Law of Sines Calculator
Find missing sides or angles of any triangle using the Law of Sines. Supports AAS (two angles + non-included side) and SSA (two sides + non-included angle) modes, including the ambiguous case.
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How to use this calculator
The Law of Sines relates each side of a triangle to the sine of its opposite angle. All three ratios are equal and can be used to find unknown sides or angles.
- 1
Choose AAS if you know two angles and any one side; choose SSA if you know two sides and the angle opposite one of them.
- 2
Enter angle A, angle B (AAS) or side b (SSA), and side a.
- 3
The calculator solves for all missing parts and reports the area. In SSA mode it may show two valid solutions (ambiguous case).
Frequently asked questions
Which values do I enter for each mode?
In AAS mode, enter angle A in the first box, angle B in the second and side a, the side opposite A, in the third. In SSA mode, enter angle A, side b and side a, where a is the side opposite A. The middle box is an angle in one mode and a side in the other, so check its label. Angles are in degrees.
What if I know two angles and the side between them?
Find the third angle first, since the angles sum to 180°, then relabel so the known side is opposite a known angle. For example, with angles 35° and 75° and the side between them known, the third angle is 70°, and the known side is opposite that third angle. Enter that angle as A, one of the others as B, and the side as a.
Why can SSA return two triangles?
Sine of an angle equals the sine of its supplement, so B and 180° − B both satisfy the equation. With A = 30°, b = 10 and a = 7 the calculator gives B = 45.584691° or 134.415309°, and two different third sides, 13.559234 and 3.761275. Both triangles fit the given data, which is called the ambiguous case.
When does it say that no triangle exists?
When b × sin A ÷ a is greater than 1, no angle B can satisfy the sine rule. With A = 40°, b = 10 and a = 5 the side is too short to reach the base line. It also reports no triangle when neither candidate for B leaves room for a positive third angle, as with an obtuse A of 120°, b = 10 and a = 9.
What happens in the right-triangle borderline?
When the side is exactly long enough to just touch the base, B is 90° and only one triangle is possible. A = 30°, b = 10 and a = 5 returns B = 90°, C = 60° and c = 8.660254. Tiny rounding differences are smoothed out so the result shows exactly 90°.
Can it solve a triangle when I know three sides, or two sides and the included angle?
No. Those cases (SSS and SAS) need the Law of Cosines. The Law of Sines needs a known side paired with its opposite angle.
What extra outputs does it provide?
AAS mode returns sides b and c, all three angles, area and perimeter. SSA mode returns, for each solution, angle B, angle C, side c and area, but no perimeter. Values are rounded to six decimal places.
Law of sines calculator — solve triangles from AAS and SSA data
The rule and what you need
The Law of Sines states that a ÷ sin A = b ÷ sin B = c ÷ sin C. To use it you need one known side together with its opposite angle, plus one more piece of information. With two angles (AAS), the answer is unique because the third angle follows from the 180° sum. With two sides and a non-included angle (SSA), the triangle may be unique, doubled or impossible.
Worked example: AAS
Given A = 40°, B = 60° and a = 10, the third angle is C = 80°. The calculator returns b = 13.472964, c = 15.320889, an area of 66.341395 and a perimeter of 38.793852. For A = 35°, B = 75° and a = 12, it gives C = 70°, b = 20.208483, c = 19.659649 and an area of 113.938577. In both cases the longest side faces the largest angle, which is a quick sanity check.
The ambiguous case in SSA
When you know angle A, the side a opposite it, and the side b, the number of triangles depends on how a compares with b and with b × sin A. For an acute A, there is no triangle if a is less than b sin A, one right triangle if they are equal, two triangles if a lies between b sin A and b, and one triangle if a is at least b. For an obtuse A, there is one triangle if a exceeds b and none otherwise.
| A | b | a | Outcome |
|---|---|---|---|
| 30° | 10 | 7 | Two triangles: B = 45.584691° or 134.415309° |
| 35° | 8 | 6 | Two triangles: B = 49.886408° or 130.113592° |
| 40° | 10 | 5 | No triangle |
| 30° | 10 | 5 | One right triangle: B = 90° |
| 50° | 6 | 9 | One triangle: B = 30.710221° |
| 120° | 10 | 9 | No triangle |
Worked example: two valid answers
For A = 30°, b = 10 and a = 7, solution 1 has B = 45.584691°, C = 104.415309°, c = 13.559234 and an area of 33.898084. Solution 2 has B = 134.415309°, C = 15.584691°, c = 3.761275 and an area of 9.403186. Both satisfy a ÷ sin A = b ÷ sin B. Extra information, such as a diagram showing whether B is acute or obtuse, tells you which one to use.
Choosing the right law
Use the Law of Sines for AAS, ASA and SSA. Use the Law of Cosines when you have three sides or two sides with the angle between them. Angles must be positive and below 180°, and in AAS mode A + B must be below 180°. Side lengths can use any unit, and the output sides, area and perimeter follow that unit.
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Results are estimates for informational purposes only and do not constitute professional financial, medical, legal, or technical advice. Read full disclaimer →