What Is the Law of Sines?
The Law of Sines (also called the sine rule) establishes a proportional relationship between the sides of any triangle and the sines of their opposite angles. Unlike basic right-triangle trigonometry that only works with 90-degree triangles, the Law of Sines applies to every triangle, whether it is acute, right, or obtuse.
This makes it one of the most versatile tools in trigonometry. Whether you are a student solving homework problems, a surveyor measuring land, or an engineer computing forces, the Law of Sines gives you a direct path from partial information to a complete triangle solution.
Here, a, b, and c are the side lengths of the triangle, and A, B, and C are the angles opposite those sides respectively. The formula can also be written in its reciprocal form:
Both forms are equivalent. The first is more convenient when solving for a side, and the second when solving for an angle.
Key Takeaway
In any triangle, the ratio of a side length to the sine of its opposite angle is constant. This constant equals the diameter of the circumscribed circle (2R).
Proof of the Law of Sines
There are multiple ways to prove the Law of Sines. The most common approach uses the altitude of a triangle to connect sides with opposite angle sines.
Proof Using the Altitude
Consider a triangle ABC with sides a, b, and c opposite angles A, B, and C respectively. Drop a perpendicular (altitude) h from vertex C to side AB (or its extension).
In the right triangle formed on the left side of the altitude:
In the right triangle formed on the right side:
Since both expressions equal h, we can set them equal:
Dividing both sides by (sin A · sin B):
By dropping an altitude from vertex A to side BC and repeating the same logic, we obtain b / sin B = c / sin C. Combining these results gives the complete Law of Sines. This proof also works for obtuse triangles when we use the supplementary angle identity sin(180 - x) = sin(x).
Connection to the Circumscribed Circle
The common ratio in the Law of Sines is actually equal to the diameter of the triangle's circumscribed circle (circumcircle):
where R is the circumradius. This elegant geometric fact connects the Law of Sines directly to the study of circles and triangle geometry.
When to Use the Law of Sines
The Law of Sines is your go-to tool whenever you know at least one side-angle pair (a side and its opposite angle). Specifically, it applies in these triangle configurations:
- AAS (Angle-Angle-Side): Two angles and a non-included side are known. Since the three angles of a triangle sum to 180 degrees, you can always find the third angle first, then use the Law of Sines to find the remaining sides. This case always yields exactly one unique solution.
- ASA (Angle-Side-Angle): Two angles and the included side are known. Again, find the third angle from the angle sum, then apply the Law of Sines for the remaining sides. This also gives one unique solution.
- SSA (Side-Side-Angle): Two sides and an angle opposite one of them are known. This is the ambiguous case, which can produce zero, one, or two valid triangles depending on the measurements.
Key Takeaway
Use the Law of Sines when you have AAS, ASA, or SSA information. For SAS or SSS configurations, use the Law of Cosines instead.
The Ambiguous Case (SSA)
The ambiguous case arises when you are given two sides and an angle that is not between those two sides (SSA configuration). Unlike AAS or ASA, the SSA setup does not always guarantee a unique triangle. Depending on the given values, there may be no solution, exactly one solution, or two distinct solutions.
Understanding Why Ambiguity Occurs
Imagine you know angle A, side a (opposite A), and side b (adjacent to A). You can fix angle A and side b in place, then try to "swing" side a from vertex B to meet the opposite side. Depending on the length of a, the swing may miss entirely, just touch the base, or intersect it at two points.
Case Analysis (When Angle A Is Acute)
Let h = b · sin A (the altitude from the known configuration). Then:
- a < h: Side a is too short to reach the opposite side. No solution exists.
- a = h: Side a exactly reaches the base, forming a right angle. One solution (a right triangle).
- h < a < b: Side a can swing to meet the base in two different locations. Two solutions exist.
- a ≥ b: The side is long enough that only one triangle configuration is possible. One solution.
Case Analysis (When Angle A Is Obtuse)
If angle A is obtuse (greater than 90 degrees):
- If a ≤ b, no triangle exists (the side opposite the obtuse angle must be the longest side).
- If a > b, exactly one solution exists.
Key Takeaway
Always check for the ambiguous case when given SSA information. Compute h = b sin A and compare it to side a to determine how many solutions exist before solving.
The Area Formula Connection
The Law of Sines is closely related to the sine area formula for triangles. When you know two sides and the included angle, the area is:
This formula follows from the same altitude construction used in the proof. Since h = b · sin A, the area is (1/2) · base · height = (1/2) · a · b · sin C. This is extremely useful because it lets you compute area without needing the altitude directly.
Worked Examples
Example 1: AAS Triangle — Find the Missing Sides
Given: In triangle ABC, angle A = 42°, angle B = 73°, and side a = 10.
Step 1: Find angle C using the angle sum property.
C = 180° − 42° − 73° = 65°
Step 2: Set up the Law of Sines ratios.
a / sin A = b / sin B = c / sin C
10 / sin 42° = b / sin 73° = c / sin 65°
Step 3: Compute the common ratio.
10 / sin 42° = 10 / 0.6691 = 14.944
Step 4: Solve for b.
b = 14.944 · sin 73° = 14.944 · 0.9563 = 14.29
Step 5: Solve for c.
c = 14.944 · sin 65° = 14.944 · 0.9063 = 13.54
Solution: C = 65°, b ≈ 14.29, c ≈ 13.54
Example 2: Ambiguous Case (SSA) — Two Solutions
Given: In triangle ABC, angle A = 35°, side a = 7, and side b = 10.
Step 1: Check for ambiguity. Compute h = b · sin A.
h = 10 · sin 35° = 10 · 0.5736 = 5.736
Step 2: Compare a to h and b.
Since h = 5.736 < a = 7 < b = 10, we are in the two-solution zone.
Step 3: Use the Law of Sines to find angle B.
sin B / b = sin A / a
sin B = 10 · sin 35° / 7 = 10 · 0.5736 / 7 = 0.8194
Step 4: Find both possible values of B.
B₁ = arcsin(0.8194) = 55.05°
B₂ = 180° − 55.05° = 124.95°
Step 5: Check that both are valid (A + B < 180°).
A + B₁ = 35° + 55.05° = 90.05° < 180° ✓
A + B₂ = 35° + 124.95° = 159.95° < 180° ✓
Step 6: Find the corresponding C values and side c for each.
Solution 1: C₁ = 89.95°, c₁ = 7 · sin 89.95° / sin 35° ≈ 12.20
Solution 2: C₂ = 20.05°, c₂ = 7 · sin 20.05° / sin 35° ≈ 4.18
Example 3: Finding the Area Using the Sine Formula
Given: In triangle ABC, a = 8, b = 11, and angle C = 52°.
Step 1: Apply the sine area formula.
Area = (1/2) · a · b · sin C
Step 2: Substitute values.
Area = (1/2) · 8 · 11 · sin 52°
Area = 44 · 0.7880 = 34.67 square units
Practice Problems
Practice 1
In triangle PQR, angle P = 48°, angle Q = 61°, and side p = 15. Find all missing parts of the triangle.
Solution:
Angle R = 180° − 48° − 61° = 71°
Common ratio: 15 / sin 48° = 15 / 0.7431 = 20.19
q = 20.19 · sin 61° = 20.19 · 0.8746 = 17.66
r = 20.19 · sin 71° = 20.19 · 0.9455 = 19.09
Practice 2
In triangle ABC, angle A = 30°, a = 6, and b = 12. How many triangle solutions exist? Solve completely.
Solution:
Compute h = b · sin A = 12 · sin 30° = 12 · 0.5 = 6.
Since a = h = 6, there is exactly one solution (a right triangle).
sin B = 12 · sin 30° / 6 = 1 → B = 90°
C = 180° − 30° − 90° = 60°
c = 6 · sin 60° / sin 30° = 6 · 0.8660 / 0.5 = 10.39
Practice 3
Two fire lookout towers are 20 km apart. From tower A the bearing to a fire is N 52° E. From tower B (due east of A) the bearing is N 38° W. Find the distance from each tower to the fire.
Solution:
The angle at A (inside the triangle) = 90° − 52° = 38°.
The angle at B (inside the triangle) = 90° − 38° = 52°.
The angle at the fire F = 180° − 38° − 52° = 90°.
Using the Law of Sines with AB = 20:
AF / sin B = 20 / sin F → AF = 20 · sin 52° / sin 90° = 15.76 km
BF / sin A = 20 / sin F → BF = 20 · sin 38° / sin 90° = 12.31 km
Practice 4
Find the area of a triangle where two sides measure 9 and 14 with an included angle of 67°.
Solution:
Area = (1/2) · 9 · 14 · sin 67°
Area = 63 · 0.9205 = 57.99 square units