Lesson Notes By Weeks and Term v5 - Grade 10
Trigonometric functions – Week 5 focus
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Trigonometric functions – Week 5 focus
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Subject: Mathematics
Class: Grade 10
Term: 2nd Term
Week: 5
Theme: General lesson support
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This week, we delve into trigonometric functions, specifically focusing on sine, cosine, and tangent functions. Trigonometry is crucial for understanding relationships between angles and sides of triangles, with far-reaching applications in fields like engineering, surveying, navigation, and even art and architecture. Think about how cellphone towers are positioned to provide coverage across South Africa, or how bridges are designed to withstand forces. Trigonometry is at the heart of these calculations. It's also foundational for understanding more advanced mathematical concepts in later grades.
2. 1. Defining Trigonometric Ratios Consider a right-angled triangle ABC, where angle B is the right angle (90°). Let's focus on angle
A. Hypotenuse: The side opposite the right angle (the longest side). In our triangle, it's side A
C. Opposite: The side opposite to angle
A. In our triangle, it's side B
C. Adjacent: The side adjacent to angle A (not the hypotenuse). In our triangle, it's side A
B. The three basic trigonometric ratios are defined as follows: Sine (sin): sin(A) = Opposite / Hypotenuse = BC / AC Cosine (cos): cos(A) = Adjacent / Hypotenuse = AB / AC Tangent (tan): tan(A) = Opposite / Adjacent = BC / AB Mnemonic: A common mnemonic to remember these ratios is SOH CAH TOA: Sine = Opposite / Hypotenuse Cosine = Adjacent / Hypotenuse Tangent = Opposite / Adjacent 2.
2. Worked Example 1: Finding Trigonometric Ratios In a right-angled triangle PQR, with angle Q = 90°, PQ = 8 cm, and PR = 17 cm, find sin(P), cos(P), and tan(P).
Step 1: Identify the sides relative to angle
P. Hypotenuse: PR = 17 cm Opposite: QR (We need to find this)
Adjacent: PQ = 8 cm Step 2: Use the Pythagorean theorem to find Q
R. PR² = PQ² + QR² 17² = 8² + QR² 289 = 64 + QR² QR² = 225 QR = 15 cm Step 3: Calculate the trigonometric ratios. sin(P) = Opposite / Hypotenuse = QR / PR = 15 / 17 cos(P) = Adjacent / Hypotenuse = PQ / PR = 8 / 17 tan(P) = Opposite / Adjacent = QR / PQ = 15 / 8 2.
3. Solving Right-Angled Triangles If we know one side and one acute angle in a right-angled triangle, or two sides, we can use trigonometric ratios to find the remaining sides and angles. 2.
4. Worked Example 2: Finding an Unknown Side A ladder leans against a wall, making an angle of 70° with the ground. The foot of the ladder is 2 meters away from the wall. How high up the wall does the ladder reach?
Step 1: Draw a diagram. This helps visualize the problem.
Step 2: Identify the knowns and unknowns.
Angle: 70° Adjacent side: 2 meters (distance from the wall)
Opposite side: Height up the wall (unknown – let's call it 'h')
Step 3: Choose the appropriate trigonometric ratio. We have the adjacent and want to find the opposite, so we use tan. tan(70°) = Opposite / Adjacent = h / 2 Step 4: Solve for the unknown. h = 2 tan(70°) h ≈ 2 2.747 h ≈ 5.49 meters Therefore, the ladder reaches approximately 5.49 meters up the wall. 2.
5. Worked Example 3: Finding an Unknown Angle A ramp is 5 meters long and rises to a height of 1 meter. What is the angle of elevation of the ramp?
Step 1: Draw a diagram.
Step 2: Identify the knowns and unknowns.
Hypotenuse: 5 meters Opposite: 1 meter Angle of elevation (θ): Unknown Step 3: Choose the appropriate trigonometric ratio. We have the opposite and hypotenuse, so we use sin. sin(θ) = Opposite / Hypotenuse = 1 / 5 = 0.2 Step 4: Solve for the unknown angle using the inverse sine function (arcsin or sin⁻¹). θ = sin⁻¹(0.2) θ ≈ 11.54° Therefore, the angle of elevation of the ramp is approximately 11.54°. 2.
6. Special Angles (30°, 45°, 60°) These angles appear frequently, and it's helpful to know their trigonometric ratios without using a calculator. We can derive these values from special triangles: 45-45-90 Triangle: Consider an isosceles right-angled triangle with sides of length
1. By the Pythagorean theorem, the hypotenuse is √2. sin(45°) = 1 / √2 = √2 / 2 cos(45°) = 1 / √2 = √2 / 2 tan(45°) = 1 / 1 = 1 30-60-90 Triangle: Start with an equilateral triangle with sides of length
2. Draw an altitude from one vertex to the midpoint of the opposite side. This divides the equilateral triangle into two 30-60-90 triangles. The altitude has length √3 (by the Pythagorean theorem). For 30°: sin(30°) = 1 / 2 cos(30°) = √3 / 2 tan(30°) = 1 / √3 = √3 / 3 For 60°: sin(60°) = √3 / 2 cos(60°) = 1 / 2 tan(60°) = √3 / 1 = √3 2.
7. Angle of Elevation and Depression Angle of Elevation: The angle between the horizontal line of sight and the line of sight upwards to an object. Imagine looking up at the top of a building – the angle between your eye level and the top of the building is the angle of elevation.
Angle of Depression: The angle between the horizontal line of sight and the line of sight downwards to an object. Imagine looking down from the top of a cliff at a boat – the angle between your eye level and the boat is the angle of depression. Guided Practice (With Solutions)
Question 1: In a right-angled triangle ABC, where angle C = 90°, AC = 5 cm, and BC = 12 cm, find sin(A), cos(A), and tan(A).
Solution: First, find the hypotenuse AB using the Pythagorean theorem: AB² = AC² + BC² = 5² + 12² = 25 + 144 =
1
6
9. Therefore, AB = √169 = 13 cm. sin(A) = Opposite / Hypotenuse = BC / AB = 12 / 13 cos(A) = Adjacent / Hypotenuse = AC / AB = 5 / 13 tan(A) = Opposite / Adjacent = BC / AC = 12 / 5 Question 2: A flagpole casts a shadow of 10 meters long. The angle of elevation of the sun is 60°. How tall is the flagpole?
Solution: Draw a diagram. Let the height of the flagpole be 'h'.