8.1Solving Oblique Triangles with the Law of Sines
Apply a/sinA = b/sinB = c/sinC to solve oblique triangles in AAS and ASA cases. Analyze the ambiguous SSA case (0, 1, or 2 solutions) and find triangle area using A = ½ab sinC.
The Law of Sines extends triangle solving beyond right triangles. It is used in navigation, surveying, and physics to find distances and angles when a right angle is not present.
Essential Question
How do you find all missing sides and angles of a triangle when no right angle is present, and how do you handle the case where two solutions might exist?
Lesson Overview
Law of Sines
a / sin A = b / sin B = c / sin C
Valid for any triangle (oblique or right). Each ratio equals the circumdiameter 2R.
When to use it
- AAS — two angles and a non-included side are known
- ASA — two angles and the included side are known
- SSA — two sides and a non-included angle (ambiguous — may give 0, 1, or 2 triangles)
Area Formula
Area = ½ · a · b · sin C
Works for any triangle; C is the angle included between sides a and b.
SSA Ambiguous Case — Decision Tree
Given angle A (opposite side a) and adjacent side b, compute h = b sin A:
If A ≥ 90°:
a > b → 1 solution
a ≤ b → 0 solutions
If A < 90°:
a < h → 0 solutions
a = h → 1 solution (right triangle)
h < a < b → 2 solutions
a ≥ b → 1 solution
Law of Sines — Labeled Triangle
a / sin A = b / sin B = c / sin C
SSA Ambiguous Case — 0, 1, or 2 Triangles
Key Vocabulary
Oblique triangle
A triangle with no right angle (all angles are acute or one is obtuse).
Example: A triangle with angles 35°, 65°, 80° is oblique.
Law of Sines
The ratio of each side to the sine of its opposite angle is constant: a/sinA = b/sinB = c/sinC.
Example: If a = 12, A = 35°, B = 65°, then b = 12·sin65°/sin35°.
AAS
Angle-Angle-Side: two angles and a non-included side are given.
Example: A = 35°, B = 65°, a = 12.
ASA
Angle-Side-Angle: two angles and the included side are given.
Example: A = 50°, C = 70°, b = 20.
SSA
Side-Side-Angle: two sides and a non-included angle are given. This is the ambiguous case.
Example: A = 30°, a = 8, b = 12.
Ambiguous case
The SSA configuration where 0, 1, or 2 triangles may satisfy the given conditions.
Example: A = 30°, a = 8, b = 12 yields two valid triangles.
Included angle
The angle formed between two given sides.
Example: In ASA, the side between angles A and C is the included side; C is included between sides a and b in the area formula.
Area formula (½ab sinC)
Area = ½ · a · b · sin C, where C is the included angle between sides a and b.
Example: a = 15, b = 22, C = 48° → Area ≈ 122.3 sq units.
Worked Examples
AAS case — Triangle with A = 35°, B = 65°, a = 12. Find b and c.
Step 1 — Find the third angle: C = 180° − 35° − 65° = 80°
Step 2 — Apply Law of Sines to find b:
b / sin B = a / sin A
b / sin 65° = 12 / sin 35°
b = 12 · sin 65° / sin 35° ≈ 12 · 0.9063 / 0.5736 ≈ 18.96
Step 3 — Apply Law of Sines to find c:
c / sin C = a / sin A
c / sin 80° = 12 / sin 35°
c = 12 · sin 80° / sin 35° ≈ 12 · 0.9848 / 0.5736 ≈ 20.60
ASA case — Triangle with A = 50°, C = 70°, b = 20. Find a, c, and the area.
Step 1 — Find the third angle: B = 180° − 50° − 70° = 60°
Step 2 — Find a using Law of Sines:
a / sin A = b / sin B
a / sin 50° = 20 / sin 60°
a = 20 · sin 50° / sin 60° ≈ 20 · 0.7660 / 0.8660 ≈ 17.69
Step 3 — Find c:
c / sin C = b / sin B
c / sin 70° = 20 / sin 60°
c = 20 · sin 70° / sin 60° ≈ 20 · 0.9397 / 0.8660 ≈ 21.70
Step 4 — Find area using ½ · a · b · sin C (C is included between a and b):
Area = ½ · 17.69 · 20 · sin 70° ≈ ½ · 17.69 · 20 · 0.9397 ≈ 166.3 sq units
SSA ambiguous case — A = 30°, a = 8, b = 12. Find all possible triangles.
Step 1 — Compute the height threshold: h = b · sin A = 12 · sin 30° = 12 · 0.5 = 6
Step 2 — Compare: A < 90°, h = 6, a = 8, b = 12. Since h < a < b → 2 solutions.
Step 3 — Find B using Law of Sines:
sin B / b = sin A / a → sin B = 12 · sin 30° / 8 = 12 · 0.5 / 8 = 0.75
B₁ = arcsin(0.75) ≈ 48.59°
B₂ = 180° − 48.59° = 131.41° (check: A + B₂ = 30° + 131.41° = 161.41° < 180° ✓)
Triangle 1: B₁ ≈ 48.59°, C₁ = 180° − 30° − 48.59° = 101.41°
c₁ = 8 · sin 101.41° / sin 30° ≈ 8 · 0.9806 / 0.5 ≈ 15.69
Triangle 2: B₂ ≈ 131.41°, C₂ = 180° − 30° − 131.41° = 18.59°
c₂ = 8 · sin 18.59° / sin 30° ≈ 8 · 0.3190 / 0.5 ≈ 5.10
SSA one solution — A = 45°, a = 10, b = 7. Find B, C, and c.
Step 1 — Compute height: h = b · sin A = 7 · sin 45° ≈ 7 · 0.7071 ≈ 4.95
Step 2 — Compare: A < 90°, a = 10 ≥ b = 7 → exactly 1 solution.
Step 3 — Find B:
sin B = b · sin A / a = 7 · sin 45° / 10 ≈ 7 · 0.7071 / 10 ≈ 0.4950
B = arcsin(0.4950) ≈ 29.67°
(B₂ = 180° − 29.67° = 150.33°; check: A + B₂ = 45° + 150.33° = 195.33° > 180° ✗ — rejected)
Step 4 — Find C: C = 180° − 45° − 29.67° = 105.33°
Step 5 — Find c:
c = a · sin C / sin A = 10 · sin 105.33° / sin 45° ≈ 10 · 0.9636 / 0.7071 ≈ 13.63
Area using ½ab sinC — a = 15, b = 22, C = 48°. Find the area.
Apply the area formula directly:
Area = ½ · a · b · sin C
Area = ½ · 15 · 22 · sin 48°
Area = ½ · 15 · 22 · 0.7431
Area = ½ · 245.22
Area ≈ 122.6 sq units
Guided Practice
AAS — B = 40°, C = 75°, b = 18. Find a.
Hint: First find A = 180° − B − C. Then use a / sin A = b / sin B.
ASA — A = 55°, B = 80°, c = 14. Find a and b.
Hint: Find C = 180° − A − B first. Then apply a / sin A = c / sin C and b / sin B = c / sin C.
SSA — Determine the number of solutions: A = 25°, a = 5, b = 9.
Hint: Compute h = b sin A = 9 sin 25°. Then compare h, a, and b to decide: 0, 1, or 2 triangles.
SSA two solutions — A = 40°, a = 9, b = 13. Find both possible values of B.
Hint: Use sin B = b sin A / a. Find B₁ = arcsin(result), then B₂ = 180° − B₁. Check that A + B₂ < 180° before accepting B₂.
Area — Two sides are 8 and 11 with an included angle of 62°. Find the area.
Hint: Use Area = ½ · a · b · sin C with a = 8, b = 11, C = 62°.
Common Mistakes
Common Mistakes
Using the Law of Sines for SAS or SSS cases.
SAS and SSS require the Law of Cosines. Law of Sines needs at least one angle–opposite-side pair.
Stopping after finding one triangle in the SSA case.
Always check for a second triangle: if sin B = k and B₁ is acute, test B₂ = 180° − B₁. Accept it only if A + B₂ < 180°.
Assuming B₂ = 180° − B₁ is always valid.
B₂ is valid only when A + B₂ < 180°. If A + B₂ ≥ 180°, the second triangle is impossible.
Rounding intermediate values (e.g., rounding sin A mid-calculation).
Keep full calculator precision throughout; round only the final answer to the required decimal places.
Math Tips
Math Tips
AAS and ASA always give exactly one triangle. SSA is the only case that can give 0, 1, or 2 triangles.
The height h = b sinA is the key threshold in the SSA case. If a < h, no triangle exists.
Always find the third angle first in AAS/ASA: C = 180° − A − B.
The area formula ½ab sinC works for any triangle — not just right triangles.
In the ambiguous case, if you find a second angle B₂ = 180° − B₁, always verify A + B₂ < 180° before accepting it.
Quick Check Quiz
Interactive Practice — 5 Questions
Which case does NOT directly use the Law of Sines?
In triangle ABC, A = 30°, a = 5, b = 10. How many solutions exist?
The area formula ½ab sinC requires:
In AAS with A = 40°, B = 70°, a = 15, what is angle C?
Using a/sinA = b/sinB: if a = 8, A = 30°, B = 45°, find b.