Classwork Series and Exercises {Mathematics – SS2}: Bearing and Distances

SSS 2 SECOND TERM MATHEMATICS WEEK SEVEN

Topic: BEARING AND DISTANCES 

Bearing can be defined as the clockwise angular movement between two distant places.

Rules for Solving Bearing and Distances

1. Taking reading in bearing starts from the North Pole in clockwise direction and ends also at the North Pole. I.e. North – North Pole reading.

2. All angles formed while taking reading in bearing is equal to 360 degrees.

3. All questions in bearing leads into the formation of a triangle.

Examples;

1. A boat sails 6km from a port X on a bearing of 065⁰ and thereafter 13km on a bearing of 136⁰. What is the distance and bearing of the boat from X?

Solution

Step 1

You will need to represent the question on a triangle.

You will actually need a ruler and pencil for drawing along with a good eraser in case you make mistake.

Take your pencil and draw a point X, take ruler and rule a straight line to another point Y and rule another straight line to point Z and finally close it up at point X again. See below;

Step 2

Use Cosine Rule to solve for y km;

 From Cosine Rule;

b² = a² + c² – 2ac Cos B

Step 3

Replace a, b, c and B with x, y, z and Y respectively;

y² = x² + z² – 2xz Cos Y

y² = 13² + 6² – 2 (13) (6) Cos 109⁰

y² = 169 + 36 – 156 (- 0.3256)

Note; (- × – = +)

y² = 205 + 156 x 0.3256

y² = 205 + 50.7886

y² = 255.7886

Step4

Take square root of both sides;

y = √255.7886

y = 15.9934km app. 16km (to the nearest km).

Step5

To find the bearing of the boat from X, we will apply Sine Rule;

From Sine Rule;

Sin X/x = Sin Y/y

Sin θ/13km = Sin 109⁰/16km

Step 6

Cross Multiply;

16km x Sin θ = 13km x Sin109⁰

16km x Sin θ = 13km x 0.9455

16km (Sin θ) = 12.2917

Step 7

Divide both sides by 16km;

Sin θ = 12.2917/16km

Sin θ = 0.7682

θ = Sin^-1 0.7682

θ = 50.1955⁰ app. 50⁰ (to the nearest degree)

Step 8

The bearing of boat from the port will be 065⁰ + θ

Which is 065⁰ + 50⁰ = 115⁰ (answer).

 2. An aircraft takes off from an airstrip at an average speed of 20km/hr on a bearing of 052⁰ for 3 hours. It then changes course and flies on a bearing of 028⁰ at an average speed of 3okm/hr for another 1¹/₂ hours. Find; a. its distance from the starting point, b. the bearing of the aircraft from the airstrip.

Solution

Step 1

Step 2

Use Cosine rule to solve for side b;

From Cosine Rule;

b² = a² + c² – 2ac Cos B

b² = 45² + 60² – 2 (45) (60) Cos 156⁰

b² = 2025 + 3600 – 5400 (- 0.9135)

b² = 5625 + 5400 x 0.9135

b² = 5625 + 4932.9

b² = 10557.9

b = √10557.9

b = 102.7516 app. 103 (nearest whole number)

b = 103km

Step 3

To find the bearing of the aircraft from the airstrip, you must first find θ;

Find θ using Sine Rule below;

From Sine Rule;

Sin A/a = Sin B/b

Sin θ/45km = Sin 156^0/103km

Step 4

Cross Multiply;

103km (Sin θ) = 45km x Sin 156⁰

Step 5

Divide both sides by 103km;

103km (sin θ)/103km = 45 x 0.4067/103km

Note; (103km cancels itself and 45 x 0.4067 gives 18.3015)

Sin θ = 18.3015/103

Sin θ = 0.1776

θ = Sin^-1 0.1776 (Note; when taking Sin to the RHS, it becomes inverse)

θ = 10.2299⁰

To find the bearing will be 052⁰ – θ

Bearing = 052⁰ – 10.2299⁰

Bearing = 41.77⁰

Practice these questions below;

1. A ship leaves port and travels 21km on a bearing of 032⁰ and then 45km on a bearing of 287⁰.

(a) Calculate its distance from the port.

(b) Calculate the bearing of the port from the ship.

2. A surveyor leaves her base camp and drives 42km on a bearing of 032⁰, she then drives 28km on a bearing of 154⁰. How far is she then from her base camp and what is her bearing from it?

3. Two ships leave port at the same time and travels at 5km/hr on a bearing of 046⁰. The other travels at 9km/hr on a bearing of 127⁰. How far apart are the ships after 2 hours?

4. A triangular field has two sides 50m and 60m long, and the angle between them is 096⁰. How long is the third side?