Lesson Notes By Weeks and Term v5 - Grade 11
Trigonometry – Week 6 focus
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Trigonometry – Week 6 focus
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Subject: Mathematics
Class: Grade 11
Term: 2nd Term
Week: 6
Theme: General lesson support
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This week, we delve deeper into the fascinating world of Trigonometry, focusing on Solving Problems in Two Dimensions. We will build upon our knowledge of trigonometric ratios, special angles, and the Laws of Sines and Cosines to tackle real-world problems involving triangles in a two-dimensional plane. Understanding trigonometry is crucial not only for succeeding in mathematics but also for many other fields, such as surveying, navigation, architecture, engineering, and even computer graphics. Imagine designing a bridge, determining the height of a building, or navigating a ship – all rely heavily on trigonometric principles.
This week's focus is on applying our trigonometric knowledge to practical problems. The core concepts we'll be utilizing include: Trigonometric Ratios: Review your SOH CAH TOA! Sine (Opposite/Hypotenuse), Cosine (Adjacent/Hypotenuse), and Tangent (Opposite/Adjacent). Remember these apply only to right-angled triangles.
Angle of Elevation: The angle formed between the horizontal line of sight and an upward line of sight to an object. Imagine looking up at the top of a building; the angle your eyes make with the horizontal is the angle of elevation.
Angle of Depression: The angle formed between the horizontal line of sight and a downward line of sight to an object. Imagine looking down at a boat from a cliff; the angle your eyes make with the horizontal is the angle of depression. Importantly, the angle of elevation from the boat to the cliff will be equal to the angle of depression from the cliff to the boat (alternate angles).
Law of Sines: For any triangle ABC, a/sin A = b/sin B = c/sin
C. This law is used when you have: Two angles and one side (AAS or ASA) Two sides and an angle opposite one of them (SSA - be careful of the ambiguous case!)
Law of Cosines: For any triangle ABC: a² = b² + c² - 2bc cos A b² = a² + c² - 2ac cos B c² = a² + b² - 2ab cos C This law is used when you have: Three sides (SSS) Two sides and the included angle (SAS)
Bearings: Used to describe directions.
There are two main types: True bearing:* Measured clockwise from North (000°T to 360°T).
Compass bearing:* Measured east or west from North or South (e.g., N 30° E, S 45° W). It's essential to convert compass bearings to true bearings for calculations.
Problem-Solving Strategy: A good approach is always: Draw a clear, labeled diagram. This is crucial. Identify the knowns and unknowns. Choose the appropriate trigonometric ratio or law. Solve for the unknowns. Interpret the solution in the context of the problem.
Example 1: A surveyor in Durban needs to determine the height of a tall building. She stands 50 meters away from the base of the building and measures the angle of elevation to the top of the building to be 72°. How tall is the building?
Diagram: Draw a right-angled triangle. The height of the building is the opposite side, the distance from the surveyor is the adjacent side, and the angle of elevation is 72°.
Knowns: Adjacent = 50m, Angle of Elevation = 72°
Unknown:* Opposite (Height of the building)
Choice: Tangent (Opposite/Adjacent)
Solution:
tan(72°) = Opposite / 50
Opposite = 50 tan(72°)
Opposite ≈ 153.88 m
Interpretation: The height of the building is approximately 153.88 meters.
Example 2: Two ships leave Cape Town harbor at the same time. Ship A travels on a bearing of 060°T at a speed of 25 km/h. Ship B travels on a bearing of 140°T at a speed of 30 km/h. How far apart are the ships after 2 hours?
Diagram: Draw a diagram showing the two ships leaving the harbor. After 2 hours, Ship A has travelled 25 km/h 2h = 50 km, and Ship B has travelled 30 km/h 2h = 60 km. The angle between their paths is 140° - 60° = 80°.
Knowns: Side a = 50 km, Side b = 60 km, Angle C = 80°
Unknown:* Side c (Distance between the ships)
Choice: Law of Cosines (c² = a² + b² - 2ab cos C)
Solution:
c² = 50² + 60² - 2 50 60 cos(80°)
c² = 2500 + 3600 - 6000 cos(80°)
c² ≈ 5060.72
c ≈ 71.14 km
Interpretation: The ships are approximately 71.14 km apart after 2 hours.
Example 3: A farmer in Limpopo wants to fence off a triangular piece of land. The sides of the land measure 150m, 200m, and 250m. What is the largest angle of the triangle?
Diagram: Draw a triangle with sides 150m, 200m, and 250m.
Knowns: a = 150m, b = 200m, c = 250m
Unknown:* Angle C (opposite the longest side, which will be the largest angle)
Choice: Law of Cosines (c² = a² + b² - 2ab cos C), rearranged to solve for cos C: cos C = (a² + b² - c²) / 2ab
Solution:
cos C = (150² + 200² - 250²) / (2 150 * 200)
cos C = (22500 + 40000 - 62500) / 60000
cos C = 0
C = cos⁻¹(0)
C = 90°
Interpretation: The largest angle of the triangle is 90°. This means the triangle is a right-angled triangle.
Guided Practice (With Solutions)
Question 1: From the top of a 30-meter high cliff, the angle of depression to a boat is 35°. How far is the boat from the base of the cliff?