Review Solutions — Trigonometry

1. Right-triangle sides and angles. Fluency

   • (a) θ=42°, hyp=18 m. Find opposite: opp = 18 sin42° = 18 × 0.6691 ≈ 12.04 m.
   • (b) adj=9, hyp=15. Find θ: cos θ = 9/15 = 0.6 ⇒ θ = cos²¹(0.6) ≈ 53.1°.
   • (c) θ=60°, opp=7. Find hyp exactly: hyp = 7/sin60° = 7/(√3/2) = 14/√3 = 14√3/3 m.
   • (d) Legs 9 and 40: hyp = √(81+1600) = √1681 = 41. Angles: 12.7° and 77.3°.

2. Angles of elevation and depression. Fluency

   • (a) Elevation 35°, 50 m away: h = 50 tan35° ≈ 35.01 m.
   • (b) 30 m cliff, depression 28°: d = 30/tan28° ≈ 56.43 m.
   • (c) 1.7 m observer, 15 m away, elevation 52°: Total = 15 tan52° + 1.7 ≈ 19.20 + 1.7 = 20.90 m.
   • (d) Kite 80 m up, string at 40°: string = 80/sin40° ≈ 124.45 m.

3. Bearings. Fluency

   • (a) 40 km, bearing 125°: 22.94 km S, 32.77 km E.
   • (b) Back-bearing of 285°: 105°.
   • (c) S35°W to three-figure: 215°.
   • (d) 6 km N then 8 km E: Bearing 053°, distance 10 km.

4. Sine rule and cosine rule. Fluency

   • (a) A=42°, B=73°, a=11. Find b: 15.71 cm.
   • (b) a=8, b=11, c=14. Find C: 93.6°.
   • (c) Sides 10, 13, angle 65°: 12.61.
   • (d) Area of triangle in (c): 58.91 sq units.

5. Trigonometry from a diagram (triangle with sides 7, 9, 11). Understanding

   • (a) Largest angle: Angle C — opposite the longest side AB = 11.
   • (b) Angle A: cos A = (49+121−81)/(2×7×11) = 89/154 ≈ 54.7°.
   • (c) Angle B: cos B = (81+121−49)/(2×9×11) = 153/198 ≈ 39.5°.
   • (d) Area: C = 180−54.7−39.5 = 85.8°. Area = ½×7×11×sin85.8° ≈ 38.40 sq units.

6. Two observers, one landmark. Understanding

   • (a) Diagram: Tower between A and B, angles of elevation 32° (from A) and 44° (from B).
   • (b) Equations: h = d tan32° and h = (200−d) tan44°.
   • (c) Solve: d ≈ 121.43 m ⇒ h ≈ 75.88 m.
   • (d) Verify: (200−121.43) × tan44° ≈ 75.88 m ✓.

7. Triangular field (sides 85, 110, 140 m). Understanding

   • (a) Largest angle: 90.8° (nearly a right angle).
   • (b) Smallest angle: 37.4°.
   • (c) Area: 4674.9 m².
   • (d) Perpendicular fence length: h = 2 × 4674.9 / 140 ≈ 66.78 m.

8. Search pattern (bearings 045° and 150°). Understanding

   • (a) Angle PBQ: 105°.
   • (b) Distance PQ: 167.6 km.
   • (c) Bearing Q from P: ≈ 194°.
   • (d) Total distance: 287.6 km.

9. Flagpole on a hill. Problem Solving

   • (a) Hilltop height above A: 80 tan18° ≈ 25.99 m.
   • (b) Top of pole above A: 80 tan27° ≈ 40.76 m.
   • (c) Pole height: 40.76 − 25.99 ≈ 14.77 m.
   • (d) Pole leans so top is above A: 90° (straight down).

10. Orienteering course (bearings 072° and 148°). Problem Solving

    • (a) Angle BAC: 76°.
    • (b) Distance BC: 5.21 km.
    • (c) Distance AD (midpoint): 3.29 km.
    • (d) Bearing of B from C: ≈ 287°.

11. Sine rule ambiguous case (a=7, b=10, A=35°). Problem Solving

    • (a) sin B: 0.8194.
    • (b) Two values of B: 55.1° or 124.9°.
    • (c) Two values of c: 12.20 or 4.20.
    • (d) Why two solutions?: SSA with a < b: the arc of radius a from B can intersect the base at two points, giving two valid triangles.

12. Exact values and proof. Problem Solving

    • (a) sin²45 + cos²45 = 1: (√2/2)² + (√2/2)² = ½ + ½ = 1 ✓.
    • (b) Exact perimeter, hyp=10, angle 30°: 15 + 5√3.
    • (c) a=b=6, C=120°. Find c exactly: 6√3.
    • (d) Area of triangle in (c): 9√3 sq units.