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Law of Sines

A free Geometry lesson from the “Triangles, Similarity, and Trigonometry” unit, with a worked example and practice problems including step-by-step solutions.

The law of sines connects the sine of each angle of a triangle to the length of the opposite side: sin(A)/a = sin(B)/b = sin(C)/c. It works for any triangle, not just right triangles. Use it when you know two angles and a side (AAS or ASA), or two sides and a non-included angle (SSA). In Triangles, Similarity, and Trigonometry, students need to read the diagram, name the relationship, choose a theorem or formula, and justify why the result follows. The expanded practice now includes fluency, transfer, cumulative review, and proof-style reasoning so Geometry feels connected instead of isolated by topic.

What you'll learn

Why it matters: Surveying, navigation, and astronomy all use the law of sines to compute distances and angles to inaccessible points using one known side and two known angles.

Worked example

Problem. In triangle ABC, angle A = 30 degrees, angle B = 45 degrees, and side a = 10. Find side b.

  1. Law of sines: sin(A)/a = sin(B)/b.
  2. sin(30)/10 = sin(45)/b.
  3. (1/2)/10 = (sqrt(2)/2)/b, so b = 10 * sqrt(2).
  4. Connect the result back to Law of Sines so the geometric relationship is explicit.

Answer: 10 * sqrt(2)

Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.

Practice problems

1. Law of sines pairs sin(A) with which side?

Choices: The side opposite angle A · The side next to A · The longest side

Show solution
  1. Opposite-angle and opposite-side go together in the law of sines.

Answer: The side opposite angle A

2. Triangle with A = 30, B = 60, a = 5. Find sin(A)/a as a fraction.

Show solution
  1. sin(30) = 1/2.
  2. (1/2)/5 = 1/10.

Answer: 1/10

3. Triangle with angles A = 30, B = 45 (so C = 105) and a = 6. Find b (the side opposite B) using sin(45) = sqrt(2)/2. Enter the coefficient in front of sqrt(2).

Show solution
  1. Use the Law of Sines: b/sin(45) = 6/sin(30).
  2. Solve for b: b = 6(sqrt(2)/2)/(1/2) = 6sqrt(2).
  3. Therefore the coefficient of sqrt(2) is 6.

Answer: 6

4. Triangle with A = 45 and B = 60 and side a = 4. Find side b. Enter the coefficient in front of sqrt(6).

Show solution
  1. sin(45)/4 = sin(60)/b.
  2. (sqrt(2)/2)/4 = (sqrt(3)/2)/b.
  3. Cross-multiply and simplify: b = 4 * sqrt(3) / sqrt(2) = 4 * sqrt(3) * sqrt(2) / 2 = 2 * sqrt(6). Coefficient = 2.

Answer: 2

5. If sin(A)/a = sin(B)/b, and a = 8, b = 12, sin(A) = 1/2, find sin(B).

Show solution
  1. (1/2)/8 = sin(B)/12.
  2. sin(B) = 12 * (1/2)/8 = 6/8 = 3/4.

Answer: 3/4

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