Bearing and Distance

Bearings and Distance

Lesson Objectives

  • Understand and calculate bearings from diagrams and word problems.
  • Apply trigonometry to determine distances and directions between points.

Lesson Introduction

Bearings describe directions using three-figure angles measured clockwise from the North. For example, East is \(090^\circ\), South is \(180^\circ\), and North-East is \(045^\circ\).

Bearings and distances problems are commonly solved using trigonometry (sine rule, cosine rule) and right-angle triangle techniques.

Worked Examples

Example 1:
A boat sails 100 km on a bearing of 045°. How far north and east has it travelled?
North: 100 \cos(45^\circ) = 70.71 \, \text{km}, \quad East: 100 \sin(45^\circ) = 70.71 \, \text{km}
Example 2:
An aircraft flies 120 km on a bearing of 120°. Find the east and south components of the flight.
East: 120 \sin(30^\circ) = 60 \, \text{km}, \quad South: 120 \cos(30^\circ) = 103.92 \, \text{km}
Example 3:
A plane flies 80 km due east then 150 km due south. Find the total distance and bearing from the start.
Use Pythagoras: d = \sqrt{80^2 + 150^2} = \sqrt{6400 + 22500} = \sqrt{28900} \approx 170 \, \text{km}
Bearing: \tan(\theta) = \frac{80}{150}, \theta \approx 28.07^\circ \Rightarrow 180^\circ + 28.07^\circ = 208.07^\circ
Example 4:
[NECO] A ship sails 60 km north, then 80 km on a bearing of 135°. Calculate its distance from the starting point. (Past Question)
Use Cosine Rule: d^2 = 60^2 + 80^2 - 2(60)(80)\cos(45^\circ) \approx 6400 \Rightarrow d \approx 80 \, \text{km}
Example 5:
A vehicle travels 100 km on a bearing of 210°, find its south and west components.
South: 100 \cos(30^\circ) = 86.60, \quad West: 100 \sin(30^\circ) = 50
Example 6:
A man walks 70 m north, then 70 m east. Find the resultant distance and bearing.
d = \sqrt{70^2 + 70^2} = \sqrt{9800} = 98.99 \, \text{m}, \theta = 45^\circ \Rightarrow \text{Bearing} = 045^\circ
Example 7:
[WAEC] A plane travels from A to B, 200 km on a bearing of 120°, then from B to C, 250 km on a bearing of 030°. Calculate AC. (Past Question)
Use Sine Rule or Cosine Rule depending on angle info available.
Example 8:
A ship moves 150 km at 210°, then 100 km at 120°. Calculate the ship’s final position from the starting point.
Break each leg into components and use vector addition to find net displacement and bearing.
Example 9:
A surveyor walks 90 m on a bearing of 315°, then 120 m on a bearing of 045°. Find the total displacement.
Use trigonometric components and vector resolution.
Example 10:
[JAMB] A pilot flies 300 km on a bearing of 240°, then 400 km on a bearing of 330°. How far and in what direction is the plane from the starting point? (Past Question)

Exercises

  1. Find the north and east components of a 150 km journey on a bearing of 030°.
  2. A ship travels 80 km east, then 100 km north. Find the resultant distance and bearing.
  3. Find the bearing of a point 50 m west and 100 m north from the origin.
  4. [WAEC] A plane flies 250 km on a bearing of 150°, then 200 km on a bearing of 060°. Find the final distance and bearing from the starting point. (Past Question)
  5. A hiker walks 60 m on a bearing of 270°, then 80 m on a bearing of 000°. Find the net displacement.
  6. A ship goes 100 km on a bearing of 135°, how far east and south is it?
  7. [NECO] A journey includes 70 km west and 120 km south. What is the total distance and bearing? (Past Question)
  8. Find the displacement vector of a person walking 40 m at 210°, then 60 m at 330°.
  9. [JAMB] A plane flies 400 km at 250°, then 300 km at 320°. Find the displacement from start. (Past Question)
  10. Determine the components of a 90 km movement on a bearing of 075°.

Conclusion/Recap

Understanding bearings helps with practical navigation and applying trigonometric rules to solve complex geometric problems. Always draw a diagram and use components to break down vectors.

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