Chapter 04: Trigonometry – Part 5: Trigonometric Applications and Examination Practice

CHAPTER 04: TRIGONOMETRY

PART 5: TRIGONOMETRIC APPLICATIONS AND EXAMINATION PRACTICE

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In this final part of Trigonometry, we apply the ideas studied in Parts 1–4 to practical and examination problems. We shall investigate heights and distances, angles of elevation and depression, bearings, non-right-angled triangles, three-dimensional applications and mixed examination questions.

Learning objectives

• Solve practical problems involving angles of elevation and depression.
• Interpret and calculate three-figure bearings.
• Apply the sine rule and cosine rule.
• Calculate the area of a triangle using trigonometry.
• Solve selected three-dimensional trigonometry problems.
• Choose appropriate methods for mixed examination questions.
• Present accurate mathematical working and answers with suitable units.

1. Trigonometry in Real-Life Applications

Trigonometry is useful whenever measurements involve angles, distances and heights. It is applied in surveying, construction, navigation, architecture, engineering and astronomy.

Some common applications include:

• Finding the height of a building without measuring it directly.
• Calculating the distance across a river.
• Determining the direction of a ship or aircraft.
• Finding inaccessible distances using measured angles.
• Calculating unknown sides and angles in triangular structures.

Important principle

Most application questions become easier when the information is converted into a clear mathematical diagram. Draw the triangle, label the known measurements and identify the unknown quantity before choosing a formula.

2. Angles of Elevation and Depression

2.1 Angle of elevation

The angle of elevation is the angle between the horizontal line through an observer and the line of sight to an object above the observer.

For example, when a person standing on level ground looks upwards towards the top of a tower, the angle between the horizontal and the line of sight is the angle of elevation.

2.2 Angle of depression

The angle of depression is the angle between the horizontal line through an observer and the line of sight to an object below the observer.

Relationship between elevation and depression

When two horizontal lines are parallel, the angle of depression from the higher point is equal to the corresponding angle of elevation from the lower point. This follows from the properties of alternate angles.

2.3 Choosing the correct trigonometric ratio

For a right-angled triangle, remember SOH–CAH–TOA:

sin θ = Opposite / Hypotenuse

cos θ = Adjacent / Hypotenuse

tan θ = Opposite / Adjacent

Worked Example 1: Height of a tower

A student stands 40 m from the foot of a vertical tower. The angle of elevation of the top of the tower is 32°. Find the height of the tower, correct to one decimal place.

Solution

Let the height of the tower be h.

The opposite side is h and the adjacent side is 40 m. Therefore, use tangent.

tan 32° = h / 40

Make h the subject:

h = 40 tan 32°

h = 24.995... m

Answer: 25.0 m

Worked Example 2: Distance from a cliff

A vertical cliff is 60 m high. The angle of depression from the top of the cliff to a boat is 12°. Find the horizontal distance of the boat from the foot of the cliff, to the nearest metre.

Solution

The angle of elevation from the boat is also 12°.

Let the horizontal distance be d.

tan 12° = 60 / d

d = 60 / tan 12°

d = 282.3... m

Answer: 282 m

3. Bearings and Navigation

A bearing describes the direction of one point from another.

True bearings are measured clockwise from the north direction and are written using three figures.

Direction	Bearing
North	000° or 360°
East	090°
South	180°
West	270°

Examples of three-figure bearings include 025°, 090°, 145° and 315°.

3.1 Back bearings

A back bearing gives the direction of the starting point from the destination.

If the bearing is less than 180°:
Back bearing = Bearing + 180°

If the bearing is 180° or greater:
Back bearing = Bearing − 180°

Worked Example 3: Finding a back bearing

The bearing of B from A is 072°. Find the bearing of A from B.

Solution

Back bearing = 072° + 180°

= 252°

Answer: 252°

Worked Example 4: Distance travelled on a bearing

A person walks 200 m on a bearing of 060° from point A to point B. Calculate how far east and how far north B is from A.

Solution

Resolve the 200 m displacement into eastward and northward components.

Eastward distance = 200 sin 60°

= 173.2 m

Northward distance = 200 cos 60°

= 100 m

Answer: 173.2 m east and 100 m north.

Examination reminder

• Always draw a north arrow at the relevant observation point.
• Measure bearings clockwise from north.
• Write three figures, for example 045° rather than 45°.
• Check whether the question asks for a bearing from A to B or from B to A.

4. The Sine Rule

The sine rule is used to solve non-right-angled triangles when a known side and its opposite angle form a pair, together with another suitable measurement.

For triangle ABC, let side a be opposite angle A, side b opposite angle B and side c opposite angle C.

a / sin A = b / sin B = c / sin C

An equivalent form is:

sin A / a = sin B / b = sin C / c

4.1 When to use the sine rule

• When two angles and one side are known.
• When two sides and a non-included opposite angle pair are known.

Worked Example 5: Finding an unknown side

In triangle ABC, angle A = 40°, angle B = 65° and side a = 12 cm. Find side b.

Solution

a / sin A = b / sin B

12 / sin 40° = b / sin 65°

b = 12 sin 65° / sin 40°

b = 16.9 cm (approximately)

Answer: 16.9 cm

Worked Example 6: Finding an angle

In a triangle, side a = 8 cm, side b = 10 cm and angle A = 30°. Find angle B, taking the acute solution.

Solution

a / sin A = b / sin B

8 / sin 30° = 10 / sin B

sin B = 10 sin 30° / 8

sin B = 5/8

B = sin⁻¹(5/8)

B = 38.7°

Answer: 38.7° for the acute solution.

Important: The ambiguous case

When using the sine rule to find an angle, a sine value between 0 and 1 may correspond to two angles between 0° and 180°: θ and 180° − θ. Check the triangle's other angles and the information in the question before selecting the answer.

5. The Cosine Rule

The cosine rule is used in non-right-angled triangles when two sides and their included angle are known, or when all three sides are known.

a² = b² + c² − 2bc cos A

b² = a² + c² − 2ac cos B

c² = a² + b² − 2ab cos C

To calculate an angle, rearrange the formula:

cos A = (b² + c² − a²) / 2bc

5.1 When to use the cosine rule

• When two sides and the included angle are known.
• When all three sides are known and an angle is required.

Worked Example 7: Finding a side

Two sides of a triangle are 5 cm and 8 cm. The included angle is 60°. Find the third side.

Solution

Let the unknown side be c.

c² = 5² + 8² − 2(5)(8)cos 60°

c² = 25 + 64 − 80(0.5)

c² = 49

c = 7 cm

Answer: 7 cm

Worked Example 8: Finding an angle

A triangle has sides 7 cm, 8 cm and 9 cm. Find the angle opposite the side of length 9 cm.

Solution

cos C = (7² + 8² − 9²)/(2 × 7 × 8)

= (49 + 64 − 81)/112

= 32/112 = 2/7

C = cos⁻¹(2/7)

C = 73.4° approximately

Answer: 73.4°

6. Area of a Triangle Using Trigonometry

When two sides and the included angle are known, the area of a triangle can be calculated without finding its perpendicular height.

Area = ½ab sin C

Similarly:

Area = ½bc sin A

Area = ½ac sin B

Worked Example 9: Area of a triangle

Find the area of a triangle whose two sides are 12 cm and 9 cm and whose included angle is 40°.

Solution

Area = ½ab sin C

= ½ × 12 × 9 × sin 40°

= 54 × 0.6428

= 34.7 cm² approximately

Answer: 34.7 cm²

7. Three-Dimensional Trigonometry

Some problems involve objects in three dimensions, such as buildings, rectangular rooms, cuboids and pyramids.

The main idea is to identify a suitable right-angled triangle within the three-dimensional figure. Often, a horizontal diagonal is found first, followed by a second triangle involving the vertical height.

Method for three-dimensional problems

1. Draw or study the three-dimensional object carefully.
2. Identify the horizontal plane and any vertical heights.
3. Find a suitable diagonal or horizontal distance using Pythagoras' theorem if necessary.
4. Identify the right-angled triangle containing the required angle or length.
5. Apply the appropriate trigonometric ratio.
6. State the answer with the correct unit.

Worked Example 10: Angle of elevation in a rectangular field

A rectangular field measures 30 m by 40 m. A vertical pole of height 20 m stands at one corner. Find the angle of elevation from the opposite corner of the field to the top of the pole.

Solution

First, find the diagonal of the rectangular field.

d² = 30² + 40²

d² = 900 + 1600 = 2500

d = 50 m

Now use the right-angled triangle formed by the diagonal and the pole.

tan θ = 20 / 50

θ = tan⁻¹(0.4)

θ = 21.8° approximately

Answer: 21.8°

8. Choosing the Correct Method

Given information	Suitable method
Right-angled triangle	SOH–CAH–TOA
Two angles and one side	Sine rule
Two sides and a non-included opposite angle	Sine rule
Two sides and included angle	Cosine rule
Three sides of a triangle	Cosine rule
Two sides and included angle for area	Area = ½ab sin C
Height and horizontal distance	Tangent ratio
Direction and distance	Bearings and trigonometric components

9. Examination Techniques and Common Mistakes

9.1 Read the question carefully

Identify what is given and what must be found. Underline important measurements and units.

9.2 Draw a diagram

Even when a diagram is not provided, sketching one can help you understand the situation and identify the correct angle.

9.3 Choose the correct formula

Do not substitute numbers until you have identified the appropriate formula.

9.4 Use the calculator correctly

Check whether the question uses degrees or radians. Set your calculator to the appropriate mode before calculating.

9.5 Avoid premature rounding

Keep several decimal places during intermediate calculations. Round only the final answer to the requested accuracy.

9.6 Include units

Lengths should have units such as cm, m or km. Areas should have square units such as cm² or m². Bearings should be written as three-figure angles.

Common mistakes

• Measuring an angle of elevation from the vertical instead of the horizontal.
• Measuring bearings anticlockwise instead of clockwise from north.
• Using the sine rule when the cosine rule is required.
• Forgetting to use the included angle in the cosine rule.
• Confusing a length with an area.
• Rounding too early.
• Giving an angle without checking whether it is geometrically possible.

10. Mixed Examination Practice

Instructions: Attempt all questions. Show complete working, use suitable units and give answers to the stated accuracy.

Section A: Right-Angled Applications

1. A ladder 10 m long rests against a vertical wall. The foot of the ladder is 6 m from the wall. Calculate the height reached by the ladder.
2. A tree casts a shadow 15 m long when the angle of elevation of the sun is 38°. Find the height of the tree, correct to one decimal place.
3. From a point 75 m away from the base of a building, the angle of elevation to its top is 28°. Find the height of the building.
4. An aircraft is observed at an angle of elevation of 25° from a point on level ground. If its vertical height is 2,000 m, find its horizontal distance from the observer.

Section B: Bearings

5. The bearing of B from A is 125°. Find the bearing of A from B.
6. A ship travels 10 km due east and then 10 km due north. Find its distance from its starting point and the bearing of its final position from the starting point.
7. A person walks 8 km on a bearing of 030°. Find the northward and eastward components of the displacement.

Section C: Sine Rule and Cosine Rule

8. In triangle ABC, A = 45°, B = 60° and side a = 12 cm. Find side b.
9. Two sides of a triangle are 9 cm and 12 cm, and their included angle is 60°. Find the third side.
10. A triangle has sides 5 cm, 6 cm and 7 cm. Find the angle opposite the longest side.
11. In a triangle, two sides are 10 cm and 14 cm and the included angle is 35°. Find its area.

Section D: Mixed Applications

12. A rectangular garden measures 24 m by 32 m. Find its diagonal.
13. A pole stands vertically at one corner of a rectangular field measuring 20 m by 21 m. If the pole is 15 m high, find the angle of elevation of its top from the opposite corner.
14. Prove that (1 − cos 2A)/sin 2A = tan A, wherever defined.
15. Solve 2 sin x = 1 for 0° ≤ x ≤ 360°.

11. Answers and Selected Solutions

Attempt the questions before opening the answer sections.

Section A: Right-Angled Applications

1. Height² = 10² − 6² = 100 − 36 = 64. Height = 8 m.
2. tan 38° = h/15. Therefore h = 15 tan 38° = 11.7 m approximately.
3. tan 28° = h/75. Therefore h = 75 tan 28° = 39.9 m approximately.
4. tan 25° = 2000/d. Therefore d = 2000/tan 25° = 4,289 m approximately.

Section B: Bearings

5. Back bearing = 125° + 180° = 305°.
6. Distance = √(10² + 10²) = √200 = 14.14 km approximately. Bearing = 045°.
7. Northward component = 8 cos 30° = 6.93 km. Eastward component = 8 sin 30° = 4 km.

Section C: Sine Rule and Cosine Rule

8. b/sin 60° = 12/sin 45°. Therefore b = 12 sin 60°/sin 45° = 14.7 cm approximately.
9. c² = 9² + 12² − 2(9)(12)cos 60° = 81 + 144 − 108 = 117. Therefore c = √117 = 10.8 cm approximately.
10. Let C be opposite the 7 cm side. cos C = (5² + 6² − 7²)/(2 × 5 × 6) = 12/60 = 0.2. Therefore C = cos⁻¹(0.2) = 78.5° approximately.
11. Area = ½ × 10 × 14 × sin 35° = 40.2 cm² approximately.

Section D: Mixed Applications

12. Diagonal² = 24² + 32² = 576 + 1024 = 1600. Diagonal = 40 m.
13. Horizontal diagonal = √(20² + 21²) = √841 = 29 m. tan θ = 15/29. Therefore θ = tan⁻¹(15/29) = 27.3° approximately.
14. Starting from the left-hand side: (1 − cos 2A)/sin 2A = [1 − (1 − 2 sin² A)]/[2 sin A cos A] = 2 sin² A/(2 sin A cos A) = sin A/cos A = tan A. Hence proved.
15. sin x = 1/2. In the interval 0° to 360°, the solutions are x = 30° and x = 150°.

12. Final Chapter Revision: Trigonometry Formula Sheet

The following formulae summarise the main ideas covered throughout Chapter 04.

Basic trigonometric ratios

sin θ = Opposite / Hypotenuse

cos θ = Adjacent / Hypotenuse

tan θ = Opposite / Adjacent

Fundamental identity

sin² θ + cos² θ = 1

Compound-angle identities

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B ∓ sin A sin B

tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)

Double-angle identities

sin 2A = 2 sin A cos A

cos 2A = cos² A − sin² A

cos 2A = 1 − 2 sin² A

cos 2A = 2 cos² A − 1

tan 2A = 2 tan A/(1 − tan² A)

Triangle formulae

a/sin A = b/sin B = c/sin C

a² = b² + c² − 2bc cos A

Area = ½ab sin C

Bearings

Back bearing = Bearing ± 180°

13. Chapter 04: Final Summary

Congratulations on completing Chapter 04: Trigonometry.

Across the five parts, we have studied:

1. Part 1: Trigonometric Functions, Angles and Graphs.
2. Part 2: Trigonometric Identities and Formulae.
3. Part 3: Trigonometric Equations and General Solutions.
4. Part 4: Compound Angles, Double Angles and Multiple Angles.
5. Part 5: Trigonometric Applications and Examination Practice.

Final Message

Trigonometry is not simply about memorising formulae. It is about recognising relationships, interpreting diagrams, choosing appropriate methods and applying mathematical reasoning accurately.

Practise regularly, show every important step and always check whether your final answer makes sense in the context of the question.

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