Lesson Notes By Weeks and Term v5 - Grade 10
Trigonometry – Week 6 focus
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Trigonometry – Week 6 focus
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Subject: Mathematics
Class: Grade 10
Term: 3rd Term
Week: 6
Theme: General lesson support
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Trigonometry is a vital branch of mathematics that deals with the relationships between the sides and angles of triangles. In Grade 10, we begin to build a strong foundation that will be crucial for more advanced topics in higher grades and for various applications in real life. This week, we will be focusing specifically on applying trigonometric ratios to solve problems involving right-angled triangles. Understanding these principles is essential for fields like surveying, navigation, engineering, and even architecture – all of which have significant relevance in South Africa.
2.1 Trigonometric Ratios (SOH CAH TOA) In a right-angled triangle, the trigonometric ratios relate the angles to the lengths of the sides.
Remember the acronym SOH CAH TOA: SOH: Sine = Opposite / Hypotenuse (sin θ = Opposite/Hypotenuse)
CAH: Cosine = Adjacent / Hypotenuse (cos θ = Adjacent/Hypotenuse)
TOA: Tangent = Opposite / Adjacent (tan θ = Opposite/Adjacent)
Important definitions: Hypotenuse: The longest side of the right-angled triangle, opposite the right angle (90°).
Opposite side: The side opposite to the angle θ.
Adjacent side: The side next to the angle θ (and not the hypotenuse). 2.2 Angle of Elevation and Angle of Depression Angle of Elevation: The angle formed between the horizontal line of sight and the upward direction of an object being viewed. Imagine a person looking up at the top of a building.
Angle of Depression: The angle formed between the horizontal line of sight and the downward direction of an object being viewed. Imagine a person standing on top of a building looking down at a car on the street. 2.3 Solving for Unknown Sides To solve for an unknown side, identify which trigonometric ratio involves the known angle and the known side. Then, set up an equation and solve for the unknown.
Example 1: A builder needs to brace a wall that is 3 meters high. He wants the brace to make an angle of 60° with the ground. How long must the brace be?
Diagram: Draw a right-angled triangle. The wall is the opposite side (3m). The brace is the hypotenuse (what we want to find). The angle is 60°.
Identify the ratio: We have the opposite and want to find the hypotenuse. So, we use sine (SOH).
Equation: sin(60°) = 3 / Hypotenuse Solve: Hypotenuse = 3 / sin(60°) ≈ 3 / 0.866 ≈ 3.46 meters Therefore, the brace must be approximately 3.46 meters long.
Example 2: From the top of a cliff 50 meters high, the angle of depression to a boat is 35°. How far is the boat from the base of the cliff?
Diagram: Draw a right-angled triangle. The cliff height is the opposite side (50m). The distance from the boat to the base is the adjacent side (what we want to find). The angle of depression is 35°. Remember the angle of depression outside the triangle is equal to the angle of elevation inside the triangle.
Identify the ratio: We have the opposite and want to find the adjacent. So, we use tangent (TOA).
Equation: tan(35°) = 50 / Adjacent Solve: Adjacent = 50 / tan(35°) ≈ 50 / 0.700 ≈ 71.43 meters Therefore, the boat is approximately 71.43 meters from the base of the cliff. 2.4 Solving for Unknown Angles To solve for an unknown angle, identify which trigonometric ratio involves the two known sides. Then, use the inverse trigonometric function (sin -1 , cos -1 , or tan -1 ) to find the angle.
Example 3: A ladder 4 meters long leans against a wall. The foot of the ladder is 1.5 meters from the wall. What angle does the ladder make with the ground?
Diagram: Draw a right-angled triangle. The ladder is the hypotenuse (4m). The distance from the wall is the adjacent side (1.5m). We want to find the angle.
Identify the ratio: We have the adjacent and the hypotenuse. So, we use cosine (CAH).
Equation: cos(θ) = 1.5 / 4 = 0.375 Solve: θ = cos -1 (0.375) ≈ 67.98° Therefore, the ladder makes an angle of approximately 67.98° with the ground. Guided Practice (With Solutions)
Question 1: A flagpole casts a shadow of 8 meters long. The angle of elevation of the sun is 50°. How tall is the flagpole?
Solution: Diagram: Draw a right-angled triangle. The shadow is the adjacent side (8m). The height of the flagpole is the opposite side (what we want to find). The angle of elevation is 50°.
Identify the ratio: We have the adjacent and want to find the opposite. So, we use tangent (TOA).
Equation: tan(50°) = Opposite / 8 Solve: Opposite = 8 tan(50°) ≈ 8 * 1.192 ≈ 9.54 meters The flagpole is approximately 9.54 meters tall.
Commentary: This problem utilizes the angle of elevation and tangent ratio effectively. Be sure to include the units.
Question 2: A ramp is built to allow wheelchair access to a building. The ramp is 5 meters long and reaches a height of 0.5 meters. What is the angle of elevation of the ramp?
Solution: Diagram: Draw a right-angled triangle. The ramp is the hypotenuse (5m). The height is the opposite side (0.5m). We want to find the angle.
Identify the ratio: We have the opposite and the hypotenuse. So, we use sine (SOH).
Equation: sin(θ) = 0.5 / 5 = 0.1 Solve: θ = sin -1 (0.1) ≈ 5.74° The angle of elevation of the ramp is approximately 5.74°.
Commentary: This is a real-world accessibility problem. Notice the small angle. Steep ramps are difficult to navigate with a wheelchair.
Question 3: A cellphone tower is supported by a cable that is anchored to the ground 15 meters from the base of the tower. The cable is 25 meters long. What is the angle the cable makes with the ground?
Solution: Diagram: Draw a right-angled triangle. The cable is the hypotenuse (25m). The distance from the base of the tower is the adjacent side (15m).