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🧭 Key Concepts: Bearings and Navigation

Compass Bearing: Expressed as N θ° E/W or S θ° E/W, where the angle is measured from North or South toward East or West.

Azimuth: Angle measured clockwise from North, ranging from 0° to 360°. N = 0°, E = 90°, S = 180°, W = 270°.

Converting bearings to azimuth: N 23° E = 23°; S 67° E = 180° − 67° = 113°; S 34° 40' E = 180° − 34°40' = 145°20'.

Resultant position: Decompose each leg into North-South (x) and East-West (y) components, sum them, then compute distance and bearing using $R = \sqrt{x^2+y^2}$ and $\theta = \arctan(|E/N|)$.

Problem 1: Two Ships Leaving Port at the Same Time

Two ships leave port at the same time. One ship sails in the direction N 23° E at 11 mph, and the second sails in the direction S 67° E at 15 mph. Find the bearing from the second ship to the first ship after one hour of travel.

Two Ships – Diagram
Solution Solution Solution Solution

After 1 hour, decompose each ship's position into N (+) and E (+) components:

Ship 1 (N 23° E, 11 mph): $\quad E_1 = 11\sin23° = 4.30$, $\quad N_1 = 11\cos23° = 10.13$

Ship 2 (S 67° E, 15 mph): $\quad E_2 = 15\sin67° = 13.81$, $\quad N_2 = -15\cos67° = -5.86$

Vector from Ship 2 to Ship 1:

$$\Delta E = 4.30 - 13.81 = -9.51 \text{ (west)}, \quad \Delta N = 10.13 - (-5.86) = 15.99 \text{ (north)}$$

The direction is northwest. Angle west of north:

$$\theta = \arctan\!\left(\frac{9.51}{15.99}\right) = \arctan(0.5947) \approx 30.7°$$
$$\text{Bearing from Ship 2 to Ship 1} \approx \boxed{\text{N } 30°42'\text{ W}}$$

Problem 2: Distance Between Shipwreck and Tower (Bearings)

From a boat sailing due north at 16.5 kph, a wrecked ship K and an observation tower T are observed in a line due east. One hour later the wrecked ship and the tower have bearings S 34°40' E and S 65°10' E, respectively. Find the distance between the wrecked ship and the tower.

Shipwreck – Diagram
Solution Solution

In 1 hour the boat travels 16.5 km north. Let $O$ = initial position, $O'$ = position after 1 hour. $K$ and $T$ are due east of $O$, so they are at $(k, 0)$ and $(t, 0)$ respectively. From $O' = (0, 16.5)$:

Distance to K (bearing S 34°40' E, south component = 16.5 km):

$$k = 16.5\tan(34°40') = 16.5\tan(34.667°) = 16.5 \times 0.6935 \approx 11.44 \text{ km}$$

Distance to T (bearing S 65°10' E, south component = 16.5 km):

$$t = 16.5\tan(65°10') = 16.5\tan(65.167°) = 16.5 \times 2.144 \approx 35.38 \text{ km}$$

Distance between K and T:

$$KT = t - k = 35.38 - 11.44 \approx \boxed{23.9 \text{ km}}$$

Problem 3: Two-Leg Journey — Distance and Bearing

A ship sails from port A on a bearing of N 40° E for 30 km to point B. It then changes course to S 50° E for 20 km to reach point C. Find: (a) the direct distance from A to C, and (b) the bearing from A to C.

Decompose each leg into North (+N) and East (+E) components:

Leg AB (N 40° E, 30 km):

$$\Delta N_{AB} = 30\cos 40° = 22.98 \text{ km}, \quad \Delta E_{AB} = 30\sin 40° = 19.28 \text{ km}$$

Leg BC (S 50° E, 20 km): bearing S 50° E means 50° from South toward East, so heading is 180°+50° = 230° from North. Components:

$$\Delta N_{BC} = -20\cos 50° = -12.86 \text{ km}, \quad \Delta E_{BC} = 20\sin 50° = 15.32 \text{ km}$$

Total displacement from A to C:

$$N_{AC} = 22.98 - 12.86 = 10.12 \text{ km}, \quad E_{AC} = 19.28 + 15.32 = 34.60 \text{ km}$$

(a) Direct distance AC:

$$AC = \sqrt{10.12^2 + 34.60^2} = \sqrt{102.4 + 1197.2} = \sqrt{1299.6} \approx \boxed{36.05 \text{ km}}$$

(b) Bearing from A to C (N component is positive, E component is positive, so NE direction):

$$\theta = \arctan\!\left(\frac{E_{AC}}{N_{AC}}\right) = \arctan\!\left(\frac{34.60}{10.12}\right) = \arctan(3.419) \approx 73.7°$$
$$\text{Bearing from A to C} = \text{N } 73.7°\text{ E} \approx \boxed{\text{N } 73.7°\text{ E}}$$