Solutions — Bearings

1. State the bearing. Fluency

   • (a) Due North: 000°.
   • (b) Due East: 090°.
   • (c) South-West: 225°. (SW is exactly halfway between South (180°) and West (270°).)
   • (d) SSE: 157.5°. (SSE is halfway between S (180°) and SE (135°): (180+135)/2 = 157.5°.)

2. Convert between notation systems. Fluency

   • (a) 045° to compass: N45°E. (45° east of North.)
   • (b) S60°W to three-figure: 240°. (South is 180°; 60° west of South: 180 + 60 = 240°.)
   • (c) N30°W to three-figure: 330°. (North is 360°/0°; 30° west of North: 360 − 30 = 330°.)
   • (d) 110° to compass: S70°E. (110° is 110 − 90 = 20° past East, or equivalently 180 − 110 = 70° east of South.)

3. Find the back-bearing. Fluency

   • (a) Bearing 060°: 060 + 180 = 240°.
   • (b) Bearing 310°: 310 − 180 = 130°.
   • (c) S40°E: S40°E = 140°; back-bearing = 140 − 180 is negative, so 140 + 180 = 320° = N40°W (or 320°).
   • (d) N25°E: N25°E = 025°; back-bearing = 025 + 180 = 205° (or S25°W).

4. North and East components. Fluency

   • (a) 10 km on 030°: North = 10 cos 30° = 10 × 0.8660 = 8.66 km N. East = 10 sin 30° = 10 × 0.5 = 5.00 km E.
   • (b) 15 km on 135°: North = 15 cos 135° = 15 × (−√2/2) ≈ 10.61 km S. East = 15 sin 135° = 15 × √2/2 ≈ 10.61 km E.
   • (c) 8 km on 210°: North = 8 cos 210° = 8 × (−√3/2) ≈ 6.93 km S. East = 8 sin 210° = 8 × (−0.5) = 4.00 km W.
   • (d) 20 km on 330°: North = 20 cos 330° = 20 × √3/2 ≈ 17.32 km N. East = 20 sin 330° = 20 × (−0.5) = 10.00 km W.

5. Read from the bearing diagram. Understanding

   • (a) Compass notation for 060°: N60°E. (60° east of North.)
   • (b) North component (10 km on 060°): North = 10 × cos 60° = 10 × 0.5 = 5 km north.
   • (c) East component: East = 10 × sin 60° = 10 × √3/2 ≈ 8.66 km east.
   • (d) Back-bearing (bearing from B to A): 060 + 180 = 240° (or S60°W).

6. Two-leg journey. Understanding

   • (a) Position after leg 1 (10 km, 000°): 8 km due North of start (bearing 000° means due north).
   • (b) Position after leg 2 (8 km N then 6 km E): 8 km north and 6 km east of the start. This forms a right triangle with legs 8 and 6.
   • (c) Straight-line distance from start to finish: d = √(8² + 6²) = √(64+36) = √100 = 10 km. (3–4–5 triple, scaled by 2.)
   • (d) Bearing from start to finish: tan θ = 6/8 = 0.75 ⇒ θ = tan²¹(0.75) ≈ 36.9°. Bearing = 037° (N37°E).

7. Find the bearing between two points. Understanding

   • (a) Bearing from P to Q: Q is 5 km north and 3 km east of P. tan θ = 3/5 = 0.6 ⇒ θ = tan²¹(0.6) ≈ 31°. Bearing = 031°.
   • (b) Back-bearing from Q to P: 031 + 180 = 211° (S31°W).
   • (c) Distance PQ: d = √(5² + 3²) = √34 ≈ 5.83 km.
   • (d) If R is 4 km due south of Q, bearing P to R: R is at (3, 5−4) = (3, 1) relative to P. tan θ = 3/1 = 3 ⇒ θ = tan²¹(3) ≈ 72°. Bearing = 072°.

8. Two ships from port. Understanding

   • (a) Angle between the two paths: 160° − 070° = 90°. The paths are perpendicular.
   • (b) Distance between the ships: d = √(30² + 40²) = √(900+1600) = √2500 = 50 km. (3–4–5 triple ×10.)
   • (c) Bearing of Ship 2 from Ship 1: Ship 1 (070°): 30 sin70° E, 30 cos70° N ≈ 28.19 km E, 10.26 km N.
     Ship 2 (160°): 40 sin160° E, 40 cos160° N ≈ 13.68 km E, −37.59 km N.
     Relative position of S2 from S1: (13.68−28.19) E, (−37.59−10.26) N = −14.51 km E, −47.85 km N.
     tan θ = 14.51/47.85 ≈ 0.303 ⇒ θ ≈ 16.9°. S2 is south-west of S1: bearing ≈ 197°.
   • (d) Why is the distance calculation simple?: Because the angle between the two paths is exactly 90°, the paths and the line joining the ships form a right triangle. Pythagoras applies directly.

9. Return journey. Problem Solving

   • (a) Final position relative to H: Leg 1 (12 km, 050°): 12 sin50° = 9.19 km E, 12 cos50° = 7.71 km N.
     Leg 2 (12 km, 140°): 12 sin140° = 7.71 km E, 12 cos140° = −9.19 km N.
     Total: (9.19+7.71) km E, (7.71−9.19) km N = 16.90 km east, 1.48 km south.
   • (b) Straight-line distance home: d = √(16.90² + 1.48²) = √(285.6+2.2) = √287.8 ≈ 16.97 km (approximately 17 km).
   • (c) Bearing home: H is west and slightly north of current position. tan θ = 16.90/1.48 ≈ 11.42 ⇒ θ ≈ 85° west of south. Bearing = 180 + 85 = 265° (approximately due west, slightly south).
   • (d) Distance saved by going direct: Two legs = 12 + 12 = 24 km. Direct = 16.97 km. Saving = 24 − 16.97 ≈ 7 km.

10. Search-and-rescue helicopter. Problem Solving

    • (a) Components B→A (025°, 80 km): North = 80 cos 25° ≈ 72.50 km N. East = 80 sin 25° ≈ 33.81 km E.
    • (b) Components A→C (115°, 60 km): North = 60 cos 115° = 60 × (−cos 65°) ≈ −25.36 km (i.e. 25.36 km S). East = 60 sin 115° = 60 × sin 65° ≈ 54.38 km E.
    • (c) Total displacement B→C: North: 72.50 − 25.36 = 47.14 km N. East: 33.81 + 54.38 = 88.19 km E.
    • (d) Direct distance B to C and bearing: d = √(47.14² + 88.19²) = √(2222 + 7777) = √9999 ≈ 100 km.
      tan θ = 88.19/47.14 ≈ 1.871 ⇒ θ = tan²¹(1.871) ≈ 61.8°. Bearing B→C = 062°. Back-bearing C→B = 062 + 180 = 242°.