Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Determine exact trig ratios for special angles without a calculator.
• Apply these ratios to solve right-angled triangles.
• Simplify expressions and solve equations with these ratios.
• Use trig ratios to solve real-life problems like heights and distances.

Lesson summary

This week, we delve into the fascinating world of Trigonometry, specifically focusing on the Trigonometric Ratios of Special Angles (0°, 30°, 45°, 60°, and 90°) and applying these ratios to solve problems. Trigonometry isn't just abstract math; it's a powerful tool used by engineers, architects, surveyors, and even navigators. Imagine calculating the height of Telkom Tower in Johannesburg without climbing it, or designing a stable ramp for wheelchair access – trigonometry makes these possible!

Lesson notes

2.1 Understanding Special Angles Special angles (0°, 30°, 45°, 60°, and 90°) are angles for which the trigonometric ratios can be expressed exactly using surds (square roots) and fractions. This stems from geometrical properties of specific triangles, primarily the equilateral triangle (for 30° and 60°) and the isosceles right-angled triangle (for 45°). 2.2 Derivation of Trigonometric Ratios 45°: Consider an isosceles right-angled triangle with sides of length 1 unit. By the Pythagorean theorem, the hypotenuse has length √(1² + 1²) = √2. sin 45° = Opposite/Hypotenuse = 1/√2 = √2/2 (Rationalizing the denominator) cos 45° = Adjacent/Hypotenuse = 1/√2 = √2/2 tan 45° = Opposite/Adjacent = 1/1 = 1 30° and 60°: Consider an equilateral triangle with sides of length 2 units. Draw a perpendicular from one vertex to the midpoint of the opposite side. This perpendicular bisects the equilateral triangle into two congruent 30°-60°-90° right-angled triangles. The length of the perpendicular is √(2² - 1²) = √

3. For 30°: sin 30° = Opposite/Hypotenuse = 1/2 cos 30° = Adjacent/Hypotenuse = √3/2 tan 30° = Opposite/Adjacent = 1/√3 = √3/3 For 60°: sin 60° = Opposite/Hypotenuse = √3/2 cos 60° = Adjacent/Hypotenuse = 1/2 tan 60° = Opposite/Adjacent = √3/1 = √3 0° and 90°: Consider a unit circle (radius = 1). As the angle approaches 0°, the opposite side approaches 0 and the adjacent side approaches

1. As the angle approaches 90°, the opposite side approaches 1 and the adjacent side approaches

0. For 0°: sin 0° = 0/1 = 0 cos 0° = 1/1 = 1 tan 0° = 0/1 = 0 For 90°: sin 90° = 1/1 = 1 cos 90° = 0/1 = 0 tan 90° = 1/0 = Undefined 2.3 Summary Table It's helpful to memorize or quickly derive these values: | Angle (θ) | sin θ | cos θ | tan θ | |---|---|---|---| | 0° | 0 | 1 | 0 | | 30° | 1/2 | √3/2 | √3/3 | | 45° | √2/2 | √2/2 | 1 | | 60° | √3/2 | 1/2 | √3 | | 90° | 1 | 0 | Undefined | 2.4 Worked

Examples: Example 1: Simplify the expression: 2sin 30° + 3cos 60° - tan 45° Solution: Substitute the known values: 2(1/2) + 3(1/2) - 1 Simplify: 1 + 3/2 - 1 Answer: 3/2 Example 2: In right-angled triangle ABC, where angle B = 90°, angle C = 30°, and AB = 5cm, find the length of A

C. Solution: Identify the relevant trigonometric ratio. We know the opposite side (AB) and want to find the hypotenuse (AC). Sine relates opposite and hypotenuse. sin C = Opposite/Hypotenuse => sin 30° = AB/AC Substitute known values: 1/2 = 5/AC Solve for AC: AC = 5 / (1/2) = 10 cm Example 3: A flagpole casts a shadow of 8 meters when the angle of elevation of the sun is 60°. Calculate the height of the flagpole.

Solution: Draw a diagram representing the situation. The flagpole forms the opposite side, the shadow forms the adjacent side, and the angle of elevation is the angle between the shadow and the line of sight to the top of the flagpole. Identify the relevant trigonometric ratio. We know the adjacent side and want to find the opposite side. Tangent relates opposite and adjacent. tan θ = Opposite/Adjacent => tan 60° = Height/8 Substitute known values: √3 = Height/8 Solve for Height: Height = 8√3 meters (approximately 13.86 meters) Guided Practice (With Solutions)

Question 1: Evaluate: cos² 30° + sin² 60° Solution: Substitute known values: (√3/2)² + (√3/2)² Simplify: 3/4 + 3/4 Answer: 6/4 = 3/2

Commentary: Remember that cos² 30° means (cos 30°)². This question reinforces the values of cos 30° and sin 60°.

Question 2: If tan θ = 1, and θ is an acute angle, find the value of sin θ + cos θ.

Solution: Recognize that tan θ = 1 when θ = 45°. Substitute θ = 45° into the expression: sin 45° + cos 45° Substitute known values: √2/2 + √2/2 Answer: 2√2/2 = √2

Commentary: This question tests your understanding of the tangent function and its relationship to the angle 45°.

Question 3: A ladder leans against a wall, making an angle of 30° with the ground. If the foot of the ladder is 4 meters away from the wall, how high up the wall does the ladder reach?

Solution: Draw a diagram. The ladder is the hypotenuse, the wall is the opposite side, and the distance from the wall is the adjacent side. We want to find the opposite side (height) and know the adjacent side. Use the tangent function. tan θ = Opposite/Adjacent => tan 30° = Height/4 Substitute known values: √3/3 = Height/4 Solve for Height: Height = (4√3)/3 meters

Commentary: This is a typical real-world application problem. Carefully identifying the opposite, adjacent, and hypotenuse is crucial. Independent Practice (Questions Only)

Calculate: (sin 60°)(cos 30°) - (sin 30°)(cos 60°)

Simplify: (tan 45° + sin 90°) / cos 0° If sin θ = √3/2, and θ is an acute angle, find the value of tan θ. In right-angled triangle PQR, where angle Q = 90°, angle R = 60°, and PQ = 7cm, find the length of PR. A kite is flying at the end of a 50-meter string. The angle of elevation of the kite is 45°. Assuming the string is taut, how high is the kite above the ground?

Evaluate: sin² 30° + cos² 30° + tan² 60° A building is 20 meters tall.