Lesson 1: The Law of Sines and the Ambiguous Case
Estimated time: 35-40 minutes
Learning Objectives
- State the Law of Sines
- Solve AAS and ASA triangles
- Recognize and handle the ambiguous case (SSA)
- Determine when SSA produces 0, 1, or 2 triangles
The Law of Sines
Law of Sines: In any triangle with sides a, b, c opposite angles A, B, C:
a/sinA = b/sinB = c/sinC
Use the Law of Sines when you know: (1) two angles and one side (AAS or ASA), or (2) two sides and an angle opposite one of them (SSA).
Solving AAS/ASA Triangles
Example 1: AAS
A = 40°, B = 60°, a = 10. Find b and c.
Step 1: C = 180° − 40° − 60° = 80°
Step 2: a/sinA = b/sinB ⇒ 10/sin40 = b/sin60 ⇒ b = 10 sin60/sin40 ≈ 13.47
Step 3: c = 10 sin80/sin40 ≈ 15.32
The Ambiguous Case (SSA)
Given two sides and an angle opposite one of them (SSA), there may be 0, 1, or 2 possible triangles. This is because the sine function can give two possible angles (since sinθ = sin(180°−θ)).
SSA possibilities:
Given a, b, and angle A (a is opposite A):
If A is acute: compare a to b sinA (the height h):
- a < h: no triangle
- a = h: one right triangle
- h < a < b: two triangles
- a ≥ b: one triangle
If A is obtuse: one triangle if a > b, no triangle otherwise.
Example 2: SSA — Two Triangles
A = 30°, a = 6, b = 10. Solve.
sinB/b = sinA/a ⇒ sinB = 10 sin30/6 = 10(0.5)/6 = 5/6 ≈ 0.8333
B = arcsin(0.8333) ≈ 56.44° or B ≈ 180° − 56.44° = 123.56°
Both are valid since A + B < 180° in each case. So two triangles.
Example 3: SSA — No Triangle
A = 50°, a = 3, b = 10.
sinB = 10 sin50/3 ≈ 2.55 > 1. Since sine cannot exceed 1, no triangle exists.
Check Your Understanding
1. In a triangle with A=35°, C=85°, b=12, find B and then a.
2. A=40°, a=5, b=8. How many triangles?
3. When does SSA give exactly one triangle?
Key Takeaways
- Law of Sines: a/sinA = b/sinB = c/sinC.
- Use for AAS, ASA (always one solution) and SSA (ambiguous).
- The ambiguous case can yield 0, 1, or 2 triangles.
- Always check if sinB > 1 (no solution) or if both B and 180−B are valid.