What Is the Law of Sines?

The Law of Sines is a foundational rule in trigonometry that connects the sides and angles of any triangle. For a triangle labeled with vertices A, B, C and opposite sides a, b, c respectively, the law states that the ratio of a side length to the sine of its opposite angle is constant:

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

This constant is equal to the diameter (2R) of the triangle’s circumscribed circle. Unlike the Law of Cosines, which works best with side-angle-side (SAS) or side-side-side (SSS) configurations, the Law of Sines shines when you know partial angle-side relationships. It applies to acute, obtuse, and right triangles alike. The elegant symmetry of the law makes it indispensable for oblique triangles—triangles without a right angle—where standard right-triangle ratios like SOHCAHTOA do not directly apply.

Derivation of the Law of Sines

Understanding where the law comes from builds intuition and helps prevent misapplication. Consider any triangle ABC. Drop an altitude from vertex B to side AC, meeting at point D. This altitude has length h and splits the original triangle into two right triangles: ABD and BCD.

In right triangle ABD, the sine of angle A is opposite over hypotenuse: sin(A) = h / c, so h = c sin(A). In right triangle BCD, sin(C) = h / a, so h = a sin(C). Equating the two expressions yields c sin(A) = a sin(C), which rearranges to a / sin(A) = c / sin(C). By repeating this process with an altitude from a different vertex (for example, from B to side AC or from C to side AB), we obtain the full set of equalities.

For obtuse triangles, the same reasoning holds because the sine of an obtuse angle equals the sine of its supplement (180° minus the angle). The altitude may fall outside the triangle, but the relationship derived from right triangles still holds when we consider the supplementary angle. This universality is what makes the Law of Sines so powerful.

When to Use the Law of Sines

The Law of Sines is the preferred method for three specific triangle configurations:

  • Angle-Angle-Side (AAS) – Two angles and a non-included side are known. For example, given angle A, angle B, and side a (opposite A), you can find side b or side c.
  • Angle-Side-Angle (ASA) – Two angles and the included side are known. For instance, given angle A, side b (between A and C), and angle C, you can find sides a and c.
  • Side-Side-Angle (SSA) – Two sides and a non-included angle are known. This case is special because it can lead to zero, one, or two possible triangles (the ambiguous case).

If you have side-angle-side (SAS) or side-side-side (SSS), the Law of Cosines is more straightforward. The Law of Sines requires an angle-side pair, so it is not a direct fit for those configurations unless you first compute a missing angle using the Law of Cosines and then switch to the Law of Sines.

Step-by-Step Application: Finding an Unknown Side

To solve for an unknown side, you need two angles and one side, or one angle and its opposite side plus another angle. The process follows a predictable pattern.

Step 1: Label the Triangle

Assign capital letters A, B, C to the vertices. The side opposite angle A is side a, opposite B is side b, and opposite C is side c. This labeling is standard for the Law of Sines.

Step 2: Identify Known Values

Write down the given angles and side lengths. Locate which angle is opposite which side. If you have two angles, compute the third using the triangle sum theorem (180° – sum of two known angles).

Step 3: Set Up the Proportion

Choose the ratio that includes your known side and its opposite angle, and equate it to the ratio of the unknown side and its opposite angle. For example, if you know side a, angle A, and angle B, and want side b:

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

Step 4: Solve Algebraically

Cross-multiply: b = a × sin(B) / sin(A). Enter the values into a calculator. Always double-check that your calculator is in the correct mode (degrees vs. radians). Round only your final answer to the required precision; keep intermediate values to four or more decimal places to avoid compounding rounding errors.

Example 1: AAS Configuration

Triangle ABC has angle A = 40°, angle B = 60°, and side a = 10 units (AAS, since we know two angles and a non-included side). Find side b.

First, compute angle C = 180° – (40° + 60°) = 80° (not needed for side b, but good practice). The proportion is:

b / sin(60°) = 10 / sin(40°)

sin(60°) ≈ 0.8660, sin(40°) ≈ 0.6428

b ≈ 10 × 0.8660 / 0.6428 ≈ 13.47 units

Side b is about 13.5 units.

Example 2: ASA Configuration

Triangle XYZ has angle X = 35°, angle Y = 70°, and side XZ = 12 units (side between X and Z, which is side y opposite angle Y). Find side x (opposite angle X).

Angle Z = 180° – (35° + 70°) = 75°. Known side is y = 12 opposite Y = 70°. The proportion for side x is:

x / sin(35°) = 12 / sin(70°)

sin(35°) ≈ 0.5736, sin(70°) ≈ 0.9397

x ≈ 12 × 0.5736 / 0.9397 ≈ 7.32 units

Notice that in ASA, the known side is included between the two known angles, but the Law of Sines still works because we can pair it with its opposite angle.

Finding an Unknown Angle with the Law of Sines

If you know two sides and an angle opposite one of them, you can solve for the other angle. Set up the proportion in the form sin(B) / b = sin(A) / a, then isolate sin(B) = b × sin(A) / a. However, the inverse sine function (arcsin) returns only acute angles between 0° and 90°. If the calculated sine value is less than 1, there may be a second possible obtuse angle (180° – acute angle). This ambiguity must be checked, especially when the known angle is acute and the given opposite side is not the longest.

Example 3: Finding an Angle (One Triangle)

Triangle PQR has side p = 8 (opposite P), side q = 10 (opposite Q), and angle P = 30°. Find angle Q.

Using the Law of Sines: sin(Q) / q = sin(P) / p → sin(Q) = q × sin(P) / p = 10 × sin(30°) / 8 = 10 × 0.5 / 8 = 0.625

Q₁ = arcsin(0.625) ≈ 38.68°

Now check the second possibility: Q₂ = 180° – 38.68° = 141.32°

For Q₁: angle R = 180° – (30° + 38.68°) = 111.32°, which is valid because all angles are positive and sum to 180°.

For Q₂: angle R = 180° – (30° + 141.32°) = 8.68°, also valid. So this SSA case yields two possible triangles. We have encountered the ambiguous case.

The Ambiguous Case (SSA) in Detail

When you know two sides and a non-included angle (SSA), the triangle may not be uniquely determined. This occurs because the given angle can be acute and the opposite side can be shorter than the other given side, allowing for two different angles for the unknown opposite angle. The number of possible triangles depends on the relationship between the known side opposite the given angle (let’s call it a), the other known side (b), and the height h = b sin(A).

Decision Rules for SSA

Given sides a, b, and angle A (opposite a):

  • If a is equal to or greater than b, and A is acute, only one triangle exists (the larger side forces the larger angle).
  • If a is less than b, compute h = b sin(A). Then:
    • If a < h: no triangle is possible (the side a is too short to reach the third vertex).
    • If a = h: exactly one right triangle (the third vertex lands exactly on the altitude).
    • If a > h: two triangles are possible (the side a can swing to create both an acute and an obtuse angle at the unknown vertex).
  • If angle A is obtuse, then side a must be the longest side for a triangle to exist. If a ≤ b, no triangle forms. If a > b, one triangle exists.

Example 4: Two Triangles from SSA

Triangle ABC with angle A = 30°, side a = 6, side b = 8. Compute h = b sin(A) = 8 × 0.5 = 4. Since a = 6 > 4, two triangles exist.

Using the Law of Sines: sin(B) = b sin(A) / a = 8 × 0.5 / 6 = 0.6667

B₁ = arcsin(0.6667) ≈ 41.81°

B₂ = 180° – 41.81° = 138.19°

For B₁: C₁ = 180° – (30° + 41.81°) = 108.19°. Side c₁ can be found with Law of Sines: c = a sin(C) / sin(A) = 6 × sin(108.19°) / 0.5 ≈ 6 × 0.9499 / 0.5 = 11.40 units.

For B₂: C₂ = 180° – (30° + 138.19°) = 11.81°. Side c₂ = 6 × sin(11.81°) / 0.5 ≈ 6 × 0.2046 / 0.5 = 2.46 units.

Both triangles satisfy the given measurements. In real-world problems, additional context (such as which side is longest or a range restriction) will determine which triangle is correct.

Practical Applications of the Law of Sines

The Law of Sines is more than an abstract formula—it is a workhorse in many STEM fields.

Surveying and Geodesy

Surveyors often need to measure distances across obstacles like lakes, ravines, or valleys. They can set up a baseline of known length and measure two angles from the endpoints to a distant point. The Law of Sines then gives the distance from each endpoint to that point, bypassing the need for direct measurement. This technique, known as triangulation, is also used in GPS and land boundary mapping.

Pilots and ship captains use bearing angles and known distances to calculate positions. For example, a ship sailing between two lighthouses can use the Law of Sines to determine how far it is from each lighthouse when only the angles between the lines of sight are known. The Math is Fun Law of Sines page provides interactive examples that illustrate these concepts.

Engineering and Architecture

Truss bridges, roof frames, and other structural designs often use triangular elements. Engineers compute the forces in each member by applying the Law of Sines to the force vectors at joints. Knowing two sides and one angle of the force triangle lets them find missing components, ensuring the structure can handle expected loads.

Astronomy

Astronomers estimate distances to stars and planets using parallax—the apparent shift in position observed from two different points (such as Earth at different times of year). The angle of parallax and the known baseline (Earth’s orbital radius) form an oblique triangle. The Law of Sines then yields the distance to the celestial object.

For quick calculations, the CalculatorSoup Law of Sines tool is a handy resource.

Common Mistakes and How to Avoid Them

  • Calculator mode: Always verify that your calculator is in degree mode when angles are given in degrees. Using radian mode by mistake will produce nonsensical results.
  • Mislabeling sides and angles: The Law of Sines pairs a side with its opposite angle. Using side a with sin(B) leads to an incorrect proportion. Label the triangle clearly before writing any equation.
  • Forgetting the ambiguous case: In SSA problems, always compute h = b sin(A) and check the number of possible triangles. Drawing a rough sketch helps visualize whether the side is long enough to reach the third vertex.
  • Propagating rounding errors: Keep intermediate results to at least four decimal places. Only round the final answer to the requested precision.
  • Using the wrong law: Do not apply the Law of Sines to SAS or SSS problems; use the Law of Cosines instead. Trying to force the Law of Sines when you lack an angle-side pair will lead to dead ends.

Practice Problem Set

Work through these problems to solidify your understanding. For each, determine whether the configuration is AAS, ASA, or SSA, and solve for the requested quantity. Check the ambiguous case when applicable.

  1. Triangle with A = 50°, B = 70°, side a = 15. Find side b.
  2. Triangle with b = 22, c = 30, angle B = 40°. Find angle C (list both possibilities).
  3. Triangle with A = 120°, a = 20, c = 15. Find angle C. Does a triangle exist?
  4. Triangle with side a = 14, side b = 16, angle A = 28°. How many triangles are possible? Find all possible angle B values.
  5. A lighthouse sees two boats at angles of 15° and 25° from its line of sight. The boats are 2 km apart. How far is the lighthouse from each boat? (Hint: draw a triangle with the boats and lighthouse as vertices.)

Solutions

  1. Angle C = 180° – (50°+70°) = 60°. Using Law of Sines: b = a sin(B) / sin(A) = 15 × sin(70°) / sin(50°) ≈ 15 × 0.9397 / 0.7660 ≈ 18.41 units.
  2. sin(C) = c sin(B) / b = 30 × sin(40°) / 22 ≈ 30 × 0.6428 / 22 ≈ 0.8765. C₁ ≈ 61.3°, C₂ ≈ 118.7°. Both are valid because B = 40°, so C₁ gives A = 78.7°, and C₂ gives A = 21.3°. Two triangles.
  3. sin(C) = c sin(A) / a = 15 × sin(120°) / 20 = 15 × 0.8660 / 20 = 0.6495. C₁ ≈ 40.5°, C₂ ≈ 139.5°. If C = 139.5°, then A + C = 259.5° > 180°, so impossible. Only C = 40.5° works, giving B = 19.5°. One triangle exists.
  4. h = b sin(A) = 16 × sin(28°) ≈ 16 × 0.4695 = 7.512. a = 14 > h, so two triangles. sin(B) = b sin(A) / a = 16 × 0.4695 / 14 ≈ 0.5366. B₁ ≈ 32.44°, B₂ ≈ 147.56°.
  5. Let the boats be points B and C, lighthouse at A. Distance BC = 2 km, angle at B = 15°, angle at C = 25°? Clarify: The problem needs more specification. As a simpler exercise: If the lighthouse sees one boat at bearing 15° and the other at 25°, and the distance between boats is known, the triangle can be solved with Law of Sines once the third angle is found.

Conclusion

The Law of Sines turns oblique triangles from intimidating to approachable. By remembering the simple ratio—side over sine of opposite angle—you unlock the ability to find unknown sides and angles from partial data. The technique requires careful identification of the given configuration (AAS, ASA, or SSA) and a watchful eye for the ambiguous case. With practice, you will be able to apply the law to real-world problems in surveying, navigation, engineering, and beyond. For a deeper theoretical discussion and historical context, review the Wikipedia article on the Law of Sines. Keep solving triangles, and the ratios will become second nature.