Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Apply the Sine Rule for AAS, ASA, and SSA cases.
• Apply the Cosine Rule for SSS and SAS cases.
• Apply the Area Rule to find the area of a triangle.
• Solve 2D problems using these rules.
• Choose the appropriate rule for any given problem.

Lesson summary

This week, we delve into the power of Trigonometry by exploring the Sine Rule, the Cosine Rule, and the Area Rule. These rules are crucial for solving problems involving triangles that are not necessarily right-angled triangles. Unlike the trigonometric ratios (sin, cos, tan) that you learned earlier, which are only applicable to right-angled triangles, these rules provide us with tools to analyze and solve any triangle, given sufficient information. Why is this important? Imagine you're a land surveyor in KwaZulu-Natal trying to determine the area of a plot of land that's not a perfect rectangle.

Lesson notes

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 its opposite angle is constant: a / sin A = b / sin B = c / sin C Alternatively, it can be written as: sin A / a = sin B / b = sin C / c When to use the Sine Rule: AAS (Angle-Angle-Side): When you know two angles and a side opposite one of them.

ASA (Angle-Side-Angle): When you know two angles and the included side. (You can easily find the third angle since the angles of a triangle add up to 180°.)

SSA (Side-Side-Angle): This is the ambiguous case! It can lead to zero, one, or two possible solutions. Be careful and always check for a second possible triangle.

Example 1 (AAS): In triangle ABC, angle A = 40°, angle B = 60°, and side a = 8 cm. Find the length of side b.

Solution: Apply the Sine Rule: a / sin A = b / sin B Substitute the given values: 8 / sin 40° = b / sin 60° Solve for b: b = (8 sin 60°) / sin 40°* Calculate: b ≈ 10.77 cm Example 2 (SSA - Ambiguous Case): In triangle ABC, side a = 10 cm, side b = 12 cm, and angle A = 30°. Find angle

B. Solution: Apply the Sine Rule: sin A / a = sin B / b Substitute the given values: sin 30° / 10 = sin B / 12 Solve for sin B: sin B = (12 sin 30°) / 10 = 0.6 Find angle B: B = sin -1 (0.6) ≈ 36.87° But wait! Since sin(180° - x) = sin(x), there's another possible angle B: B' = 180° - 36.87° ≈ 143.13° Now we must check if this second angle is valid. If B' + A 2 = b 2 + c 2 - 2bc cos A b 2 = a 2 + c 2 - 2ac cos B c 2 = a 2 + b 2 - 2ab cos C When to use the Cosine Rule: SSS (Side-Side-Side): When you know the lengths of all three sides and want to find an angle.

SAS (Side-Angle-Side): When you know two sides and the included angle and want to find the third side.

Example 3 (SAS): In triangle ABC, side b = 5 cm, side c = 8 cm, and angle A = 77°. Find the length of side a.

Solution: Apply the Cosine Rule: a 2 = b 2 + c 2 - 2bc cos A Substitute the given values: a 2 = 5 2 + 8 2 - 2(5)(8) cos 77° Calculate: a 2 ≈ 25 + 64 - 80 0.2250 ≈ 71 Solve for a: a ≈ √71 ≈ 8.43 cm Example 4 (SSS): In triangle ABC, side a = 7 m, side b = 9 m, and side c = 5 m. Find the angle

C. Solution: Apply the Cosine Rule (solving for cos C): c 2 = a 2 + b 2 - 2ab cos C => cos C = (a 2 + b 2 - c 2 ) / 2ab Substitute the given values: cos C = (7 2 + 9 2 - 5 2 ) / (2 7 9)

Calculate: cos C = (49 + 81 - 25) / 126 = 105 / 126 ≈ 0.8333 Solve for C: C = cos -1 (0.8333) ≈ 33.56° 2.3 The Area Rule The Area Rule provides a way to calculate the area of a triangle when you know two sides and the included angle: Area = (1/2) ab sin C Area = (1/2) ac sin B Area = (1/2) bc sin A When to use the Area Rule: When you know two sides and the included angle (SAS).

Example 5: In triangle ABC, side a = 12 cm, side b = 15 cm, and angle C = 50°. Find the area of the triangle.

Solution: Apply the Area Rule: Area = (1/2) ab sin C Substitute the given values: Area = (1/2) 12 15 sin 50°* Calculate: Area ≈ 90 0.7660 ≈ 68.94 cm 2 * Guided Practice (With Solutions)

Question 1: In triangle PQR, angle P = 110°, angle Q = 35°, and side r = 15 cm. Find the length of side p.

Solution: Identify the given information: AAS (Angle-Angle-Side). This indicates the Sine Rule.

Find angle R: R = 180° - P - Q = 180° - 110° - 35° = 35° Apply the Sine Rule: p / sin P = r / sin R Substitute values: p / sin 110° = 15 / sin 35° Solve for p: p = (15 sin 110°) / sin 35° ≈ (15 0.9397) / 0.5736 ≈ 24.57 cm Question 2: In triangle XYZ, side x = 8 m, side y = 5 m, and angle Z = 60°. Find the length of side z.

Solution: Identify the given information: SAS (Side-Angle-Side). This indicates the Cosine Rule.

Apply the Cosine Rule: z 2 = x 2 + y 2 - 2xy cos Z Substitute values: z 2 = 8 2 + 5 2 - 2(8)(5) cos 60° = 64 + 25 - 80 0.5 = 49 Solve for z: z = √49 = 7 m Question 3: In triangle ABC, side a = 9 cm, side b = 7 cm, and angle C = 40°. Find the area of triangle AB

C. Solution: Identify the given information: SAS (Side-Angle-Side). This indicates the Area Rule.

Apply the Area Rule: Area = (1/2) ab sin C Substitute values: Area = (1/2) 9 7 sin 40° ≈ 31.5 0.6428 ≈ 20.25 cm 2 Question 4: In triangle KLM, side k = 13m, side l = 14m and side m = 15m. Calculate the size of angle

K. Solution Identify the given information: SSS (Side-Side-Side). This indicates the Cosine Rule.

Apply the Cosine Rule (solving for cos K): k 2 = l 2 + m 2 - 2lm cos K => cos K = (l 2 + m 2 - k 2 ) / 2lm Substitute values: cos K = (14 2 + 15 2 - 13 2 ) / (2 14 15) = (196 + 225 - 169) / 420 = 252/420 = 0.6 Solve for K: K = cos -1 (0.6) ≈ 53.13° Independent Practice (Questions Only) In triangle DEF, angle D = 65°, angle E = 45°, and side f = 12 cm. Find the length of side d. In triangle GHI, side g = 10 m, side h = 7 m, and angle I = 30°. Find the length of side i. In triangle JKL, side j = 5 km, side k = 8 km, and angle L = 120°. Find the area of triangle JKL. In triangle MNO, side m = 6 cm, side n = 8 cm, and side o = 10 cm. Find the size of angle M.