# Understanding Bearings in Mathematics JSS 3 Mathematics

**Topic: Understanding Bearings in Mathematics**

**What is a Bearing?**- A bearing tells us the direction of one place from another using angles 🧭.

**How to Read a Bearing?**- Bearings are measured clockwise from north.

**Examples:**- North is 0°, east is 90°, south is 180°, and west is 270°.

**Practice Bearing Directions:**- If a tree is to the north, the bearing is 0°.
- If a bird is to the east, the bearing is 90°.

**Using Bearings in Maps:**- Maps use bearings to show directions of places.

**Drawing Bearings:**- Draw a line from one place to another to show the bearing angle.

**Summary:**- Bearings help us understand which way one place is from another using angles.

- A bearing of x from Y is 196 degree, calculate the bearing of Y
- To find the bearing of Y from X, we subtract 180° from the given bearing of X from Y, which is 196°.Bearing of Y from X = 196° – 180° = 16°
Therefore, the bearing of Y from X is 16 degrees.

### Here are five worked examples on bearings:

**Example 1:**- Given: The bearing of A from B is 120°.
- Find: The bearing of B from A.
- Solution: Bearing of B from A = 120° – 180° =
**-60°**(since we’re moving in the opposite direction)

**Example 2:**- Given: The bearing of X from Y is 250°.
- Find: The bearing of Y from X.
- Solution: Bearing of Y from X = 250° – 180° =
**70°**

**Example 3:**- Given: The bearing of P from Q is 30°.
- Find: The bearing of Q from P.
- Solution: Bearing of Q from P = 30° – 180° =
**-150°**(again, opposite direction)

**Example 4:**- Given: The bearing of M from N is 320°.
- Find: The bearing of N from M.
- Solution: Bearing of N from M = 320° – 180° =
**140°**

**Example 5:**- Given: The bearing of R from S is 90°.
- Find: The bearing of S from R.
- Solution: Bearing of S from R = 90° – 180° =
**-90°**(opposite direction)

- Evaluation :

### A girl moves 2km due west and then 4km due north. Find her bearing from 5he starting point

- To find the girl’s bearing from the starting point after moving 2 km due west and then 4 km due north, we can use trigonometry to calculate the angle.Let’s denote the starting point as O, the point after moving west as A, and the final point after moving north as B.
Given: OA (West) = 2 km AB (North) = 4 km

First, we find the angle OAB using trigonometry:

$tan(θ)=ABOA $ $tan(θ)=42 =21 $

Using the inverse tangent (arctan) function to find the angle: $θ=arctan(21 )≈26.5_{∘}$

The bearing is measured clockwise from the north direction. Since the angle we found is with respect to the east direction (due to moving west first), we subtract it from 90° to get the bearing from the starting point:

Bearing = 90° – 26.57° = 63.43°

Therefore, the girl’s bearing from the starting point is approximately 63.43 degrees.

### More Worked Examples

**Example 1:**A girl moves 3 km due east and then 4 km due south. Find her bearing from the starting point.- First, we find the angle using trigonometry: $tan(θ)=34 $, so $θ=arctan(34 )≈53.1_{∘}$.
- Bearing = 180° – 53.13° = 126.87°

**Example 2:**A car moves 5 km due north and then 12 km due east. Find its bearing from the starting point.- $tan(θ)=512 $, so $θ=arctan(512 )≈67.3_{∘}$.
- Bearing = 90° + 67.38° = 157.38°

**Example 3:**A hiker moves 8 km due south and then 15 km due west. Find the hiker’s bearing from the starting point.- $tan(θ)=815 $, so $θ=arctan(815 )≈61.9_{∘}$.
- Bearing = 270° + 61.93° = 331.93°

**Example 4:**An airplane moves 20 km due west and then 16 km due north. Find its bearing from the starting point.- $tan(θ)=1620 =45 $, so $θ=arctan(45 )≈51.3_{∘}$.
- Bearing = 90° – 51.34° = 38.66°

**Example 5:**A boat moves 10 km due east and then 8 km due south. Find its bearing from the starting point.- $tan(θ)=108 =54 $, so $θ=arctan(54 )≈38.6_{∘}$.
- Bearing = 180° + 38.66° = 218.66°

These examples demonstrate how to calculate bearings after moving in different directions.

- After moving 6 km due east and then 8 km due south, the bearing from the starting point is _______. a) 45° b) 135° c) 225° d) 315°
- If a cyclist moves 10 km due north and then 15 km due east, the bearing from the starting point is _______. a) 26.57° b) 63.43° c) 116.57° d) 153.43°
- After moving 5 km due west and then 12 km due north, the bearing from the starting point is _______. a) 60° b) 120° c) 240° d) 300°
- A car moves 8 km due south and then 6 km due west. The bearing from the starting point is _______. a) 14.04° b) 75.96° c) 105.96° d) 255.96°
- After moving 7 km due east and then 24 km due south, the bearing from the starting point is _______. a) 67.38° b) 112.62° c) 202.62° d) 247.38°
- If a boat moves 15 km due west and then 20 km due south, the bearing from the starting point is _______. a) 38.66° b) 141.34° c) 218.66° d) 321.34°
- After moving 12 km due north and then 5 km due east, the bearing from the starting point is _______. a) 21.80° b) 68.20° c) 111.80° d) 158.20°
- A plane moves 16 km due south and then 9 km due west. The bearing from the starting point is _______. a) 29.15° b) 60.85° c) 119.15° d) 150.85°
- After moving 20 km due east and then 15 km due north, the bearing from the starting point is _______. a) 39.81° b) 50.19° c) 130.19° d) 139.81°
- A hiker moves 18 km due west and then 12 km due south. The bearing from the starting point is _______. a) 33.69° b) 56.31° c) 123.69° d) 146.31°
- After moving 25 km due north and then 20 km due east, the bearing from the starting point is _______. a) 34.38° b) 55.62° c) 124.38° d) 145.62°
- A cyclist moves 30 km due east and then 40 km due south. The bearing from the starting point is _______. a) 36.87° b) 53.13° c) 126.87° d) 143.13°
- After moving 10 km due west and then 8 km due north, the bearing from the starting point is _______. a) 38.66° b) 51.34° c) 141.34° d) 148.66°
- A boat moves 12 km due south and then 16 km due west. The bearing from the starting point is _______. a) 33.69° b) 56.31° c) 123.69° d) 146.31°
- After moving 15 km due east and then 20 km due north, the bearing from the starting point is _______. a) 38.66° b) 51.34° c) 141.34° d) 148.66°