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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).

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).

Answer: 10 * sqrt(2)

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. Warm-up: First identify exactly what the question is asking: Law of sines pairs sin(A) with which side?
  2. Compare each answer choice with the calculation or rule, and eliminate choices that do not satisfy the condition.
  3. Opposite-angle and opposite-side go together in the law of sines.
  4. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

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. Warm-up: First identify exactly what the question is asking: Triangle with A = 30, B = 60, a = 5. Find sin(A)/a as a fraction.
  2. Use the relevant geometric relationship first, then set up an equation from the angle measures or side relationships.
  3. sin(30) = 1/2.
  4. (1/2)/5 = 1/10.
  5. Check the result by substituting or estimating: the response should match 1/10 and make sense in the original problem.

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. sin(30)/6 = sin(45)/b.
  2. (1/2)/6 = (sqrt(2)/2)/b.
  3. Cross multiply: b/2 = 6 * sqrt(2)/2 = 3*sqrt(2)... wait let me redo: b * (1/2) / 6 = sqrt(2)/2, so b * (1/12) = sqrt(2)/2, b = 12 * sqrt(2)/2 = 6 * sqrt(2). Coefficient 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. Core Practice: First identify exactly what the question is asking: If sin(A)/a = sin(B)/b, and a = 8, b = 12, sin(A) = 1/2, find sin(B).
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. (1/2)/8 = sin(B)/12.
  4. sin(B) = 12 * (1/2)/8 = 6/8 = 3/4.
  5. Check the result by substituting or estimating: the response should match 3/4 and make sense in the original problem.

Answer: 3/4

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