University:
The University Of Western Ontario
Course:
MATH 3020A | Introduction to Abstract Algebra
Academic year:

2024
Views:

413

Pages:

4
Author:

Kyuwon J.

Triangle Ambiguity Cases Notes: Solving Triangles Using Law of Sines 1. The Law of Sines states: a b c = = sin(A) sin(B) sin(C) 2. **Ambiguity Case** (SSA triangles): When two sides and a non-included angle are given, there may be: • No solution (if sin(B) > 1). • One solution (if ∠B = 90◦ ). • Two solutions (if sin(B) is valid and the triangle can be formed in two ways). 3. To resolve ambiguity: • Compute sin(B). • Use inverse sine to find ∠B1 = arcsin(sin(B)). • If applicable, find ∠B2 = 180◦ − ∠B1 . 4. Use the sum of angles property: ∠C = 180◦ − ∠A − ∠B 5. Finally, use the Law of Sines to solve for the remaining side: b= a · sin(B) . sin(A) — Problem 1: Solve △ABC Question: Given a = 5.9 m, b = 7.8 m, ∠A = 36◦ , find the length of side c. 1 Solution: Using the Law of Cosines c2 = a2 + b2 − 2ab cos(∠A) Substitute: c2 = 5.92 + 7.82 − 2(5.9)(7.8) cos(36◦ ) c2 ≈ 34.81 + 60.84 − 2(5.9)(7.8)(0.809) c2 ≈ 95.65 − 74.53 ≈ 21.12 √ c ≈ 21.12 ≈ 4.60 m. Solution: Using the Law of Sines Question: Use the Law of Sines to verify the length of side c. Using the Law of Sines: c a = sin(A) sin(C) First, find ∠B: b · sin(A) 7.8 · sin(36◦ ) = a 5.9 7.8 · 0.5878 ≈ 0.7773 sin(B) ≈ 5.9 ∠B = arcsin(0.7773) ≈ 50.99◦ sin(B) = Now, ∠C = 180◦ − ∠A − ∠B: ∠C ≈ 180◦ − 36◦ − 50.99◦ ≈ 93.01◦ Using the Law of Sines: c= 5.9 · sin(93.01◦ ) a · sin(C) = sin(A) sin(36◦ ) c≈ 5.9 · 1 ≈ 10.02 m. 0.5878 — Problem 2: Ambiguous Case for △ABC Question: Given a = 2.4 cm, c = 3.2 cm, ∠A = 28◦ , determine two possible measures for ∠C and the length of side b. 2 Solution 1. Use the Law of Sines to find sin(C): sin(C) = 3.2 · sin(28◦ ) c · sin(A) = a 2.4 sin(C) ≈ 2. Solve for ∠C: 3.2 · 0.4695 ≈ 0.626 2.4 ∠C1 = arcsin(0.626) ≈ 38.75◦ ∠C2 = 180◦ − ∠C1 = 141.25◦ 3. Solve for ∠B: ∠B1 = 180◦ − ∠A − ∠C1 ≈ 113.25◦ ∠B2 = 180◦ − ∠A − ∠C2 ≈ 10.75◦ 4. Solve for b using the Law of Sines: b= For ∠B1 : a · sin(B) sin(A) b1 = 2.4 · sin(113.25◦ ) ≈ 4.70 cm sin(28◦ ) b2 = 2.4 · sin(10.75◦ ) ≈ 0.95 cm sin(28◦ ) For ∠B2 : — Problem 3: Ambiguity in △DEF Question: Given d = 3 cm, e = 5 cm, ∠D = 30◦ , determine two possible measures for ∠E and the length of side f . Solution 1. Use the Law of Sines to find sin(E): sin(E) = 2. Solve for ∠E: e · sin(D) 5 · 0.5 = ≈ 0.833 d 3 ∠E1 = arcsin(0.833) ≈ 56.02◦ ∠E2 = 180◦ − ∠E1 ≈ 123.98◦ 3. Solve for ∠F : ∠F1 = 180◦ − ∠D − ∠E1 ≈ 93.98◦ ∠F2 = 180◦ − ∠D − ∠E2 ≈ 26.02◦ 3 4. Solve for f : For ∠F1 : f1 = For ∠F2 : f2 = 3·1 d · sin(F1 ) ≈ = 6 cm sin(D) 0.5 3 · sin(26.02◦ ) 3 · 0.438 ≈ ≈ 2.63 cm ◦ sin(30 ) 0.5 — 4