Lesson Notes By Weeks and Term v5 - Grade 11
Trigonometry – Week 8 focus
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Trigonometry – Week 8 focus
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Subject: Mathematics
Class: Grade 11
Term: 2nd Term
Week: 8
Theme: General lesson support
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This week, we delve deeper into trigonometry, building upon your knowledge of trigonometric ratios, special angles, and the Cartesian plane. Trigonometry is not just an abstract mathematical concept; it's a powerful tool used in various real-world applications. For example, surveyors use trigonometry to determine distances and heights of buildings and land features, which is essential for infrastructure development and urban planning in South Africa. Engineers use it to design bridges and other structures, ensuring their stability and safety. Architects use it to create aesthetically pleasing and functional designs.
2.1 The Sine Rule: The sine rule states that in any triangle ABC, the ratio of the length of a side to the sine of the angle opposite that side is constant. Mathematically, this is expressed as: ``` a/sinA = b/sinB = c/sinC ``` Where: a, b, c are the lengths of the sides opposite angles A, B, and C, respectively.
When to use the Sine Rule: You can use the sine rule when you know: Two angles and one side (AAS or ASA). Two sides and a non-included angle (SSA - Ambiguous Case; be careful here!). The ambiguous case might result in 0, 1, or 2 possible triangles.
Example 1: A surveyor in KwaZulu-Natal needs to determine the distance across a river. She stands at point A on one side of the river and sights a tree at point C on the opposite bank. She then walks 50 meters along the riverbank to point B and measures the angle ABC to be 62° and the angle BAC to be 48°. Calculate the distance AC across the river.
Solution: Draw a diagram (This is crucial for visualising the problem).
Identify the knowns: AB = 50m, ∠ABC = 62°, ∠BAC = 48°.
Find the unknown: AC (let's call it 'b').
Find the missing angle: ∠ACB = 180° - 62° - 48° = 70°.
Apply the Sine Rule: ``` b/sinB = a/sinA = c/sinC b/sin(62°) = 50/sin(70°) b = (50 * sin(62°)) / sin(70°) b ≈ 46.9 m ``` Therefore, the distance AC across the river is approximately 46.9 meters. 2.2 The Cosine Rule: The cosine rule relates the lengths of the sides of a triangle to the cosine of one of its angles.
There are three forms of the Cosine Rule: ``` a² = b² + c² - 2bc*cosA b² = a² + c² - 2ac*cosB c² = a² + b² - 2ab*cosC ``` When to use the Cosine Rule: You can use the cosine rule when you know: Two sides and the included angle (SAS). Three sides (SSS).
Example 2: Two hikers start at the same point in the Drakensberg mountains. One hiker walks 5 km on a bearing of 060°, and the other walks 4 km on a bearing of 340°. Calculate the distance between the two hikers.
Solution: Draw a diagram (Showing the bearings and the hikers' paths).
Identify the knowns: Let hiker 1's position be A, hiker 2's position be B, and the starting point be C. Then, AC = 5 km, BC = 4 km. The angle ACB is the difference between 360° - 340° and 60°, which is 80°.
Find the unknown: AB (let's call it 'c').
Apply the Cosine Rule: ``` c² = a² + b² - 2ab*cosC c² = 4² + 5² - 2 4 5 * cos(80°) c² = 16 + 25 - 40 * cos(80°) c² ≈ 34.06 c ≈ √34.06 ≈ 5.84 km ``` Therefore, the distance between the two hikers is approximately 5.84 km.
Example 3 (Finding an angle): A triangular garden plot has sides of length 7m, 8m, and 10m. Calculate the size of the largest angle.
Solution: The largest angle will be opposite the longest side, which is 10m. Let's call this angle
C. Using the cosine rule: ``` c² = a² + b² - 2ab*cosC 10² = 7² + 8² - 278*cosC 100 = 49 + 64 - 112*cosC 100 = 113 - 112*cosC -13 = -112*cosC cosC = 13/112 C = arccos(13/112) C ≈ 83.33° ``` Therefore, the largest angle is approximately 83.33 degrees. 2.3 The Area Rule: The area rule allows you to calculate the area of a triangle when you know two sides and the included angle. ``` Area = (1/2) ab sinC Area = (1/2) bc sinA Area = (1/2) ac sinB ``` Where: a, b, c are the lengths of the sides. A, B, C are the angles opposite sides a, b, c, respectively.
Example 4: A farmer in the Western Cape wants to calculate the area of a triangular piece of land. He measures two sides to be 120 meters and 150 meters, and the angle between them is 75°. Calculate the area of the land.
Solution: Identify the knowns: a = 120m, b = 150m, C = 75°.
Apply the Area Rule: ``` Area = (1/2) ab sinC Area = (1/2) 120 150 * sin(75°) Area ≈ 8693.30 m² ``` Therefore, the area of the land is approximately 8693.30 square meters. Guided Practice (With Solutions)
Question 1: In triangle PQR, PQ = 8 cm, QR = 6 cm, and angle PQR = 50°. Calculate the length of P
R. Solution: Identify knowns: PQ = 8 cm, QR = 6 cm, ∠PQR = 50°.
Identify unknown: PR (let's call it 'p').
Apply Cosine Rule: Since we know two sides and the included angle, we use the cosine rule. ``` p² = q² + r² - 2qr*cosP p² = 6² + 8² - 2 6 8 * cos(50°) p² = 36 + 64 - 96 * cos(50°) p² ≈ 38.28 p ≈ √38.28 ≈ 6.19 cm ```
Commentary: The cosine rule is appropriate here because we have two sides and the included angle (SAS). We square root the final result to get the actual length of P
R. Question 2: In triangle ABC, angle A = 105°, angle B = 35°, and side c = 12 cm. Find the length of side a.
Solution: Identify knowns: ∠A = 105°, ∠B = 35°, c = 12 cm.
Identify unknown: a.
Find missing angle: ∠C = 180° - 105° - 35° = 40°.
Apply Sine Rule: ``` a/sinA = c/sinC a/sin(105°) = 12/sin(40°) a = (12 * sin(105°)) / sin(40°) a ≈ 18.06 cm ```
Commentary: The sine rule is the correct approach as we have two angles and one side. First, determine all the angles, then apply the Sine rule.
Question 3: Calculate the area of triangle XYZ if XY = 15 cm, YZ = 10 cm, and angle XYZ = 65°.
Solution: Identify knowns: XY = 15 cm, YZ = 10 cm, ∠XYZ = 65°.