Title: Bearings and Distances – Understanding Direction and Measurement in Mathematics
What is Bearings and Distances?
Bearings and distances are mathematical techniques used to determine the direction and the measurement between two points. Bearings usually refer to angles measured clockwise from the north. Distances involve the use of geometry and trigonometry to compute how far apart places or objects are. These tools are widely used in navigation, surveying, and geographical mapping.
Why Study Bearings and Distances?
Studying bearings and distances is important for real-world navigation problems such as marine, air, and road travel. It helps develop spatial awareness and strengthens trigonometric understanding. Recently, researchers like Dr. Amina Wahid (University of Manchester) and Prof. Ken Li (University of Singapore) have published work on algorithms involving bearings for autonomous vehicle navigation and geospatial mapping.
To explore Bearings and Distances more deeply, students should consider joining Skorminda—our specialized learning platform offering in-depth math tutorials, personalized practice, and real-time mentorship from expert educators.
Compass Bearings
Compass bearings are directions given in relation to the cardinal points (N, S, E, W) and are often measured in degrees from North going clockwise.
📘 Easy Questions:
1. A boat sails east from a port. What is the compass bearing?
🧮 Solution: East is 90°, hence the bearing = 090°.
2. A plane travels southeast. What is the bearing?
🧮 Solution: Southeast lies between East and South = 135°.
📙 Medium Questions:
3. A ship moves on a bearing of 045°. How many degrees is this from North?
🧮 Solution: It’s 45° from North clockwise.
4. A student cycles in a direction of 210°. Which compass direction is this closest to?
🧮 Solution: 210° lies between South (180°) and Southwest (225°), so it’s South-southwest.
📗 Hard Questions:
5. A drone flies 10 km on a bearing of 060°. Then it flies 8 km on a bearing of 150°. Find the total displacement using vectors.
🧮 Solution:
Using vector breakdowns:
First vector:
\( x_1 = 10 \cos(60^\circ) = 5 \)
\( y_1 = 10 \sin(60^\circ) = 8.66 \)
Second vector:
\( x_2 = 8 \cos(150^\circ) = -6.93 \)
\( y_2 = 8 \sin(150^\circ) = 4 \)
Total displacement:
\( x = 5 – 6.93 = -1.93 \)
\( y = 8.66 + 4 = 12.66 \)
Displacement = \( \sqrt{(-1.93)^2 + (12.66)^2} \approx 12.8 \) km
6. A plane flies from point A to B (bearing 080°, 120 km), then from B to C (bearing 200°, 150 km). Find the distance from A to C.
🧮 Solution: Use cosine rule with vector diagrams and bearings.
Draw triangle ABC and compute angle using bearings.
Angle = 200° – 80° = 120°
Then use cosine rule:
\( AC^2 = 120^2 + 150^2 – 2 \cdot 120 \cdot 150 \cdot \cos(120^\circ) \approx 44063 \)
So, \( AC \approx 209.9 \) km
True Bearings
True bearings are measured clockwise from due North and written as a three-figure angle (e.g., 065°, 250°).
📘 Easy Questions:
1. What is the true bearing of due South?
🧮 Solution: 180°
2. What is the true bearing of due West?
🧮 Solution: 270°
📙 Medium Questions:
3. A hiker walks on a true bearing of 135°. Which direction is he going in relation to North?
🧮 Solution: Southeast (between East and South)
4. A plane travels on a bearing of 315°. Describe its direction.
🧮 Solution: Northwest (between North and West)
📗 Hard Questions:
5. Two towns are located such that Town B is 60 km from Town A on a bearing of 230°. Find the position of Town B from Town A in Cartesian coordinates.
🧮 Solution:
Convert bearings:
\( x = 60 \cos(230^\circ) \approx -38.59 \)
\( y = 60 \sin(230^\circ) \approx -45.97 \)
Coordinates: (–38.59, –45.97)
6. A boat travels from Point P on a bearing of 110° for 100 km and then on a bearing of 200° for 80 km. Find the total displacement using vector addition.
🧮 Solution:
First vector:
\( x_1 = 100 \cos(110^\circ) \approx -34.2 \)
\( y_1 = 100 \sin(110^\circ) \approx 94 \)
Second vector:
\( x_2 = 80 \cos(200^\circ) \approx -75.2 \)
\( y_2 = 80 \sin(200^\circ) \approx -27.4 \)
Add vectors:
\( x = -109.4 \), \( y = 66.6 \)
Displacement = \( \sqrt{(-109.4)^2 + (66.6)^2} \approx 128.2 \) km
Distance using Trigonometry
Trigonometry helps calculate distances using right-angled triangle relationships, predominantly using sine, cosine, and tangent functions with known angles and sides.
📘 Easy Questions:
1. Right triangle with hypotenuse = 10 m, angle = 30°. Find adjacent side.
🧮 Solution:
\( \cos(30^\circ) = \frac{adjacent}{10} \Rightarrow adjacent = 10 \cos(30^\circ) = 8.66 \) m
2. Find the distance to a tower, if its angle of elevation is 45° and your eye level is 20 meters from base.
🧮 Solution:
\( \tan(45^\circ) = \frac{height}{distance} = 1 \)
So, \( distance = 20 \) m
📙 Medium Questions:
3. A plane flies 100 km on a bearing of 120°. Find the eastward and northward components.
🧮 Solution:
Eastward = \( 100 \cos(120^\circ) = -50 \) km
Northward = \( 100 \sin(120^\circ) = 86.6 \) km
4. An island is 50 km at a bearing of 045° from a lighthouse. Find coordinates relative to lighthouse.
🧮 Solution:
\( x = 50 \cos(45^\circ) = 35.36 \)
\( y = 50 \sin(45^\circ) = 35.36 \)
Coordinates: (35.36, 35.36)
📗 Hard Questions:
5. Two navigators travel from origin: one on 60°, 80 km, another on 135°, 130 km. Find how far apart they are.
🧮 Solution: Use cosine rule.
Angle between: \( 135° – 60° = 75° \)
\( d = \sqrt{80^2 + 130^2 – 2*80*130*\cos(75^\circ)} \approx 125.6 \) km
6. A triangle has: side AB = 90 km, AC = 80 km, angle BAC = 40°. Find BC.
🧮 Solution:
Use cosine rule:
\( BC^2 = 90^2 + 80^2 – 2 \cdot 90 \cdot 80 \cdot \cos(40^\circ) \approx 3421.3 \)
So, \( BC \approx 58.5 \) km
Bearings in Navigation
Bearings in navigation help determine direction across geographical points using angles from North. This is essential in map reading and route planning.
📘 Easy Questions:
1. A ship follows a 270° bearing. Which direction is this?
🧮 Solution: Due West
2. What bearing shows due South-East?
🧮 Solution: 135°
📙 Medium Questions:
3. Plane A is flying 300°, Plane B is 75°—how angularly separated are they?
🧮 Solution: Difference = \( 360 – 300 + 75 = 135^\circ \)
4. A boat sails 100 km at bearing 020°, then 120 km at 160°. Find final position.
🧮 Solution:
Split paths into components and add:
Total displacement approx. \( \sim 153.6 \) km
📗 Hard Questions:
5. From town A, town B lies 150 km on bearing 60°. Town C is 180 km from A on 145°. Find distance between B and C.
🧮 Solution:
Use cosine rule:
Angle = \( 145° – 60° = 85° \)
\( d = \sqrt{150^2 + 180^2 – 2 \cdot 150 \cdot 180 \cdot \cos(85^\circ)} \approx 229.5 \) km
6. An aircraft travels 200 km on 35°, then 250 km on 120°. How far from start?
🧮 Solution: Resolve vectors and apply Pythagoras:
Final displacement = \( \approx 310.8 \) km
Join Skorminda
Join Skorminda today and master Bearings and Distances with expert guidance, interactive simulations, and real-world problems aligned with your curriculum! Our specialized courses make Math both fun and practical.
Happy Learning!
📚 Learn More @ www.skorminda.com
No Comments