Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Apply the Sine Rule to find unknown sides and angles.
• Apply the Cosine Rule to find unknown sides and angles.
• Calculate the area of any triangle using the Area Rule.
• Choose the correct rule to solve 2D problems.

Lesson summary

This week, we delve deeper into the world of Trigonometry, building upon the foundational knowledge you gained previously. Our focus shifts to understanding and applying the Sine, Cosine, and Area rules in solving two-dimensional problems. Trigonometry is not just abstract mathematics; it's a powerful tool used in various fields, from surveying and navigation to architecture and engineering. Imagine town planners using trigonometry to design optimal road layouts or architects calculating roof angles for weather resistance and aesthetic appeal. Even mobile phone signal strength calculations rely on trigonometric principles!

Lesson notes

2.1 The Sine Rule: The Sine Rule establishes a relationship between the sides of a triangle and the sines of their opposite angles.

It is particularly useful when you know: Two angles and one side (AAS or ASA) Two sides and an angle opposite one of them (SSA - ambiguous case, discussed later)

The Sine Rule states: `a / sin A = b / sin B = c / sin C` Where: a, b, and c are the side lengths of the triangle. A, B, and C are the angles opposite sides a, b, and c, respectively. Why does it work? The Sine Rule arises from the fact that the ratio of a side length to the sine of its opposite angle is constant for any triangle. This can be proven by drawing the height of the triangle and using basic trigonometric ratios (SOH CAH TOA) in the resulting right-angled triangles.

Example 1: A farmer wants to fence a triangular piece of land. He knows one side is 150m long, and the angles at either end of that side are 55° and 70°. How much fencing will he need in total?

Solution: Draw a diagram: (Crucial for visualization!)

Identify givens: Let a = 150m, A = 55°, B = 70°. We need to find b and c.

We can also find C: C = 180° - 55° - 70° = 55°.

Apply the Sine Rule: `150 / sin 55° = b / sin 70°` `b = (150 sin 70°) / sin 55°` `b ≈ 169.7 m` Apply Sine rule again: `150 / sin 55° = c / sin 55°` Since sin 55 = sin 55, therefore c = 150 m.

Calculate the total fencing: 150m + 169.7m + 150m = 469.7m The Ambiguous Case (SSA): When given two sides and a non-included angle, there might be zero, one, or two possible triangles. Always consider this possibility when using the Sine Rule in this scenario! To determine the number of possible solutions, carefully sketch the triangle and analyze the height. 2.2 The Cosine Rule: The Cosine Rule relates the sides and angles of a triangle and is particularly useful when you know: Two sides and the included angle (SAS) Three sides (SSS)

The Cosine Rule comes in three forms: `a² = b² + c² - 2bc cos A` `b² = a² + c² - 2ac cos B` `c² = a² + b² - 2ab cos C` Why does it work? The Cosine Rule is a generalization of the Pythagorean theorem. When angle A (or B, or C) is 90°, cos A = 0, and the formula reduces to the Pythagorean theorem. The `- 2bc cos A` term accounts for the "slant" of the triangle when it is not a right triangle.

Example 2: A surveyor measures the distance between two points, A and B, as 200m and 250m, respectively, from a third point

C. The angle at point C is 60°. What is the distance between points A and B?

Solution: Draw a diagram.

Identify givens: a = 250m, b = 200m, C = 60°. We need to find c.

Apply the Cosine Rule: `c² = a² + b² - 2ab cos C` `c² = 250² + 200² - 2 250 200 cos 60°` `c² = 62500 + 40000 - 50000` `c² = 52500` `c = √52500 ≈ 229.1 m` Example 3: The side lengths of a triangle are 7cm, 8cm and 5cm. Determine the size of the largest angle.

Solution: Draw a diagram. Identify the largest angle will be opposite the longest side i.e 8 cm.

Therefore, we need to find angle

B. Apply the cosine rule: b² = a² + c² - 2acCosB 8² = 7² + 5² - 2(7)(5)CosB 64 = 49 + 25 -70CosB -10 = -70CosB CosB = 1/7 B = Cos⁻¹(1/7) B = 81.79° 2.3 Area of a Triangle: The area of a triangle can be calculated using the formula: `Area = (1/2)ab sin C` Where: a and b are the lengths of two sides of the triangle. C is the angle included between sides a and b. Why does it work? This formula is derived from the standard formula for the area of a triangle (1/2 base height) by expressing the height in terms of one of the sides and the included angle using the sine function.

Example 4: A triangular garden bed has sides of length 8m and 10m, with an angle of 75° between them. What is the area of the garden bed?

Solution: Identify givens: a = 8m, b = 10m, C = 75°.

Apply the Area Formula: `Area = (1/2) 8 10 sin 75°` `Area ≈ 38.64 m²` Guided Practice (With Solutions)

Question 1: In triangle PQR, PQ = 12cm, PR = 8cm, and angle QPR = 40°. Calculate the area of triangle PQ

R. Solution: Identify the appropriate formula: We have two sides and the included angle, so we use the area formula: Area = (1/2)ab sin

C. Substitute values: Area = (1/2) 12 8 sin 40° Calculate: Area ≈ 30.86 cm² Question 2: In triangle ABC, AB = 5cm, BC = 7cm, and angle BAC = 60°. Calculate the length of A

C. Solution: Identify the appropriate formula: We have two sides and a non-included angle. We need to find the side opposite the known angle, therefore we use Sine Rule to find angle

C. Applying sine rule: 7/sin 60 = 5/sin C sinC = (5sin 60) /7 C = sin⁻¹((5sin 60) /7) C= 38.21° Now we can calculate angle B as follows: B = 180 - (60 + 38.21) B = 81.79° Apply sine rule again to find AC (b): 7/sin 60 = b/sin 81.79 b = (7sin 81.79) / sin 60 b = 8 cm (AC)

Question 3: In triangle XYZ, XY = 10m, YZ = 15m, and XZ = 12m. Calculate the size of the largest angle.

Solution: Identify the appropriate formula: We have three sides, so we use the Cosine Rule. The largest angle is opposite the longest side (YZ).

Therefore, we will calculate angle X.