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# 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 1560/103km

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?

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