Law of sines (sine rule)
The law of sines says each side of a triangle divided by the sine of its opposite angle gives the same value: a / sin A = b / sin B = c / sin C. Use it when you know an angle and its opposite side, plus one more side or angle.
| Use it to | Formula |
|---|---|
| Find a side | b = a × sin B / sin A |
| Find an angle | sin B = b × sin A / a |
| Reciprocal form | sin A / a = sin B / b = sin C / c |
| Circumradius link | a / sin A = 2R |
What each letter means
- a, b, c
- The side lengths
- A, B, C
- The angles opposite sides a, b, and c
- R
- The radius of the circle through all three vertices
Law of sines (sine rule) calculator
The solver opens with two angles and a side. Choose “Two sides + opposite angle (SSA)” to explore the ambiguous case.
What do you know?
Each side must be shorter than the other two combined.
Follow the calculation
The idea
Use the sine rule when you know an angle and its opposite side, together with another side or angle. The ratio of a side to the sine of its opposite angle is the same throughout a triangle.
When to use it
- Two angles and any side (AAS or ASA): find the remaining sides.
- Two sides and an angle opposite one of them (SSA): find another angle, then check for a second triangle.
- Not for SAS or SSS: no side is paired with its known opposite angle, so start with the law of cosines.
The ambiguous case (SSA) at a glance
Given angle A, its opposite side a, and another side b, compare a with the height h = b sin A.
| Angle A | Condition | Number of triangles |
|---|---|---|
| Acute | a < h | None |
| Acute | a = h | One right triangle |
| Acute | h < a < b | Two |
| Acute | a ≥ b | One |
| Right or obtuse | a ≤ b | None |
| Right or obtuse | a > b | One |
Worked examples
1Two angles and a side (AAS)
- Known: A = 40°, B = 60°, a = 10.
- C = 180° − 40° − 60° = 80°.
- b = a sin B / sin A = 10 × sin 60° / sin 40° ≈ 13.473.
- c = a sin C / sin A = 10 × sin 80° / sin 40° ≈ 15.321.
Answer: C = 80°, b ≈ 13.473, c ≈ 15.321.
2Find an angle
- Known: a = 8, b = 6, A = 50°. Find B.
- sin B = b sin A / a = 6 × sin 50° / 8 ≈ 0.5745.
- B = arcsin(0.5745) ≈ 35.07°. The other candidate, 144.93°, is too large because 50° + 144.93° > 180°.
- C = 180° − 50° − 35.07° ≈ 94.93°.
Answer: One triangle: B ≈ 35.07° and C ≈ 94.93°.
3The ambiguous case: two triangles
- Known: a = 7, b = 10, A = 30°. The height is h = 10 sin 30° = 5.
- Because 5 < 7 < 10, two triangles are possible.
- sin B = 10 × sin 30° / 7 ≈ 0.7143, so B ≈ 45.58° or B ≈ 134.42°.
- Then C ≈ 104.42° or C ≈ 15.58°, and c = a sin C / sin A ≈ 13.559 or 3.761.
Answer: Two valid triangles: c ≈ 13.559 or c ≈ 3.761.
Why the law of sines works
- Draw the height h from vertex C to side c (or its extension).
- In the two right triangles formed, h = b sin A and h = a sin B.
- Set them equal: b sin A = a sin B, so a / sin A = b / sin B.
- Repeat with a height from another vertex to include c / sin C.
Common mistakes
- Pairing a side with the wrong angle. Each side must be matched with the angle opposite it.
- Forgetting the second SSA solution, 180° − B, when it still fits in the triangle.
- Using the law of sines for SAS or SSS, where no side–opposite-angle pair is known.
- Mixing degree and radian calculator modes.
Make the formula move.
Open the full tool for a larger workspace.
Keep exploring
Law of sines (sine rule): common questions
Wondering about the why? Start here.
What is the law of sines used for?
It finds missing sides and angles when you know at least one side together with its opposite angle. Typical cases are two angles and a side (AAS or ASA) and two sides with a non-included angle (SSA).
Open the triangle solverWhat is the ambiguous case of the law of sines?
In SSA, the known angle may not fix one triangle. Because sin B = sin(180° − B), there can be zero, one, or two valid triangles. Compare the side opposite the known angle with the height h = b sin A to decide.
Explore both SSA solutionsHow do you find an angle using the law of sines?
Rearrange to sin B = b sin A / a, then take the inverse sine. Check whether 180° − B also fits: it does when A + (180° − B) is still less than 180°.
Can the law of sines be used on right triangles?
Yes. With C = 90°, sin C = 1, so a / sin A = c, which gives sin A = a / c: the familiar opposite-over-hypotenuse definition.
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