Lesson Notes By Weeks and Term v5 - Grade 10
Trigonometry – Week 10 focus
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Trigonometry – Week 10 focus
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Subject: Mathematics
Class: Grade 10
Term: 1st Term
Week: 10
Theme: General lesson support
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Trigonometry is a fundamental branch of mathematics that explores the relationships between angles and sides of triangles. In Grade 10, we build upon the basic trigonometric ratios (sine, cosine, and tangent) that you may have encountered before and introduce the use of right-angled triangles to solve problems. This week, we focus on applying these trigonometric ratios to calculate unknown sides and angles in right-angled triangles and solving problems in two dimensions.
2.1 Trigonometric Ratios In a right-angled triangle, the side opposite the right angle is called the hypotenuse. Relative to a specific acute angle θ (theta), the other two sides are called the opposite and adjacent sides. The trigonometric ratios are defined as follows: Sine (sin θ): The ratio of the length of the opposite side to the length of the hypotenuse. `sin θ = Opposite / Hypotenuse` Cosine (cos θ): The ratio of the length of the adjacent side to the length of the hypotenuse. `cos θ = Adjacent / Hypotenuse` Tangent (tan θ): The ratio of the length of the opposite side to the length of the adjacent side. `tan θ = Opposite / Adjacent` A helpful mnemonic to remember these ratios is SOH CAH TOA: Sine = Opposite / Hypotenuse Cosine = Adjacent / Hypotenuse Tangent = Opposite / Adjacent 2.2 Finding Unknown Sides If we know one side and one acute angle in a right-angled triangle, we can use trigonometric ratios to find the other sides.
Example 1: A flagpole casts a shadow of 5 meters long. The angle of elevation of the sun is 60°. How tall is the flagpole?
Solution:* Draw a diagram. The flagpole is the opposite side, the shadow is the adjacent side, and the angle of elevation is 60°. We need to find the height (opposite) and we know the adjacent side.
Therefore, we use the tangent ratio: `tan θ = Opposite / Adjacent` Substitute the known values: `tan 60° = Opposite / 5` Solve for the opposite side: `Opposite = 5 * tan 60°` Using a calculator, `tan 60° ≈ 1.732` `Opposite ≈ 5 * 1.732 ≈ 8.66` Therefore, the flagpole is approximately 8.66 meters tall.
Example 2: A ladder leaning against a wall makes an angle of 70° with the ground. The foot of the ladder is 2 meters away from the wall. How long is the ladder?
Solution:* Draw a diagram. The ladder is the hypotenuse, the distance from the wall is the adjacent side, and the angle with the ground is 70°. We need to find the hypotenuse and we know the adjacent side.
Therefore, we use the cosine ratio: `cos θ = Adjacent / Hypotenuse` Substitute the known values: `cos 70° = 2 / Hypotenuse` Solve for the hypotenuse: `Hypotenuse = 2 / cos 70°` Using a calculator, `cos 70° ≈ 0.342` `Hypotenuse ≈ 2 / 0.342 ≈ 5.85` Therefore, the ladder is approximately 5.85 meters long. 2.3 Finding Unknown Angles If we know two sides in a right-angled triangle, we can use inverse trigonometric functions (arcsin, arccos, arctan) to find the unknown angles. Your calculator will have buttons labeled sin⁻¹, cos⁻¹, and tan⁻¹, which represent these inverse functions.
Example 3: In a right-angled triangle, the opposite side is 4 cm and the adjacent side is 3 cm. Find the angle θ.
Solution:* We know the opposite and adjacent sides.
Therefore, we use the tangent ratio: `tan θ = Opposite / Adjacent` Substitute the known values: `tan θ = 4 / 3` To find θ, we use the inverse tangent function: `θ = tan⁻¹(4/3)` Using a calculator, `tan⁻¹(4/3) ≈ 53.13°` Therefore, the angle θ is approximately 53.13°. 2.4 Angles of Elevation and Depression Angle of Elevation: The angle formed between the horizontal line of sight and the line of sight to an object above the horizontal. Imagine looking up at the top of a building.
Angle of Depression: The angle formed between the horizontal line of sight and the line of sight to an object below the horizontal. Imagine looking down at a boat from the top of a cliff. 2.5 Solving 2D Problems Many real-world problems can be modeled using right-angled triangles. When solving these problems, it is crucial to draw a clear diagram, identify the relevant angles and sides, and choose the appropriate trigonometric ratio to use.
Example 4: From the top of a cliff 50 meters high, the angle of depression to a boat is 30°. How far is the boat from the base of the cliff?
Solution:* Draw a diagram. The height of the cliff is the opposite side, the distance of the boat from the base is the adjacent side, and the angle of depression is 30°. Note that the angle of depression from the cliff to the boat is equal to the angle of elevation from the boat to the top of the cliff (alternate angles are equal). We need to find the adjacent side and we know the opposite side.
Therefore, we use the tangent ratio: `tan θ = Opposite / Adjacent` Substitute the known values: `tan 30° = 50 / Adjacent` Solve for the adjacent side: `Adjacent = 50 / tan 30°` Using a calculator, `tan 30° ≈ 0.577` `Adjacent ≈ 50 / 0.577 ≈ 86.65` Therefore, the boat is approximately 86.65 meters away from the base of the cliff. Guided Practice (With Solutions)
Question 1: In a right-angled triangle ABC, where angle B = 90°, AB = 8 cm, and angle C = 30°, find the length of B
C. Solution:* Draw a diagram. BC is the adjacent side to angle C, and AB is the opposite side.
We use the tangent ratio: `tan C = AB / BC` Substitute the values: `tan 30° = 8 / BC` Solve for BC: `BC = 8 / tan 30°` Using a calculator, `tan 30° ≈ 0.577` `BC ≈ 8 / 0.577 ≈ 13.86` cm Therefore, the length of BC is approximately 13.86 cm.