Enter any 3 known values (at least one side) and instantly solve the whole triangle — every side, every angle, area, perimeter, heights and radii — with a to-scale diagram. 📐
SSS • SAS • ASA • AAS • SSA
Fill in exactly 3 values (at least one side). Side a is opposite angle A, b opposite B, c opposite C.
Tip: two sides + the angle opposite one of them (SSA) can have two valid triangles — both are shown.
SSS, SAS, ASA, AAS and the tricky ambiguous SSA case — including both solutions when two triangles exist.
A live diagram of your solved triangle with labeled vertices and sides, so you can sanity-check the shape instantly.
Beyond sides and angles: area by Heron’s formula, perimeter, all three heights, inradius and circumradius.
Every triangle is fully determined by three independent measurements, as long as at least one of them is a side. Three angles alone fix only the shape, not the size — there are infinitely many similar triangles with angles 30°-60°-90°. This calculator takes whichever 3 values you know, classifies the case, and solves the rest using two classical tools taught in every US geometry and trigonometry course: the law of sines and the law of cosines.
The law of sines says the ratio of each side to the sine of its opposite angle is constant: a / sin A = b / sin B = c / sin C = 2R, where R is the circumradius. It is the natural tool when you know an angle and its opposite side (ASA, AAS, SSA). The law of cosines, c² = a² + b² − 2ab·cos C, generalizes the Pythagorean theorem and handles the cases the law of sines cannot start: three sides (SSS) or two sides with the included angle (SAS). When C = 90°, cos C = 0 and the formula collapses to the familiar c² = a² + b².
Once all three sides are known, the area comes from Heron’s formula: with semi-perimeter s = (a + b + c) / 2, Area = √(s(s−a)(s−b)(s−c)). For a 3-4-5 right triangle, s = 6 and Area = √(6·3·2·1) = 6. From the area, everything else follows: each height is h = 2·Area / base, the inradius is r = Area / s, and the circumradius is R = abc / (4·Area). The calculator also classifies the triangle — equilateral, isosceles or scalene by sides, and acute, right or obtuse by its largest angle.
| Case | You know | Solved with | Solutions |
|---|---|---|---|
| SSS | 3 sides | Law of cosines | 1 (if triangle inequality holds) |
| SAS | 2 sides + included angle | Law of cosines | 1 |
| ASA / AAS | 2 angles + 1 side | Angle sum + law of sines | 1 |
| SSA | 2 sides + non-included angle | Law of sines (ambiguous case) | 0, 1 or 2 |
The SSA “ambiguous case” deserves its reputation. Knowing two sides and an angle opposite one of them is like swinging a door of fixed length toward a wall: it can miss the wall entirely (no triangle), just touch it (one right triangle), or cross it in two places (two distinct triangles). Algebraically, sin B = b·sin A / a may exceed 1 (no solution), equal 1 (one solution) or be less than 1 — giving both an acute B and its supplement 180° − B as candidates. This calculator checks both and displays every valid triangle, which is exactly what trigonometry teachers expect on homework in 2026 — and what surveyors, carpenters and navigators rely on in the field.
Which triangle formula applies depends entirely on what you already know. Right triangles yield to Pythagoras and basic trigonometry, but general triangles need the law of cosines or the law of sines. The law of cosines is the generalisation of Pythagoras: when the included angle is 90°, its cosine is zero and the correction term vanishes, leaving a² + b² = c² exactly. The law of sines is simpler to apply but carries the ambiguous case — with two sides and a non-included angle, two different triangles can satisfy the same data, so the law of cosines is safer when you have a choice.
Law of cosines: c² = a² + b² − 2ab·cos(C)Law of sines: a/sin(A) = b/sin(B) = c/sin(C)Area (two sides + included angle) = ½ab·sin(C)Heron: Area = √[s(s−a)(s−b)(s−c)], where s = (a+b+c)/2where:
Assumptions: Angles must be in the unit your calculator expects — degrees here, radians in most programming languages. A valid triangle requires each side to be shorter than the sum of the other two.
Find the third side and the area from side-angle-side data.
ResultThird side 7.241 — area 24.822
Notice the sanity check: with a 52° angle the opposite side (7.24) is shorter than the 9-side, as it must be, since larger angles face longer sides. Had the angle been 90°, the correction term would be zero and c would be √130 = 11.40.