Triangle Calculator: SSS, SAS, ASA, AAS & SSA

Solve any triangle from known sides and angles. See every step, compare SSA solutions, and learn when to use the laws of sines or cosines.

Worked example — Triangle Solver

Case	SSS

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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When One Dataset Splits Reality 🖖

Three measurements are not always three facts. In the SSA case the same numbers describe two different triangles, because sine cannot tell an angle from its supplement: sin(θ) = sin(180° − θ). One set of inputs, two triangles, and nothing in the arithmetic prefers either.

Why you always need a length 🖖

Three known values usually pin down a triangle, but at least one must be a side. Three angles alone (AAA) fix the shape yet not the size: a tiny triangle and a giant one can share identical angles. The tool leans on two rules: the law of cosines when you know all three sides (SSS) or two sides and the angle between them (SAS), and the law of sines for the rest.

A hidden circle sets the ratio 🖖

In the law of sines, a/sin(A) = b/sin(B) = c/sin(C) has a value worth knowing on its own. It equals 2R, the diameter of the unique circle passing through all three corners (the circumscribed circle). So every triangle secretly carries a circle, and the sine law is really measuring its size. Double the circumradius and every side-to-sine ratio doubles with it.

Triangle Theory — Case by Case

Which triangle case do you have?

Start with the values marked as known. Their pattern—not the triangle's appearance—chooses the method. A solvable triangle needs at least one known side.

SSS - three sides c² = a² + b² − 2ab cos C

SAS - two sides and included angle c² = a² + b² − 2ab cos C

ASA/AAS - two angles and one side C = 180° − A − B; a / sin A = c / sin C

SSA - ambiguous case sin B = b sin A / a; B′ = 180° − B

AAA — three angles are not enough A + B + C = 180° ⇒ ∞ △

SSS - three sides

What you know: All three side lengths: a, b, and c.

Method: c² = a² + b² − 2ab cos C

Worked example: a = 3, b = 4, c = 5 produces a right triangle.

Open this case: SSS

SSS - three sides. SSS: three fixed side lengths determine one triangle when the triangle inequality holds. All three side lengths: a, b, and c. — SSS: three fixed side lengths determine one triangle when the triangle inequality holds.

SAS - two sides and included angle

What you know: Two sides and the angle included between them.

Method: c² = a² + b² − 2ab cos C

Worked example: a = 7, b = 5, C = 60° fixes the opposite side c.

Open this case: SAS

SAS - two sides and included angle. SAS: the highlighted sides and their included angle determine a unique triangle. Two sides and the angle included between them. — SAS: the highlighted sides and their included angle determine a unique triangle.

ASA/AAS - two angles and one side

What you know: Two angles and any one side (ASA or AAS).

Method: C = 180° − A − B; a / sin A = c / sin C

Worked example: A = 40°, B = 75°, c = 8 gives C = 65° before solving the sides.

Open this case: ASA

ASA/AAS - two angles and one side. ASA/AAS: two known angles fix the shape; one known side fixes its scale. Two angles and any one side (ASA or AAS). — ASA/AAS: two known angles fix the shape; one known side fixes its scale.

SSA - ambiguous case

What you know: Two sides and a non-included angle opposite one of them.

Method: sin B = b sin A / a; B′ = 180° − B

Worked example: a = 7, b = 9, A = 40° creates two valid triangles.

Open this case: SSA ambiguous

SSA - ambiguous case. SSA: one set of givens can intersect in two places and create two valid triangles. Two sides and a non-included angle opposite one of them. — SSA: one set of givens can intersect in two places and create two valid triangles.

AAA — three angles are not enough

What you know: All three angles, but no side length.

Method: A + B + C = 180° ⇒ ∞ △

Worked example: 30°, 60°, 90° describes infinitely many triangles of different sizes.

AAA needs a side — open ASA/AAS

AAA — three angles are not enough. AAA: equal angles can describe both a small and a large similar triangle. All three angles, but no side length. — AAA: equal angles can describe both a small and a large similar triangle.

Problems solved in full

Solving the 3-4-5 triangle from the three sides alone 5 steps
The 3-4-5 triangle has a right angle, and nobody has to be told which corner it is in. Solve it from the three sides alone, then find the two circles hiding in it. This is the SSS case.
1. With three sides and no angles, the only way in is the cosine law, rearranged to give the cosine of the angle opposite the side you leave on the left.
  
  $cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c} = \frac{16 + 25 - 9}{40} = 0.8 ⟹ A = 36.869 9^{\circ}$
2. Repeat for the second angle. Nothing here is an approximation yet — 0.8 and 0.6 are exact, and the decimals only appear when the arccosine is taken.
  
  $cos B = \frac{a ^{2} + c ^{2} - b ^{2}}{2 a c} = \frac{9 + 25 - 16}{30} = 0.6 ⟹ B = 53.130 1^{\circ}$
3. The third angle is free, from the angle sum. Run the cosine law on it anyway and the numerator is 9 + 16 − 25 = 0, so the cosine is exactly zero: this triangle is right-angled because 3² + 4² happens to equal 5², and for no other reason.
  
  $C = 18 0^{\circ} - A - B = 9 0^{\circ}, cos C = \frac{9 + 16 - 25}{24} = 0$
4. Heron's formula needs only the perimeter and the three sides. It returns 6, which is also ½ × 3 × 4 — the two formulas must agree, and here you can watch them do it.
  
  $s = \frac{12}{2} = 6, K = 6 \cdot 3 \cdot 2 \cdot 1 = 6 = \frac{1}{2} ab$
5. Two radii the panel never mentions. In a right triangle the hypotenuse is a diameter of the circumscribed circle, so R = 2.5. The inscribed circle has r = (a + b − c)/2 = 1, and the area equals r × s, which is 6 again.
  
  $R = \frac{c}{2} = 2.5, r = \frac{a + b - c}{2} = 1, K = r s = 1 \times 6 = 6$
Answer

The tool prints A = 36.8699°, B = 53.1301°, C = 90° and a perimeter of 12. Step 3 is the one to keep: Pythagoras is not a separate theorem sitting beside the cosine law, it is the cosine law evaluated at 90°, where the −2ab·cos C term vanishes. Everything the right angle is famous for comes from that single term going to zero, which is why the cosine law works on every triangle and Pythagoras works on one kind.

$cos C = 0 ⟺ a^{2} + b^{2} = c^{2} : Pythagoras is the cosine law at 9 0^{\circ}$
Two sides of 7 and 5 with a 60° angle between them 5 steps
Two sides of 7 and 5 with a 60° angle between them. Find the third side and both remaining angles — and take the second angle from the cosine law rather than the sine law, on purpose. This is the SAS case.
1. The known angle sits between the two known sides, which is exactly the arrangement the cosine law is built for: it returns the side opposite that angle.
  
  $c^{2} = a^{2} + b^{2} - 2 ab cos C = 49 + 25 - 2 (35) (\frac{1}{2}) = 39 ⟹ c = 6.245$
2. At 60° the cosine is exactly ½, so the whole −2ab·cos C term collapses to −ab and the expression becomes a² + b² − ab. That is a useful shape to recognise; 60° is the only angle where it appears.
  
  $cos 6 0^{\circ} = \frac{1}{2} ⟹ c^{2} = a^{2} + b^{2} - ab$
3. Now for angle A. The sine law would work and is easier to type, but arcsine returns an angle in the first quadrant and its supplement is equally valid — so it would hand back two candidates and no way to choose. The cosine law returns a signed cosine, and cosine is one-to-one across the whole range an angle of a triangle can take.
  
  $cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c} = \frac{25 + 39 - 49}{62.45} = 0.2402 ⟹ A = 76.102 1^{\circ}$
4. The last angle comes from the angle sum, which cannot be ambiguous.
  
  $B = 18 0^{\circ} - 6 0^{\circ} - 76.102 1^{\circ} = 43.897 9^{\circ}$
5. The area needs neither of the angles you just found, nor the third side: two sides and the angle between them are already enough.
  
  $K = \frac{1}{2} ab sin C = \frac{1}{2} (7) (5) sin 6 0^{\circ} = 15.155$
Answer

The tool prints c = 6.245, A = 76.1021° and B = 43.8979°. Step 3 is why SAS is a safe case and SSA is not. The cosine law can never be ambiguous, because cos θ takes each value once between 0° and 180°; the sine law always can be, because sin θ takes each value twice. When two sides and an included angle are given, the triangle is already determined and the arithmetic simply reads it off. The next problem is what happens when it is not.

$SAS is never ambiguous: cos θ is one-to-one on (0^{\circ}, 18 0^{\circ})$
The SSA case of a = 7, b = 9, A = 40° 5 steps
a = 7, b = 9, A = 40° — and the panel says two solutions while printing one triangle. Find the other one. This is the SSA case.
1. Drop a perpendicular from the far vertex onto the line the known angle sits on. Its length is h = b·sin A = 5.7851, and the side a has to reach that line: too short and there is no triangle, exactly h and there is one, longer than h and there are two — until a exceeds b, when the second closes off. Here h < a < b, so two.
  
  $h = b sin A = 9 sin 4 0^{\circ} = 5.7851, h < a < b$
2. The sine law gives the sine of B, and arcsine returns the acute branch.
  
  $sin B = \frac{b sin A}{a} = \frac{9 sin 4 0 ^{\circ}}{7} = 0.826441 ⟹ B_{1} = 55.734 9^{\circ}$
3. The obtuse branch has the same sine. It survives only if it still leaves room for a third angle, and 40° + 124.2651° is under 180°, so it does.
  
  $B_{2} = 18 0^{\circ} - 55.734 9^{\circ} = 124.265 1^{\circ}, A + B_{2} = 164.2 7^{\circ} < 18 0^{\circ}$
4. Complete the first triangle: third angle, then the third side by the sine law, then the area.
  
  $C_{1} = 84.265 1^{\circ}, c_{1} = \frac{a sin C _{1}}{sin A} = 10.8356, K_{1} = 31.3423$
5. Complete the second the same way. Same a, same b, same A — and a triangle less than a third the size.
  
  $C_{2} = 15.734 9^{\circ}, c_{2} = \frac{a sin C _{2}}{sin A} = 2.9532, K_{2} = 8.5424$
Answer

The panel reports SSA (2 solutions) and prints h = 5.7851, B = 55.7349°, c = 10.8356 and an area of 31.3423 — the first branch. The second, with c = 2.9532 and area 8.5424, is left for you, and that is the honest state of the case: the data does not determine the triangle, so no calculator can hand you the answer. Anything measured this way — a bearing and two distances, a radar range with an off-axis angle — carries the same fork, and the way out is never more arithmetic. It is one more measurement, or a physical reason why one of the two is impossible.

$sin θ = sin (18 0^{\circ} - θ) : two triangles, and the panel prints one$

Learning path

The triangle rules, and where they come from

Leads to Circle angle theorems

References (1)

The circle hiding behind the law of sines, and why one dataset can describe two triangles: G. Van Brummelen, "Sines, cosines, and their relatives", chapter 2 of Trigonometry: A Very Short Introduction. Oxford University Press, 2020.

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

SSS - Three sides and no angles, so the cosine law is the only way in. The 3-4-5 comes back with a right angle, and the reason sits in the arithmetic: 9 + 16 − 25 is zero, so the cosine of that angle is exactly zero.
SAS - Two sides with the angle between them. The cosine law has to go first, because the sine law needs a side and its opposite angle to start from and there is not one yet.
ASA - Two angles hand you the third for free, so 40° and 75° leave 65°. Everything after that is one sine-law ratio, and the single given side sets the scale for the whole triangle.
SSA ambiguous - The case where three measurements are not three facts. The missing side comes out at either 10.836 or 2.953 from the same inputs, because sine cannot tell 55.7° from 124.3° and both close a triangle.