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Objectives:
• Calculate sin, cos, and tan for angles between –360° and 360°.
• Apply special angles (30°, 45°, 60°, 75°, 90°, 105°, 135°) to solve problems.
• Solve problems involving angles of elevation and depression.
• Solve problems involving bearings.
• Apply trigonometric formulae to find areas of triangles.
• Analyze and solve problems using sine and cosine graphs.
Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles. It also involves the study of periodic functions such as sine and cosine, which are vital in modeling waves and oscillations.
In a right triangle, the trigonometric ratios are defined as follows:
These are remembered by the mnemonic SOH-CAH-TOA.
Special angles such as 30°, 45°, 60°, 75°, 90°, 105°, and 135° have known trigonometric values that are useful for solving problems quickly. For example:
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The angle of elevation is the angle between the horizontal and the line of sight upward to an object. The angle of depression is the angle between the horizontal and the line of sight downward. Diagrams help illustrate these concepts.
Bearings are used to describe direction in navigation. They are typically measured in degrees clockwise from the north direction. For example, a bearing of 045° points northeast.
The area of a triangle can be found by several methods. One common formula is:
Area = ½ × base × height
Another useful formula when two sides and the included angle are known is:
Area = ½ab sin θ
This section also covers methods to solve for missing side lengths or angles in triangles.
The sine and cosine functions produce smooth, periodic curves. Their key features include:
Some key identities include:
Example 1: Calculate sin 30°, cos 30°, and tan 30°.
Solution:
sin 30° = 1/2, cos 30° = √3/2, tan 30° = (1/2)/(√3/2) = 1/√3.
Example 2: Find the angle of elevation if a 10 m tall object is seen from 15 m away on level ground.
Solution:
tan θ = opposite/adjacent = 10/15 = 2/3, so θ = tan−1(2/3) ≈ 33.69°.
Example 3: A ship is observed with an angle of depression of 5°. Find the horizontal distance if the observer is 20 m above sea level.
Solution:
tan 5° ≈ 0.0875, so distance = height/tan 5° = 20/0.0875 ≈ 228.57 m.
Example 4: Calculate the area of a triangle with sides 8 m and 10 m with an included angle of 45° using ½ab sinθ.
Solution:
Area = ½ × 8 × 10 × sin 45° = 40 × (√2/2) ≈ 28.28 m².
Example 5: Given a bearing of 120° from North, find the directional components (using sine and cosine).
Solution:
Northward component = cos 120° = –0.5, Eastward component = sin 120° = 0.866 (scaled appropriately).
Example 6: Find the value of cos 45° using the unit circle.
Solution:
cos 45° = √2/2 ≈ 0.707.
Example 7: Calculate sin 75° using the angle sum formula: sin(45° + 30°).
Solution:
sin 75° = sin 45° cos 30° + cos 45° sin 30° = (√2/2 × √3/2) + (√2/2 × 1/2) = (√6 + √2)/4.
Example 8: Determine tan 105° using the angle sum formula: tan(60° + 45°).
Solution:
tan 105° = (tan 60° + tan 45°) / (1 − tan 60° tan 45°) = (√3 + 1)/(1 − √3×1). Simplify as needed.
Example 9: Find the area of a triangle given sides 7 m, 9 m and an included angle of 60° using ½ab sinθ.
Solution:
Area = ½ × 7 × 9 × sin 60° = 31.5 × (√3/2) ≈ 27.28 m².
Example 10: Using the graphs of sine and cosine, determine the period of y = 2 cos(3x - π/4) + 1.
Solution:
Period = 2π/|3| = 2π/3.
Bearings describe direction as an angle measured clockwise from north. They are used to find distances and positions on maps.
Solution:
A bearing of 045° indicates a direction of northeast.
Solution:
For a distance of 1000 km at a bearing of 120°, the north component = 1000 × cos 120° = 1000 × (–0.5) = –500 km (i.e. 500 km south), and the east component = 1000 × sin 120° ≈ 1000 × 0.866 = 866 km.
Solution:
If a point is located 500 km away at a bearing of 210°, resolve its components using trigonometry.
In triangle RSQ, the given data are: Angle R = 65°, Angle S = 65°, Side RS = 54 km, and Side SQ = 80 km. Determine the distance between points R and Q.
Solution:
First, note that the included angle between sides RS and SQ is angle S = 65°. Using the Law of Cosines:
RQ² = RS² + SQ² - 2 × RS × SQ × cos(65°)
Substitute the values:
RQ² = 54² + 80² - 2 × 54 × 80 × cos(65°)
= 2916 + 6400 - 8640 × 0.4226 (cos 65° ≈ 0.4226)
= 9316 - 3652.59 ≈ 5663.41
RQ ≈ √5663.41 ≈ 75.26 km.
Given in triangle ABC: AC = 22 km, BC = 18 km, Angle A = 25°, and Angle B = 23°. Find the distance AB.
Solution:
First, find angle C:
Angle C = 180° - (25° + 23°) = 132°
Now apply the Law of Cosines:
AB² = AC² + BC² - 2 × AC × BC × cos(C)
Substitute the values:
AB² = 22² + 18² - 2 × 22 × 18 × cos(132°)
cos(132°) = -cos(48°) ≈ -0.6691
= 484 + 324 - 792 × (-0.6691)
= 808 + 529.47 ≈ 1337.47
AB ≈ √1337.47 ≈ 36.57 km.
Trigonometry is essential in fields such as engineering, architecture, navigation, and physics. It is used to model periodic phenomena, determine distances, and solve real-world problems involving heights and bearings.
Practice with unit circles and special triangles to quickly recall trig values. Use graphing calculators for complex angles and always check your work by verifying with identities.
Review all key concepts from trigonometric ratios, special angles, angle of elevation and depression, bearings, triangle area formulas, and the graphs of sine and cosine. Test your understanding with the quiz below.
Click the "Start Quiz" button to begin. You will have 15 minutes to answer 30 questions.
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Edwin Ogie Library - Trigonometry E‑Book http://www.edwinogie-library.comLatest updates on trigonometric ratios, special angles, triangle solutions, and trig graphs. New Lesson on Trigonometry http://www.edwinogie-library.com/trigonometryExplore our comprehensive e‑book covering trigonometric functions, identities, and applications.
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