Lesson Notes By Weeks and Term v5 - Grade 11
Trigonometry – Week 10 focus
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Trigonometry – Week 10 focus
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Subject: Mathematics
Class: Grade 11
Term: 2nd Term
Week: 10
Theme: General lesson support
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This week, we delve deeper into trigonometry, focusing on solving two-dimensional problems using trigonometric ratios and rules (Sine, Cosine, and Area rules). This builds upon your understanding of right-angled trigonometry from previous grades and introduces powerful tools to tackle non-right-angled triangles. These concepts are crucial not only for further studies in mathematics and related fields like engineering and physics but also have practical applications in surveying, navigation, and architecture. Imagine being able to calculate the height of a building or the distance across a river without physically measuring it – trigonometry makes this possible.
2.1 The Sine Rule The Sine Rule states the relationship between the sides of a triangle and the sines of their opposite angles. For any triangle ABC, with sides a, b, and c opposite angles A, B, and C respectively: a / sin A = b / sin B = c / sin C Alternatively, this can be written as: sin A / a = sin B / b = sin C / c The Sine Rule is useful when you have: Two angles and one side (AAS or ASA) Two sides and a non-included angle (SSA) - be aware of the ambiguous case (we will cover this later)* Example 1: In triangle PQR, PQ = 8cm, angle P = 50°, and angle R = 72°. Calculate the length of side Q
R. Solution: We are given two angles and a side, so we can use the Sine Rule. Let QR = p. p / sin P = r / sin R p / sin 50° = 8 / sin 72° p = (8 sin 50°) / sin 72°* p ≈ 6.43 cm Therefore, QR ≈ 6.43 cm.
Example 2 (Ambiguous Case): In triangle ABC, AB = 12cm, BC = 8cm, and angle A = 30°. Find the possible values of angle
C. Solution: We are given two sides and a non-included angle (SSA), so we can use the Sine Rule. sin C / AB = sin A / BC sin C / 12 = sin 30° / 8 sin C = (12 sin 30°) / 8 sin C = 0.75 C = arcsin(0.75) C ≈ 48.59° However, since sin(180° - x) = sin(x), there is another possible solution: C = 180° - 48.59° C ≈ 131.41° We need to check if both solutions are valid. If C ≈ 131.41°, then B ≈ 180° - 30° - 131.41° ≈ 18.59°. This is a valid angle, so both solutions are possible.
Therefore, angle C ≈ 48.59° or angle C ≈ 131.41°. The ambiguous case arises because with SSA, two different triangles can be formed with the given information. 2.2 The Cosine Rule The Cosine Rule relates the sides and angles of a triangle. For any triangle ABC, with sides a, b, and c opposite angles A, B, and C respectively: a² = b² + c² - 2bc cos A b² = a² + c² - 2ac cos B c² = a² + b² - 2ab cos C The Cosine Rule is useful when you have: Two sides and the included angle (SAS) Three sides (SSS) You can also rearrange the Cosine Rule to solve for an angle: cos A = (b² + c² - a²) / 2bc cos B = (a² + c² - b²) / 2ac cos C = (a² + b² - c²) / 2ab Example 3: In triangle XYZ, XY = 7cm, YZ = 9cm, and angle Y = 40°. Calculate the length of side X
Z. Solution: We are given two sides and the included angle, so we can use the Cosine Rule. Let XZ = x. x² = 7² + 9² - 2 7 9 cos 40°* x² = 49 + 81 - 126 cos 40°* x² ≈ 33.28 x ≈ √33.28 x ≈ 5.77 cm Therefore, XZ ≈ 5.77 cm.
Example 4: In triangle DEF, DE = 5cm, EF = 7cm, and DF = 8cm. Calculate the size of angle
D. Solution: We are given three sides, so we can use the Cosine Rule to find the angle. cos D = (5² + 8² - 7²) / (2 5 8) cos D = (25 + 64 - 49) / 80 cos D = 40 / 80 cos D = 0.5 D = arccos(0.5) D = 60° Therefore, angle D = 60°. 2.3 The Area Rule The area of any triangle can be calculated using the following formula if you know two sides and the included angle: Area = (1/2) ab sin C = (1/2) bc sin A = (1/2) ac sin B Example 5: In triangle KLM, KL = 10cm, KM = 12cm, and angle K = 60°. Calculate the area of triangle KL
M. Solution: We are given two sides and the included angle, so we can use the Area Rule. Area = (1/2) KL KM sin K Area = (1/2) 10 12 sin 60°* Area = 60 (√3 / 2)* Area ≈ 51.96 cm² Therefore, the area of triangle KLM ≈ 51.96 cm². 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 a bird in a tree.
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 from a cliff at a boat in the sea. Important
Note: Problems involving angles of elevation and depression often require you to visualize right-angled triangles and use the basic trigonometric ratios (SOH CAH TOA) in addition to the Sine, Cosine, and Area rules. Guided Practice (With Solutions)
Question 1: In triangle ABC, angle A = 35°, angle B = 68°, and side c = 15cm. Calculate the length of side a.
Solution: We have two angles and a side (AAS).
We can use the Sine Rule: a / sin A = c / sin C First, find angle C: C = 180° - A - B = 180° - 35° - 68° = 77° Now apply the Sine Rule: a / sin 35° = 15 / sin 77° a = (15 sin 35°) / sin 77°* a ≈ 8.81 cm Therefore, the length of side a is approximately 8.81 cm.
Question 2: In triangle PQR, p = 7cm, q = 10cm, and angle R = 55°. Calculate the length of side r.
Solution: We have two sides and the included angle (SAS).
We can use the Cosine Rule: r² = p² + q² - 2pq cos R r² = 7² + 10² - 2 7 10 cos 55°* r² = 49 + 100 - 140 cos 55°* r² ≈ 68.64 r ≈ √68.64 r ≈ 8.28 cm Therefore, the length of side r is approximately 8.28 cm.
Question 3: A flagpole is standing vertically on level ground. From a point 25 meters away from the base of the flagpole, the angle of elevation to the top of the flagpole is 32°. Calculate the height of the flagpole.
Solution: This is a right-angled triangle problem. Let the height of the flagpole be h.