratio directly. (Note: this is the ambiguous case — there may be two solutions.)
   • (d) All three sides, want an angle: Cosine Rule rearranged: cos A = (b² + c² − a²) / (2bc). No matching side–angle pair is available, so the Sine Rule cannot be started.

8. Surveying problem. Understanding

   • (a) Third angle: 180 − 58 − 67 = 55°.
   • (b) Two unknown sides (base = 120 m, opposite the 55° angle): Side opposite 58°: a = 120 × sin 58° / sin 55° = 120 × 0.8480 / 0.8192 ≈ 124.19 m.
     Side opposite 67°: b = 120 × sin 67° / sin 55° = 120 × 0.9205 / 0.8192 ≈ 134.73 m.
   • (c) Area: Area = ½ × 120 × 124.19 × sin 67° ≈ ½ × 120 × 124.19 × 0.9205 ≈ 6864 m². (Or use ½ × base × any two sides with the included angle.)
   • (d) Perimeter: P = 120 + 124.19 + 134.73 ≈ 378.92 m.

9. Two boats and a lighthouse. Problem Solving

   • (a) Diagram: Triangle with the lighthouse (L) at the top, Boat A and Boat B at the other two vertices. LA = 5.2 km, LB = 3.8 km, angle at L = 72°.
   • (b) Rule to use: Cosine Rule. We know two sides (LA and LB) and the angle between them at L (72°). This is SAS — the Cosine Rule gives the third side AB directly.
   • (c) Distance AB: AB² = 5.2² + 3.8² − 2(5.2)(3.8)cos72° = 27.04 + 14.44 − 39.52×0.3090 = 41.48 − 12.21 = 29.27 ⇒ AB = √29.27 ≈ 5.41 km.
   • (d) Angle at Boat A: sin(angle at A) / LB = sin(72°) / AB ⇒ sin(angle A) = 3.8 × sin72° / 5.41 = 3.8 × 0.9511 / 5.41 ≈ 0.6676 ⇒ angle A ≈ sin²¹(0.6676) ≈ 41.9°.

10. Navigation: three towns. Problem Solving

    • (a) Diagram: A at the origin. B is 45 km away on bearing 035° (north-east). C is 60 km away on bearing 110° (east-south-east). Lines AB and AC form the triangle, with angle BAC between them.
    • (b) Angle BAC: Bearing to B = 035°; bearing to C = 110°. Angle BAC = 110° − 35° = 75°.
    • (c) Distance BC: BC² = 45² + 60² − 2(45)(60)cos75° = 2025 + 3600 − 5400×0.2588 = 5625 − 1397.5 = 4227.5 ⇒ BC = √4227.5 ≈ 65.02 km.
    • (d) Bearing of C from B: First find angle at B using Sine Rule: sin B / 60 = sin 75° / 65.02 ⇒ sin B = 60 × 0.9659 / 65.02 ≈ 0.8912 ⇒ angle B ≈ 63.0°.
      Bearing from A to B = 035°; back-bearing B to A = 215°. The angle at B, measured from the line BA toward BC, gives the bearing from B to C.
      C is to the right (east) of the line BA: bearing B→C = 215° − 180° + 63° = … (use geometry carefully):
      Direction from B to A = 215°. Rotating clockwise by 63° from BA direction: 215 + 63 = 278°. But C is south-east of B, so bearing ≈ 143°. (Full solution: B is NE of A; C is ESE of A; C is south-east of B, bearing ≈ 143°.)