Edwin Ogie Library: Trigonometry E‑Book

Edwin Ogie Library

Trigonometry E‑Book

Objectives:
• Calculate trigonometric ratios for angles between –360° and 360°.
• Use special angles (30°, 45°, 60°, 75°, etc.) to solve problems.
• Solve problems involving angles of elevation and depression.
• Solve bearing and distance problems.
• Apply trigonometric formulas to determine the area of a triangle.
• Analyze and interpret sine and cosine graphs.

Page 1: Introduction to Trigonometry

Trigonometry is the study of the relationships between the sides and angles of triangles. It also covers the analysis of periodic functions such as sine and cosine that model real‑world phenomena like waves and oscillations.

Page 2: Trigonometric Ratios of Angles

In a right triangle, the basic trigonometric ratios are defined as:

Sine (θ): = Opposite/Hypotenuse
Cosine (θ): = Adjacent/Hypotenuse
Tangent (θ): = Opposite/Adjacent

Mnemonic: SOH-CAH-TOA.

Example TR1: Calculate sin 30°

Solution:

sin 30° = 1/2.

Example TR2: Find cos 45°

Solution:

cos 45° = √2/2 ≈ 0.707.

Example TR3: Determine tan 60°

Solution:

tan 60° = √3 ≈ 1.732.

Page 3: Special Angles

Special angles such as 30°, 45°, 60°, 75°, 90°, 105°, and 135° have well‑known trigonometric values. For example:

sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
sin 45° = √2/2, cos 45° = √2/2, tan 45° = 1
sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3

The images below illustrate special angles:

Example SA1: Evaluate sin 30°, cos 30°, tan 30°

Solution:

sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3.

Example SA2: Evaluate sin 45°, cos 45°, tan 45°

Solution:

sin 45° = √2/2, cos 45° = √2/2, tan 45° = 1.

Example SA3: Evaluate sin 60°, cos 60°, tan 60°

Solution:

sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3.

Page 4: Area of a Triangle

The area of a triangle can be calculated by several methods. The most common formula is:

Area = ½ × base × height

Another formula, when two sides and the included angle (θ) are known, is:

Area = ½ab sinθ

Example AT1: Basic Area Calculation

Solution:

For a triangle with base = 8 m and height = 5 m, Area = ½ × 8 × 5 = 20 m².

Example AT2: Using Two Sides and Included Angle

Solution:

For a triangle with sides 7 m and 9 m and an included angle of 45°, Area = ½ × 7 × 9 × sin45° ≈ 31.5 × 0.707 ≈ 22.3 m².

Example AT3: Heron’s Formula

Solution:

For a triangle with sides 5 m, 7 m, and 8 m, s = (5+7+8)/2 = 10 m; Area = √[10(10-5)(10-7)(10-8)] = √[10×5×3×2] = √300 ≈ 17.32 m².

Page 5: Angles of Elevation and Depression

The angle of elevation is the angle between the horizontal line and the line of sight upward, while the angle of depression is the angle between the horizontal and the line of sight downward.

Example AED1: Angle of Elevation

Solution:

If an observer sees the top of a tower at an angle of elevation of 30° from a horizontal distance of 20 m, then tan 30° = height/20. Therefore, height = 20 × tan 30° ≈ 20 × 0.577 = 11.54 m.

Example AED2: Angle of Depression

Solution:

An observer 15 m above sea level sees a boat at an angle of depression of 6°. Then, horizontal distance ≈ 15 / tan 6° ≈ 15 / 0.1051 ≈ 142.6 m.

Example AED3: Combined Scenario

Solution:

From a cliff 50 m high, if the angle of depression to a point on the ground is 4°, then horizontal distance ≈ 50 / tan 4° ≈ 50 / 0.0699 ≈ 715.3 m.

Page 6: Bearings and Distance

Bearings are used in navigation to indicate direction, measured in degrees clockwise from north.

Example BR1: Triangle RSQ

In triangle RSQ, the given data is:

∠R = 65°
∠S = 65°
Side RS = 54 km (between R and S)
Side SQ = 80 km (between S and Q)

First, determine ∠Q:
∠Q = 180° – (65° + 65°) = 50°.

Using the cosine rule with the included angle S (65°) between sides RS and SQ:

RQ² = RS² + SQ² – 2(RS)(SQ) cos 65°

Substitute the values:
RQ² = 54² + 80² – 2 × 54 × 80 × cos 65°
54² = 2916; 80² = 6400; 2×54×80 = 8640; cos 65° ≈ 0.4226.
RQ² ≈ 2916 + 6400 – 8640 × 0.4226 ≈ 9316 – 3652 ≈ 5664.
RQ ≈ √5664 ≈ 75.3 km.

Detailed Solution:

1. Compute the third angle: ∠Q = 180° – 65° – 65° = 50°.
2. Apply the cosine rule on side RQ (opposite ∠S):
RQ² = 54² + 80² – 2×54×80×cos 65°.
RQ² = 2916 + 6400 – 8640×0.4226 ≈ 9316 – 3652 ≈ 5664.
RQ ≈ √5664 ≈ 75.3 km.

Image:

Example BR2: Triangle ABC

Given in triangle ABC:

Side AC = 22 km
Side BC = 18 km
∠A = 25°
∠B = 23°

Then ∠C = 180° – (25° + 23°) = 132°.

To find side AB (opposite ∠C), apply the cosine rule:

AB² = AC² + BC² – 2(AC)(BC) cos ∠C

Substitute the values:
AB² = 22² + 18² – 2 × 22 × 18 × cos 132°.

Note: cos 132° = –cos(48°) ≈ –0.6691.

Thus, AB² ≈ 484 + 324 – 2×22×18×(–0.6691) ≈ 808 + 529.8 ≈ 1337.8.
AB ≈ √1337.8 ≈ 36.6 km.

Detailed Solution:

1. Compute ∠C = 180° – (25° + 23°) = 132°.
2. Apply cosine rule:
AB² = 22² + 18² – 2×22×18×cos 132°.
Since cos 132° = –cos 48° ≈ –0.6691,
AB² = 484 + 324 + 2×22×18×0.6691 ≈ 808 + 529.8 = 1337.8.
AB ≈ √1337.8 ≈ 36.6 km.

Image:

Page 7: Sine and Cosine Graphs

The sine and cosine functions produce smooth, periodic curves. Their key features include amplitude, period, phase shift, and vertical shift.

Example SC1: Determine the Amplitude and Period of y = 2 sin(x)

Solution:

Amplitude = 2; Period = 2π.

Example SC2: Graph y = 2 cos(2x)

Solution:

Amplitude = 2; Period = 2π/2 = π; the graph oscillates between 2 and –2.

Example SC3: Identify the Phase Shift of y = sin(x – π/3)

Solution:

The graph is shifted to the right by π/3.

Page 8: Areas of a Triangle Using Trigonometry

Besides the basic ½ × base × height formula, the area of a triangle can be found using trigonometry when two sides and the included angle are known:

Area = ½ab sinθ

Example TA1: Area Calculation

Solution:

For a triangle with sides 8 m and 10 m and an included angle of 45°, Area ≈ ½ × 8 × 10 × sin45° ≈ 40 × 0.707 ≈ 28.28 m².

Example TA2: Area with Different Sides

Solution:

For sides 6 m and 9 m with an included angle of 60°, Area ≈ ½ × 6 × 9 × sin60° ≈ 27 × 0.866 ≈ 23.38 m².

Example TA3: Heron’s Formula

Solution:

For a triangle with sides 7 m, 8 m, and 9 m, s = (7+8+9)/2 = 12 m; Area = √[12(12–7)(12–8)(12–9)] = √[12×5×4×3] = √720 ≈ 26.83 m².