Lesson Notes By Weeks and Term v5 - Grade 11
Trigonometry (sine, cosine and area rules) – Week 5 focus
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Trigonometry (sine, cosine and area rules) – Week 5 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 11
Term: 3rd Term
Week: 5
Theme: General lesson support
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This week, we delve deeper into trigonometry, specifically focusing on the Sine, Cosine, and Area Rules. These rules are essential tools for solving problems involving non-right-angled triangles. They extend our trigonometric knowledge beyond the familiar SOH CAH TOA, allowing us to calculate unknown sides and angles in any triangle, a skill crucial in various fields. Think about land surveying, construction, navigation, and even sports (calculating angles of trajectories) – these rules underpin many practical applications. In the South African context, these skills are vital for infrastructure development, land management, and ensuring accuracy in various professions.
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. Mathematically, it is expressed as: a / sin(A) = b / sin(B) = c / sin(C)
Where: a, b, and c are the lengths of the sides of the triangle. A, B, and C are the angles opposite to sides a, b, and c, respectively.
When to use the Sine Rule: The Sine Rule is most useful when you have: Two angles and one side (AAS or ASA) Two sides and a non-included angle (SSA) – Be cautious of the ambiguous case here!
Example 1: Finding a side In triangle PQR, angle P = 48°, angle Q = 72°, and side r = 15 cm. Find the length of side p.
Solution: Identify what you have: We have two angles (P and Q) and one side (r). We want to find side p.
Apply the Sine Rule: p / sin(P) = r / sin(R)
Find angle R: Since the angles in a triangle add up to 180°, R = 180° - 48° - 72° = 60° Substitute and solve: p / sin(48°) = 15 / sin(60°) p = (15 * sin(48°)) / sin(60°) p ≈ (15 * 0.7431) / 0.8660 p ≈ 12.86 cm Example 2: Finding an angle In triangle ABC, side a = 10 m, side b = 7 m, and angle A = 65°. Find the measure of angle
B. Solution: Identify what you have: We have two sides (a and b) and a non-included angle (A). We want to find angle
B. Apply the Sine Rule: a / sin(A) = b / sin(B)
Substitute and solve: 10 / sin(65°) = 7 / sin(B) sin(B) = (7 * sin(65°)) / 10 sin(B) ≈ (7 * 0.9063) / 10 sin(B) ≈ 0.6344 Find angle B: B = arcsin(0.6344) B ≈ 39.38° The Ambiguous Case (SSA): When using the Sine Rule with two sides and a non-included angle, there might be two possible triangles, one triangle, or no triangle at all. Always consider the possibility of two solutions and check if both solutions are valid (i.e., all angles are positive and sum to 180°). 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. It's extremely useful when you can't use the Sine Rule.
Forms of the Cosine Rule: a² = b² + c² - 2bc cos(A) b² = a² + c² - 2ac cos(B) c² = a² + b² - 2ab cos(C)
When to use the Cosine Rule: The Cosine Rule is most useful when you have: Three sides (SSS) and want to find an angle. Two sides and the included angle (SAS) and want to find the third side.
Example 3: Finding a side In triangle XYZ, side x = 8 cm, side y = 5 cm, and angle Z = 43°. Find the length of side z.
Solution: Identify what you have: We have two sides (x and y) and the included angle (Z). We want to find side z.
Apply the Cosine Rule: z² = x² + y² - 2xy * cos(Z)
Substitute and solve: z² = 8² + 5² - 2 8 5 * cos(43°) z² = 64 + 25 - 80 * 0.7314 z² = 89 - 58.512 z² = 30.488 Find side z: z = √30.488 z ≈ 5.52 cm Example 4: Finding an angle In triangle DEF, side d = 7 m, side e = 9 m, and side f = 5 m. Find the measure of angle
E. Solution: Identify what you have: We have three sides (d, e, and f). We want to find angle
E. Apply the Cosine Rule (rearranged to solve for cos(E)): cos(E) = (d² + f² - e²) / (2df)
Substitute and solve: cos(E) = (7² + 5² - 9²) / (2 7 5) cos(E) = (49 + 25 - 81) / 70 cos(E) = -7 / 70 cos(E) = -0.1 Find angle E: E = arccos(-0.1) E ≈ 95.74° 2.3 The Area Rule The area of any triangle can be calculated using the following formula: Area = ½ a b sin(C) = ½ b c sin(A) = ½ a c * sin(B)
Where: a, b, and c are the lengths of the sides of the triangle. A, B, and C are the angles opposite to sides a, b, and c, respectively. This formula is particularly useful when you know two sides and the included angle.
Example 5: Finding the area In triangle KLM, side k = 12 cm, side l = 10 cm, and angle M = 55°. Find the area of triangle KL
M. Solution: Identify what you have: We have two sides (k and l) and the included angle (M). We want to find the area.
Apply the Area Rule: Area = ½ k l * sin(M)
Substitute and solve: Area = ½ 12 10 * sin(55°) Area = 60 * 0.8192 Area ≈ 49.15 cm² Guided Practice (With Solutions)
Question 1: In triangle ABC, angle A = 30°, angle B = 70°, and side a = 8 cm. Find the length of side b.
Solution: Identify: We have two angles (A and B) and one side (a). We need to find side b. This is suitable for the Sine Rule.
Sine Rule: a / sin(A) = b / sin(B)
Substitute: 8 / sin(30°) = b / sin(70°)
Solve for b: b = (8 sin(70°)) / sin(30°) = (8 0.9397) / 0.5 = 15.0352 cm Answer: b ≈ 15.04 cm Question 2: In triangle PQR, p = 7cm, q = 10cm and angle R = 50°. Find the length of side r.
Solution: Identify: We have two sides (p and q) and the included angle (R). We need to find side r. This is perfect for the Cosine Rule.
Cosine Rule: r² = p² + q² - 2pq*cos(R)
Substitute: r² = 7² + 10² - 2(7)(10)cos(50°) = 49 + 100 - 140 0.6428 = 149 - 89.992 = 59.008 Solve for r: r = √59.008 ≈ 7.68 cm Answer: r ≈ 7.68 cm Question 3: Calculate the area of triangle XYZ if x = 5 cm, y = 8 cm, and angle Z = 60°.
Solution: Identify: Two sides and the included angle. We need to use the area rule.