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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: 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: 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: 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):

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

Problem: Bearing Components

A survey line is 250 m long on bearing N 35° E. Find its latitude and departure.

$$Lat=250\cos35^\circ=204.8\text{ m}$$
$$Dep=250\sin35^\circ=143.4\text{ m}$$

Answer: Latitude = 204.8 m north; departure = 143.4 m east.

Problem: Distance from Latitude and Departure

A traverse line has latitude 96 m south and departure 128 m east. Find its length.

$$L=\sqrt{96^2+128^2}=160\text{ m}$$

Answer: The line length is 160 m.

Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q259

MSTE - Geometry and Trigonometry / Bearings and Distances / Engr. Janclyde Espinosa (Clidez)

A ship takes a sighting on two buoys. At a certain instant, the bearing of buoy A is N 44.23° W, and that of buoy B is N 62.14° E. The distance between the buoys is 3.60 km, and the bearing of B from A is N 87.87° E. Find the distance of the ship from each buoy B.

Answer:

  1. 2.78km
  2. 1.63km
  3. 3.45km
  4. 7.82km
At the ship, the angle between the sight lines is $44.23^\circ+62.14^\circ=106.37^\circ$. From A, the bearing to the ship is S44.23°E, or azimuth $135.77^\circ$. The bearing from A to B is $87.87^\circ$, so angle A is:
$135.77^\circ-87.87^\circ=47.90^\circ$
Using the sine law with $AB=3.60$ km:
$\frac{SB}{\sin47.90^\circ}=\frac{3.60}{\sin106.37^\circ}$
$\boxed{SB=2.78\text{ km}}$

Question Bank: t361

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

Two flies flew at the same time from the same point in different directions. One fly flew at 0.7 m/s and the other fly at 0.5 m/s. If the angle between their paths is $85^\circ$, find their distance after 2.5 seconds.

  1. 2.06 m
  2. 2.34 m
  3. 1.81 m
  4. 3 m
After $2.5$ s the flies are at distances $d_1=0.7(2.5)=1.75$ m and $d_2=0.5(2.5)=1.25$ m, with $85^\circ$ between them.
By the Law of Cosines: $d^2=1.75^2+1.25^2-2(1.75)(1.25)\cos85^\circ=4.625-0.381=4.244$.
$d=\sqrt{4.244}=2.06$ m.
$\boxed{2.06\text{ m}}$

Question Bank: t390

MSTE - Geometry and Trigonometry / Trigonometry / Gemini mapped Chapter 1 to 3

An Air Force pilot making preliminary aerial survey, first observed that when he flew directly over a straight road connecting two coastal towns X and Y, the angles made by these towns with the vertical were $30^\circ$ and $60^\circ$, respectively. When he flew back directly over the same road for the second observation, his altimeter indicated that he was 1,000 meters higher and noted that both towns made and angle of $45^\circ$ with the vertical.

What is the distance between the towns, in kilometers.

  1. 13.678
  2. 14.928
  3. 15.456
  4. 12.834

How high is the airplane when the first observation was made?

  1. 6834 m
  2. 5678 m
  3. 6464 m
  4. 6154 m

How high is the airplane when the second observation was made?

  1. 7464 m
  2. 7834 m
  3. 6678 m
  4. 7154 m

Part 1.

Let $h$ be the first altitude. With towns on opposite sides, $XY=h(\tan30^\circ+\tan60^\circ)=2.3094\,h$.
At the second pass (1000 m higher), both towns subtend $45^\circ$, so $XY=2(h+1000)$.
$2.3094h=2h+2000\Rightarrow h=6464$ m, giving $XY=2(7464)=14928$ m $=14.928$ km.
$\boxed{14.928}$

Part 2.

From the previous part, equating the two expressions for $XY$: $2.3094h=2(h+1000)$.
$0.3094h=2000\Rightarrow h=6464$ m.
$\boxed{6464\text{ m}}$

Part 3.

The second observation was made 1000 m higher than the first.
$h_2=6464+1000=7464$ m.
$\boxed{7464\text{ m}}$

Question Bank: t2153

MSTE - Geometry and Trigonometry / Trigonometry / Besavilla CE Pre-Board Math & Surveying

Two aircraft leave an airfield at the same time. One travels due north at an average speed of 300 kph and the other due west at an average speed of 220 kph. Calculate their distance apart after 4 hrs.

  1. 1235 km.
  2. 1612 km.
  3. 1312 km.
  4. 1526 km.
  5. 1488 km.
After 4 hours, the northbound aircraft travels $300(4)=1200$ km, and the westbound aircraft travels $220(4)=880$ km.
Their paths are perpendicular, so use the Pythagorean theorem.
$d=\sqrt{1200^2+880^2}$
$d=\sqrt{2{,}214{,}400}$
$\boxed{d\approx 1488\text{ km}}$