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Free Triangle Calculator – Solve Every Case, Including Both Ambiguous Triangles

Solve any triangle from SSS, SAS, ASA, AAS or SSA and get sides, angles, area, altitudes, medians, inradius and circumradius. The SSA ambiguous case shows both valid triangles.

Written & reviewed by Helperzy Editorial Team · Updated July 2026

SSS · SAS · ASA · AAS · SSABoth ambiguous trianglesLaw of CosinesHeron's FormulaFree

The SSA ambiguous case shows both triangles. When two sides and a non-included angle can form two different triangles, this solver lists both instead of quietly picking one. Try side a = 7, side b = 10, angle A = 30°.

Enter any 3 values (1 must be a side)

opposite angle A

opposite angle B

opposite angle C

opposite side a

opposite side b

opposite side c

Try:

Solved case

SSA (ambiguous — two triangles)

Two different triangles satisfy these values. Both are shown below — swinging side b left or right joins side c in two places. Most free calculators show only one of them.

Triangle 1 of 2

Side a

7

Side b

10

Side c

13.5592

Angle A

30°

Angle B

45.5847°

Angle C

104.4153°

Area (sq units)

33.8981

Perimeter (units)

30.5592

Altitudes, medians and circle radii for this triangle
QuantityTo aTo bTo c
Altitude (units)9.68526.77965
Median (units)11.38769.56175.342

Inradius r (units)

2.2185

Circumradius R (units)

7

By angle

Obtuse

By sides

Scalene

Step-by-step working

  1. Case SSA — ambiguous, this is triangle 1 of 2. Start with the law of sines.
  2. sin B / b = sin A / a → sin B = 10 × sin 30° / 7 = 0.7143
  3. arcsin gives 45.5847°, and 180° − 45.5847° = 134.4153° has the same sine. Both keep the angle sum under 180°, so both are real triangles.
  4. Here B = 45.5847°, so the third angle = 180° − 30° − 45.5847° = 104.4153°
  5. Law of sines again: c = a · sin(third angle) / sin A = 13.5592

Triangle 2 of 2

Side a

7

Side b

10

Side c

3.7613

Angle A

30°

Angle B

134.4153°

Angle C

15.5847°

Area (sq units)

9.4032

Perimeter (units)

20.7613

Altitudes, medians and circle radii for this triangle
QuantityTo aTo bTo c
Altitude (units)2.68661.88065
Median (units)6.6952.56398.424

Inradius r (units)

0.9058

Circumradius R (units)

7

By angle

Obtuse

By sides

Scalene

Step-by-step working

  1. Case SSA — ambiguous, this is triangle 2 of 2. Start with the law of sines.
  2. sin B / b = sin A / a → sin B = 10 × sin 30° / 7 = 0.7143
  3. arcsin gives 45.5847°, and 180° − 45.5847° = 134.4153° has the same sine. Both keep the angle sum under 180°, so both are real triangles.
  4. Here B = 134.4153°, so the third angle = 180° − 30° − 134.4153° = 15.5847°
  5. Law of sines again: c = a · sin(third angle) / sin A = 3.7613

Methods follow the law of cosines and law of sines as stated on Wolfram MathWorld, Heron's formula for area, and the ambiguous-case treatment taught by Math is Fun. Lengths are unit-agnostic — enter all sides in the same unit and areas come out in that unit squared.

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Every angle, area and drawing is computed in your browser. Nothing is uploaded.

How to Use Triangle Calculator

1

Enter Any Three Values

Type three of the six measurements — sides a, b, c or angles A, B, C — making sure at least one is a side length. Each angle is opposite the matching lower-case side, so angle A always sits across from side a.

2

Read the Case and Both Solutions

The solver names the case it recognised, such as SSS or SAS, and returns every remaining side and angle. In the SSA ambiguous case it lists Triangle 1 and Triangle 2, because two different triangles genuinely satisfy those inputs.

3

Check the Working and the Drawing

Open the step-by-step panel to see which law was applied with your own numbers substituted, then compare the to-scale diagram against your sketch. Turn on radians if your coursework needs them, and copy the full solution in one click.

How Triangle Solving Works and Why the Ambiguous Case Matters

A triangle is fully determined by any three of its six measurements — three sides a, b, c and three angles A, B, C — provided at least one of the three is a side. Give this solver three values and it returns the other three plus the area, perimeter, all three altitudes, all three medians, the inradius, the circumradius and the triangle's classification. Students checking geometry homework, carpenters cutting a roof brace, surveyors closing a traverse and machinists laying out a fixture all need the same thing: the missing measurements and enough working to trust them. The solver names which of the five standard cases it recognised — SSS, SAS, ASA, AAS or SSA — because the case decides which law applies and how many answers exist. Two relations do the heavy lifting. The law of cosines, as stated on Wolfram MathWorld, gives c² = a² + b² − 2ab·cos C, where a and b are two sides and C is the angle between them; rearranged as cos A = (b² + c² − a²) / (2bc), it turns three known sides into three angles. The law of sines says a / sin A = b / sin B = c / sin C, which pairs each side with the angle opposite it and lets you scale a triangle once one side-angle pair is known. Area comes from Heron's formula √(s(s−a)(s−b)(s−c)), where s = (a + b + c) / 2 is the semi-perimeter, or from ½ab·sin C when you have two sides and the included angle. Medians use m_a = ½√(2b² + 2c² − a²), altitudes use h_a = 2·Area / a, the inradius is Area / s and the circumradius is a / (2·sin A). Enter a = 3, b = 4, c = 5 and the law of cosines returns A = 36.8699°, B = 53.1301° and C = 90°, with Heron giving an area of exactly 6, an inradius of 1 and a circumradius of 2.5. Now try the case that separates a careful solver from a careless one: a = 7, b = 10, angle A = 30°. The law of sines gives sin B = 10 × sin 30° / 7 = 0.714286, and arcsin returns 45.5847° — but 180° − 45.5847° = 134.4153° has the identical sine. Both keep the angle sum under 180°, so **two real triangles exist**. The first has B = 45.5847°, C = 104.4153°, c = 13.5592 and area 33.8981. The second has B = 134.4153°, C = 15.5847°, c = 3.7613 and area 9.4032. Same three inputs, two completely different triangles, and this tool shows you both. That second triangle is not a curiosity. A student handed an SSA problem is usually being tested on exactly this: recognising zero, one or two solutions. A calculator that reports only the acute branch will mark half those problems wrong without warning. Outside the classroom, a surveyor working from two measured distances and one bearing needs to know that a second position fits the data equally well before committing to a boundary. A roof framer given a rafter length, a rise and one pitch angle faces the same fork. And anyone verifying a CAD constraint benefits from seeing that the geometry is under-constrained rather than discovering it later. When the second branch pushes the angle sum past 180°, this solver says so explicitly and reports a single triangle — as with 12.4, 7.6 and 125°, where the alternative angle of 149.9° cannot coexist with 125°. A sanity check worth running first: the triangle inequality requires the sum of any two sides to exceed the third. Sides of 1, 2 and 5 fail it, and this tool refuses them by name — "1 + 2 = 3 is not greater than 5" — instead of returning a confident wrong number. Three angles alone are rejected too, since they fix the shape but not the size. In SSA, when the required sine exceeds 1, the short side cannot reach, and the tool says so rather than printing NaN. Intermediates keep full precision and rounding happens only on display, so a right angle reads as exactly 90°. Sides are unit-agnostic: use one unit throughout and areas come out squared. Nothing you type is uploaded.

Triangle Calculator Formula & Method

Law of cosines (Wolfram MathWorld): c² = a² + b² − 2ab·cos C cos A = (b² + c² − a²) / (2bc) Law of sines: a / sin A = b / sin B = c / sin C = 2R Area: Heron = √(s(s−a)(s−b)(s−c)) where s = (a + b + c) / 2 SAS = ½ab·sin C Where: a, b, c = side lengths (any consistent length unit) A, B, C = the angles opposite a, b, c (degrees; radians = degrees × π/180) s = semi-perimeter, R = circumradius Derived quantities: altitude to a h_a = 2·Area / a median to a m_a = ½·√(2b² + 2c² − a²) inradius r = Area / s circumradius R = a / (2·sin A) SSA ambiguous case: sin B = b·sin A / a B₁ = arcsin(sin B) and B₂ = 180° − B₁ have the same sine. Each Bᵢ is a real triangle when A + Bᵢ < 180°. If both qualify, TWO triangles exist. If sin B > 1, no triangle exists — side a is too short to reach. Validity: all sides > 0, all angles in (0°, 180°), a + b > c for every pairing, A + B + C = 180°. Rounding: full double precision throughout; values rounded only for display.

Examples: Triangle Calculator

Input

SSS: a = 3, b = 4, c = 5

Result

A = 36.8699°, B = 53.1301°, C = 90° · area 6 · perimeter 12 · inradius 1 · circumradius 2.5 · right scalene

cos A = (4² + 5² − 3²) / (2 × 4 × 5) = 32/40 = 0.8, so A = 36.8699°. Heron gives s = 6 and area = √(6 × 3 × 2 × 1) = 6, matching calculator.net's published Heron example.

Input

SSA (ambiguous): a = 7, b = 10, A = 30°

Result

Two triangles. #1: B = 45.5847°, C = 104.4153°, c = 13.5592, area 33.8981. #2: B = 134.4153°, C = 15.5847°, c = 3.7613, area 9.4032

sin B = 10 × sin 30° / 7 = 0.714286. Both 45.5847° and its supplement 134.4153° keep the angle sum under 180°, so both are real triangles — the published two-solution result for these inputs.

Input

SSA (ambiguous): b = 8, c = 13, B = 31°

Result

Two triangles. #1: C = 56.818°, A = 92.182°, a = 15.52. #2: C = 123.182°, A = 25.818°, a = 6.76

Reproduces the Math is Fun worked SSA example exactly: sin C = 13 × sin 31° / 8 = 0.836937, giving C = 56.8° or 123.2° and side a of 15.52 or 6.76.

Input

SAS: a = 9, b = 7, C = 30°

Result

c = 4.5696 · area 15.75 · A = 100.0094° · B = 49.9906° · obtuse scalene

Area = ½ × 9 × 7 × sin 30° = 15.75, matching calculator.net's published SAS area example, and c comes from the law of cosines: c² = 81 + 49 − 126·cos 30°.

Input

Impossible SSS: a = 1, b = 2, c = 5

Result

Rejected — "1 + 2 = 3 is not greater than 5, so no triangle can have these three sides"

The triangle inequality requires the sum of any two sides to exceed the third. Rather than returning a NaN area, the tool names the failing pair so you know exactly which value to change.

Frequently Asked Questions – Triangle Calculator

When you know two sides and an angle not between them, the law of sines can produce two valid angles, since an angle and its supplement share the same sine. If both keep the angle sum under 180°, two genuinely different triangles fit your data and both are correct.