Enter any three measurements that fix a triangle. You get the missing sides and angles, the area, perimeter and heights, what type it is, and a drawing to scale.
How the triangle is solved
Every triangle obeys two rules that link its sides (a, b, c) to the angles opposite them (A, B, C):
Law of cosines: c² = a² + b² − 2ab·cos C
Law of sines: a ÷ sin A = b ÷ sin B = c ÷ sin C
Angles: A + B + C = 180°
- SSS (three sides): the law of cosines gives each angle, for example A = arccos((b² + c² − a²) ÷ 2bc).
- SAS (two sides and the angle between): the law of cosines gives the third side, then the angles.
- ASA and AAS (two angles and a side): the third angle is 180° minus the other two, then the law of sines gives the sides.
- SSA (two sides and an angle not between them): the law of sines gives sin B, which can have two answers. See below.
The area uses Heron's formula, which needs only the three sides:
s = (a + b + c) ÷ 2 Area = √(s(s − a)(s − b)(s − c))
Each height is twice the area divided by its base (ha = 2 × Area ÷ a). A triangle is right if one angle is 90°, obtuse if one is more than 90°, and acute otherwise. It's equilateral if all sides match, isosceles if two do, and scalene if none do.
Worked example (SAS)
Two sides are a = 5 and b = 7, with C = 60° between them.
- c² = 5² + 7² − 2 × 5 × 7 × cos 60° = 25 + 49 − 35 = 39, so c = √39 ≈ 6.245.
- A = arccos((7² + 39 − 5²) ÷ (2 × 7 × 6.245)) ≈ 43.9°, and B = 180° − 60° − 43.9° ≈ 76.1°.
- s = (5 + 7 + 6.245) ÷ 2 ≈ 9.1225, and the area is √(s(s − 5)(s − 7)(s − c)) ≈ 15.155. That matches ½ × 5 × 7 × sin 60° = 15.155.
The ambiguous case (SSA)
With two sides and an angle that isn't between them, side a can swing like a gate and touch the base line in two places, one place or none. Take a = 3, b = 4 and A = 40°: sin B = 4 × sin 40° ÷ 3 ≈ 0.8571, so B ≈ 58.99° or 180° − 58.99° = 121.01°. Both leave room for C, so two triangles fit and the calculator shows both. If sin B comes out above 1, side a is too short to reach and there's no triangle.
When this doesn't apply
- Three angles alone (AAA) don't fix a size: every similar triangle has the same angles. Add at least one side.
- These are flat (plane) triangles. Triangles on a globe, such as between three cities, follow spherical rules instead.
- Three sides only make a triangle if the two shorter ones add up to more than the longest.
Sources
How we checked this
Before every update, this calculator is run in a real browser with these known answers, and it must match all of them:
- SSA 4, 8, 30°: a equals the height, one right triangle, B = 90°, c = 4√3 ✓
- SSA 4, 8, 30°: type is right, c = 6.92820323 ✓
- Sides 3, 4, 6 read as 'an obtuse' triangle ✓
- Sides 3, 4, 5: right triangle, area 6, perimeter 12 ✓
- SAS 5, 7 and 60° gives c = √39 ≈ 6.244998 ✓
- AAS 30°, 45°, a = 10 gives b = 10√2 ≈ 14.1421356 ✓
- ASA in radians: A = 0.5, B = 1, c = 10 gives a ≈ 4.8062952 ✓
- SSA 3, 4, 40° is the ambiguous case: B ≈ 58.987° or 121.013° ✓
- SSA 2, 10, 30° has no triangle (sin B = 2.5) ✓
- Sides 1, 2, 3 can't form a triangle ✓
Last checked October 26, 2026. Spotted something wrong? The math is done in your browser with the formula shown on this page, so you can check any result by hand.
Questions people ask
How do I find the missing side of a triangle?
If you know two sides and the angle between them, use the law of cosines: c² = a² + b² − 2ab·cos C. If you know two angles and one side, find the third angle (180° minus the other two) and use the law of sines. For a right triangle, the Pythagorean theorem is quicker.
How do I find the angles of a triangle from three sides?
Use the law of cosines rearranged: A = arccos((b² + c² − a²) ÷ 2bc). Do the same for B, then C = 180° − A − B. For sides 3, 4 and 5 the angles are 36.87°, 53.13° and 90°.
Why does SSA sometimes give two triangles?
Knowing two sides and an angle that isn't between them lets the third side swing to two different positions. The sine of the unknown angle has two possible angles, B and 180° − B, and both can work.
How do I find the area of a triangle with three sides?
Use Heron's formula: work out s = (a + b + c) ÷ 2, then Area = √(s(s − a)(s − b)(s − c)). For 3, 4, 5: s = 6 and the area is √(6 × 3 × 2 × 1) = 6.
Can a triangle have sides 1, 2 and 3?
No. 1 + 2 is only equal to 3, not more, so the sides would lie flat in a line. The two shorter sides must add up to more than the longest.