A ship sails 6km from a port on a bearing 070° and then 8km on a bearing of 040°. Find the distance from the port.

Analyze the question: The ship makes two legs of a journey. Leg 1 is 6 km on bearing 070°, and leg 2 is 8 km on bearing 040°. We need the straight-line distance from the port to the final position.

Concept: Resolve each leg into East and North components (bearings are measured clockwise from North):

• East = d sin θ
• North = d cos θ

Solution:

Leg 1 (6 km, 070°):

• East = 6 sin 70° = 6(0.9397) = 5.638
• North = 6 cos 70° = 6(0.3420) = 2.052

Leg 2 (8 km, 040°):

• East = 8 sin 40° = 8(0.6428) = 5.142
• North = 8 cos 40° = 8(0.7660) = 6.128

Totals:

• East = 5.638 + 5.142 = 10.780
• North = 2.052 + 6.128 = 8.180

Distance = √(10.780² + 8.180²) = √(116.21 + 66.91) = √183.12 ≈ 13.53 km

Check with the cosine rule: The angle between the two legs at the turning point is 180° − (70° − 40°) = 150°. d² = 6² + 8² − 2(6)(8)cos 150° = 36 + 64 + 96(0.8660) = 100 + 83.14 = 183.14, so d ≈ 13.53 km.

Answer: B. 13.53 km

Tip: When a ship's two bearings are close, the legs are almost in the same direction, so the answer should be close to 6 + 8 = 14 km. That immediately rules out A, C, D and E. Always do this quick sanity check in bearing questions.