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Triangle Calculator

SSS, SAS, ASA, AAS and the ambiguous SSA case — which can have two answers.

Updated

What do you know?

cm

cm

cm

The solved triangle

6cm²

A scalene right triangle. The three angles sum to exactly 180°, which is a check rather than a coincidence.

Side a
3 cm
Side b
4 cm
Side c
5 cm
Angle A
36.87°
Angle B
53.13°
Angle C
90.00°
Area
6 cm²
Perimeter
12 cm
Inradius
1 cm
Circumradius
2.5000 cm

The area comes from a rearrangement of Heron's formula that stays accurate on long thin triangles. The textbook version subtracts nearly equal numbers there and loses most of its significant digits.

In short

Which triangle measurements can the calculator solve?

Choose 1 of 6 modes: SSS, SAS, ASA, AAS, base and height, or SSA. Complete triangles return sides, angles, area, perimeter, radii, height and classifications. SSA may return 0, 1 or 2 triangles; base and height return area without inventing a triangle.

Three angles fix shape, not size.

How to use the triangle calculator

Choose a mode before entering measurements

SSS uses three sides. SAS uses two sides and the angle between them. ASA uses two angles and their included side, while AAS uses two angles and a non-included side. SSA handles two sides with an angle that is not between them and must be treated separately.

  1. 1

    SSS

    Check the triangle inequality, then derive stable area and all three angles.

  2. 2

    SAS

    Use the included angle to calculate the opposite side and area.

  3. 3

    ASA or AAS

    Find the third angle by subtraction and scale the sides with the law of sines.

  4. 4

    SSA

    Solve a quadratic for every positive third-side candidate and validate each triangle.

  5. 5

    Base and height

    Return the area while leaving unsupported sides and angles undetermined.

Three side lengths must pass the triangle inequality: the sum of any two sides must exceed the third. Lengths 2, 3 and 6 fail because 2 + 3 ≤ 6. Rather than allowing a square root to become NaN, the calculator returns zero area and explains that the lengths cannot close.

For sides 7, 7 and 0.00000002, the exact area used here is 0.00000007. The textbook semiperimeter calculation returns 0.00000006999999957 in ordinary floating-point arithmetic, while the stable rearrangement returns 0.00000007 without losing the seventh significant figure.

The two-solution SSA case
Inputs
A = 30°, a = 8, b = 12
Quadratic discriminant
a² − b²sin²A = 64 − 36 = 28
Third sides
c = 6√3 ± 2√7
First angle pair
B = 48.59°, C = 101.41°
Second angle pair
B = 131.41°, C = 18.59°
Valid results
2 different triangles, both obtuse

The two possible B angles are supplements: 48.59° + 131.41° = 180°.

The SSA solver treats the third side as the unknown in a law-of-cosines quadratic. This avoids relying on inverse sine near its worst-conditioned boundary. For A = 30°, a = 6 and b = 12, the discriminant is zero and the single result has B = 90°, not a slightly acute approximation.

0, 1 or 2

SSA results

Every valid branch is returned

180°

Angle total

The third angle is found by subtraction

Area only

Base and height

The full triangle remains undetermined

For a completed triangle, two angles come from the law of cosines and the third is calculated as 180° minus their sum. This prevents three independently rounded inverse-cosine results from displaying an angle total such as 59.99° + 60.00° + 60.00°.

Base and perpendicular height determine area but not the triangle. The apex may slide along a line parallel to the base while preserving both measurements. Side lengths, angles, perimeter, inradius and circumradius therefore remain unavailable rather than being filled with assumed values.

Do

  • Keep each lowercase side opposite its matching uppercase angle.
  • Check all SSA branches before deciding how many triangles exist.
  • Use the triangle inequality before interpreting an SSS result.
  • Treat base and height as an area-only mode.

Don't

  • Assume every SSA input has one solution.
  • Describe the standard A = 30°, a = 8, b = 12 pair as one acute and one obtuse triangle.
  • Assume three angles determine the scale of a triangle.
  • Use a sloping side as a perpendicular height.

Completed results also include height to side a, inradius, circumradius and two classifications. The angle classification is right, acute or obtuse. The side classification is equilateral, isosceles or scalene. These labels describe the solved geometry and are not inferred for base-and-height-only input.

Solve the quadratic directly

Use the quadratic formula calculator when the underlying polynomial and both algebraic roots are the main result.

Open quadratic calculator

Worked triangle cases showing supported modes, stable SSS arithmetic and every SSA outcome

Mode and inputsCheckable workingResult
SSS: a = 3, b = 4, c = 53² + 4² = 5²; K = 3 × 4 ÷ 2; P = 3 + 4 + 5A = 36.87°, B = 53.13°, C = 90°; K = 6; P = 12; right scalene
SSS: 7, 7, 0.00000002K = ¼√((14.00000002)(0.00000002)(0.00000002)(13.99999998))Stable K = 0.00000007; textbook result = 0.00000006999999957
SSS: 2, 3, 62 + 3 = 5 ≤ 6, so the shortest two sides cannot closeK = 0; invalid triangle with an explanation
SAS: a = 5, C = 60°, b = 7c² = 5² + 7² − 2 × 5 × 7 × cos 60° = 39; K = 5 × 7 × sin 60° ÷ 2c = √39; K = 35√3/4; P = 12 + √39; acute scalene
ASA: A = 45°, B = 45°, c = 10C = 180° − 45° − 45° = 90°; a = b = 10sin45° = 5√2a = b = 5√2; K = 25; P = 10 + 10√2; right isosceles
AAS: A = 30°, B = 90°, a = 5C = 60°; b = 5/sin30° = 10; c = 10sin60° = 5√3a = 5; b = 10; c = 5√3; K = 25√3/2; right scalene
Base and height: base = 12, h = 5K = 12 × 5 ÷ 2K = 30; sides, angles, perimeter and radii undetermined
SSA: A = 30°, a = 8, b = 12c = 12cos30° ± √(8² − 12²sin²30°) = 6√3 ± 2√72 triangles: (B, C) = (48.59°, 101.41°) or (131.41°, 18.59°); both obtuse
SSA: A = 30°, a = 6, b = 12a² − b²sin²A = 36 − 36 = 0; c = 12cos30° = 6√31 triangle: B = 90°, C = 60°; K = 18√3; right scalene
SSA: A = 30°, a = 5, b = 12a² − b²sin²A = 25 − 36 = −110 triangles; the quadratic has no real third side
Angles are in degrees; displayed decimal angles are rounded to 2 places, while exact radicals are retained where useful.

Why SSA can produce two triangles

With A and its opposite side a fixed, another side b can sometimes swing into two positions. The corresponding angle B is acute in one position and its obtuse supplement in the other. Both candidates must still leave a positive third angle.

For A = 30°, a = 8 and b = 12, the resulting B values are 48.59° and 131.41°. The first triangle is obtuse at C = 101.41°, while the second is obtuse at B = 131.41°. Neither completed triangle is acute.

Why the SSS area formula is rearranged

The familiar semiperimeter form is mathematically correct, but its floating-point evaluation can lose precision when one side is tiny compared with the other two. Intermediate subtraction then removes matching leading digits before the remaining factors are multiplied.

Kahan’s rearrangement sorts the side lengths and groups the factors to reduce that cancellation. It also allows a failed triangle inequality to be identified before taking a square root, producing zero with an explanation instead of NaN.

Why three angles do not determine size

Triangles with the same three angles are similar, but they may have any positive scale. A triangle with sides 3, 4 and 5 has the same angles as one with sides 6, 8 and 10, while its perimeter and area are different.

The formula, worked line by line

Sides a, b and c lie opposite angles A, B and C. A complete solution uses the selected mode to recover all missing sides and angles before calculating area, perimeter, height, radii and classifications.

The area symbol K is used below to avoid confusing area with angle A. In SSS mode, the sides are additionally sorted as x ≥ y ≥ z before the stable area expression is evaluated.

K = ¼√((x + (y + z))(z − (x − y))(z + (x − y))(x + (y − z)))
A = arccos((b² + c² − a²)/(2bc))
B = arccos((a² + c² − b²)/(2ac))
C = 180° − A − B
a² = b² + c² − 2bc cos A
a/sin A = b/sin B = c/sin C
SSA: c = b cos A ± √(a² − b²sin²A)
Base and height: K = bh/2
P = a + b + c
hₐ = 2K/a
r = K/s, where s = P/2
R = abc/(4K)
One set of measurements, two trianglesWith angle A of 30 degrees, side a of 8 and side b of 12, two different triangles fit. In one, angle B is 48.59 degrees; in the other it is 131.41 degrees. The two add to 180.THE AMBIGUOUS SSA CASEB = 48.59°c = 15.68, area 47.05B = 131.41°c = 5.1, area 15.3A = 30°, a = 8, b = 12 — and 48.59° + 131.41° = 180°
The ambiguous case drawn twice: angle A of 30°, side a of 8 and side b of 12 give two different triangles, whose values of angle B are 48.59° and 131.41° and add to 180°.

Kahan’s area expression is algebraically equivalent to Heron’s formula but arranges the factors to reduce damaging cancellation in needle-shaped triangles. The triangle inequality is checked before its square root is evaluated.

The SSA quadratic can have a negative, zero or positive discriminant. After geometric validation, those cases produce zero, one or as many as two distinct triangle solutions.

Two angles are calculated from side relationships and the third by subtraction from 180°. This makes the displayed angle total exact under the calculator’s rounding convention.

Questions people ask

Sources

Where the constants and formulas on this page come from. Each line names the figure it backs.

  1. Supports the relationships among triangle sides, angles, area, radii and classifications.

    TriangleWolfram MathWorld

  2. Supports the classical relationship between three side lengths and triangle area.

    Heron's FormulaWolfram MathWorld

  3. Supports solving side-angle relationships through the law of cosines.

    Law of CosinesWolfram MathWorld

  4. Supports the stable rearrangement used for the area and angles of needle-like triangles.

    Miscalculating Area and Angles of a Needle-like TriangleW. Kahan, University of California, Berkeley