Law of Sines- Complete Reference Table and Practice Problems

What Is the Law of Sines?

The Law of Sines is a trigonometric relationship that connects the sides of any triangle to the sines of its angles. It's the tool you reach for when you have partial triangle information and need to find missing pieces.

The formula is straightforward:

a/sin(A) = b/sin(B) = c/sin(C)

Where a, b, c are the side lengths and A, B, C are their opposite angles. You can use any two ratios—the third is just confirmation.

When to Use the Law of Sines

This law works in specific situations:

⚠️ Warning: SSA can produce the ambiguous case. Sometimes you'll get two possible triangles. We'll cover this below.

Law of Sines Complete Reference Table

ScenarioGiven InformationLaw of Sines Use
Find missing sideTwo angles + one sideSet up ratio with known side, solve for unknown
Find missing angleTwo sides + one angle (not included)Set up ratio, use inverse sine
Ambiguous caseTwo sides + angle opposite one known sideCheck if h = a·sin(B) comparison yields 0, 1, or 2 solutions
Area calculationTwo sides + included angleUse: Area = ½ab·sin(C)

The Ambiguous Case Explained

When you know side a, side b, and angle A (but A is not between a and b), you might have zero, one, or two valid triangles. Here's how to determine which:

Step 1: Calculate the Height

h = b · sin(A)

Step 2: Compare a to h

This is why the Law of Sines isn't always a clean solve. The ambiguous case trips up most students.

How to Apply the Law of Sines

Example Problem

Given: Angle A = 35°, side a = 7, side b = 10
Find: Angle B

Step 1: Set up the ratio

7/sin(35°) = 10/sin(B)

Step 2: Cross-multiply

7 · sin(B) = 10 · sin(35°)

Step 3: Solve for sin(B)

sin(B) = (10 · 0.574) / 7
sin(B) = 5.74 / 7
sin(B) = 0.82

Step 4: Find the angle

B = sin⁻¹(0.82) ≈ 55°

Step 5: Check for the ambiguous case

h = 10 · sin(35°) ≈ 5.74
Since 7 > 5.74 and 7 < 10, there could be a second solution at 180° - 55° = 125°.

Practice Problems

Problem 1:
Triangle ABC has A = 50°, a = 12, b = 15. Find angle B.

Problem 2:
Triangle ABC has A = 30°, B = 70°, and a = 8. Find side b.

Problem 3:
Triangle ABC has a = 6, b = 9, and A = 25°. Determine how many triangles are possible.

Solutions

Problem 1:
sin(B)/15 = sin(50°)/12
sin(B) = 15 · 0.766 / 12 = 0.9575
B ≈ 73° (also potentially 107° since a < b)

Problem 2:
First find C: 180° - 50° - 70° = 60°
8/sin(50°) = b/sin(70°)
b = 8 · 0.940 / 0.766 ≈ 9.8

Problem 3:
h = 9 · sin(25°) ≈ 3.8
Since 3.8 < 6 < 9, two triangles are possible.

Law of Sines vs. Law of Cosines

FeatureLaw of SinesLaw of Cosines
Best forSSA, ASA, AAS casesSAS, SSS cases
Known infoAt least one angle-side pairTwo sides + included angle, or all three sides
Ambiguous caseYes (SSA)No
Formulaa/sin(A) = b/sin(B) = c/sin(C)c² = a² + b² - 2ab·cos(C)

Use Law of Sines when you have an angle and its opposite side. Use Law of Cosines when the angle is夹在 between two known sides (SAS) or when you only know three sides (SSS).

Common Mistakes to Avoid

Quick Reference Cheat Sheet

Formula: a/sin(A) = b/sin(B) = c/sin(C)

When solving for an angle:
sin(B) = (b · sin(A)) / a
B = sin⁻¹(result)

When solving for a side:
b = (sin(B) · a) / sin(A)

For area:
Area = ½ · side₁ · side₂ · sin(included angle)

That's the Law of Sines. The formula is simple. The execution is where people lose points. Pay attention to the ambiguous case and always verify your angle-side pairings.