Solutions — Radian Measure, Arcs and Sectors

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1. Q1 — Degrees to radians

   (a) 60° × π/180 = π/3   (b) 150° × π/180 = 5π/6   (c) 225° × π/180 = 5π/4   (d) 315° × π/180 = 7π/4   (e) 270° × π/180 = 3π/2

2. Q2 — Radians to degrees

   (a) π/4 × 180/π = 45°   (b) 5π/6 × 180/π = 150°   (c) 4π/3 × 180/π = 240°   (d) 7π/4 × 180/π = 315°   (e) 5π/2 × 180/π = 450° (one full revolution + 90°)

3. Q3 — Arc length

   (a) l = 5 × π/3 = 5π/3 cm ≈ 5.24 cm   (b) l = 12 × 2π/3 = 8π cm ≈ 25.13 cm   (c) l = 9 × 1.4 = 12.6 cm   (d) 60° = π/3 rad; l = 7 × π/3 = 7π/3 cm ≈ 7.33 cm

4. Q4 — Area of sector

   (a) A = ½ × 36 × π/2 = 9π cm² ≈ 28.27 cm²   (b) A = ½ × 100 × π/5 = 10π cm² ≈ 31.42 cm²   (c) A = ½ × 16 × 2.5 = 20 cm²   (d) 45° = π/4 rad; A = ½ × 64 × π/4 = 8π cm² ≈ 25.13 cm²

5. Q5 — Finding unknowns

   (a) θ = l/r = 15/5 = 3 rad = 3 × 180/π ≈ 171.9°   (b) r = l/θ = 8π ÷ (2π/3) = 8π × 3/(2π) = 12 cm   (c) θ = 2A/r² = 2(27)/36 = 54/36 = 1.5 rad   (d) r = √(2A/θ) = √(100/(π/4)) = √(400/π) = 20/√π ≈ 11.28 cm

6. Q6 — Perimeter of a sector

   (a) P = r(θ+2) = 7(π/3+2) = 7π/3+14 = (7π/3+14) cm ≈ 21.33 cm   (b) P = rθ+2r = 9θ+18 = 30, so 9θ = 12, θ = 4/3 rad. Area = ½(81)(4/3) = 54 cm².   (c) P = r(π/2+2) = 40, so r = 40/(π/2+2) = 40/(π/2+2). Numerically: π/2+2 ≈ 3.571, r ≈ 40/3.571 ≈ 11.2 cm. Area = ½r²×π/2 = ½(125.4)(π/2) ≈ 98.6 cm².

7. Q7 — Clock hands

   (a) The minute hand completes 2π rad in 60 min. In 20 min: θ = (20/60) × 2π = 2π/3 rad.   (b) Arc length: l = rθ = 14 × 2π/3 = 28π/3 cm ≈ 29.3 cm.   (c) In 45 min: θ = (45/60) × 2π = 3π/2. Area = ½ × 14² × 3π/2 = ½ × 196 × 3π/2 = 147π ≈ 461.8 cm².

8. Q8 — Windscreen wiper

   (a) 120° = 120 × π/180 = 2π/3 rad.   (b) Large sector: A = ½ × 50² × 2π/3 = ½ × 2500 × 2π/3 = 2500π/3 ≈ 2618.0 cm².   (c) Small sector: A = ½ × 10² × 2π/3 = ½ × 100 × 2π/3 = 100π/3 ≈ 104.7 cm².   (d) Area cleaned = 2500π/3 − 100π/3 = 2400π/3 = 800π ≈ 2513 cm².

9. Q9 — Irrigation system

   (a) 240° = 240 × π/180 = 4π/3 rad.   (b) A = ½ × 80² × 4π/3 = ½ × 6400 × 4π/3 = 12800π/3 ≈ 13 404 m².   (c) Cost = 13 404 × $0.02 = $268.08.   (d) 300° = 5π/3 rad. New area = ½ × 6400 × 5π/3 = 16000π/3. Ratio: (16000π/3) ÷ (12800π/3) = 16000/12800 = 1.25. Increase = 25%.

10. Q10 — Segment area

    (a) Sector area: A = ½ × 10² × π/2 = 25π ≈ 78.54 cm².   (b) Triangle area: ½r²sinθ = ½ × 100 × sin(π/2) = ½ × 100 × 1 = 50 cm².   (c) Minor segment = sector − triangle = 25π − 50 ≈ 78.54 − 50 = 28.54 cm².   (d) Total circle area = π × 100 = 100π. Major segment = 100π − 28.54 ≈ 314.16 − 28.54 = 285.6 cm².