SOLVETUTORMATH SOLVER

Instrument MI-01-579 · Mathematics

SSS Triangle Calculator

Three sides are all a triangle needs to reveal every one of its angles. Enter them, and this sheet solves the whole triangle at once.

Instrument MI-01-579
Sheet 1 OF 1
Rev A
Verified
Type 05 — Trigonometry SER. 2026-01579

Angle A (opposite side a)

38.21321070 deg

A = acos((b² + c² − a²) ⁄ 2bc)

60.00000000 Angle B (opposite side b) (deg)
81.78678930 Angle C (opposite side c) (deg)
The working Every figure verified twice
  1. angleA = acos((7^2 + 8^2 − 5^2) ⁄ (2·7·8)) = 0.66694634
  2. angleB = acos((5^2 + 8^2 − 7^2) ⁄ (2·5·8)) = 1.04719755
  3. angleC = π − acos((7^2 + 8^2 − 5^2) ⁄ (2·7·8)) − acos((5^2 + 8^2 − 7^2) ⁄ (2·5·8)) = 1.42744876
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

SSS (side-side-side) is one of the classic triangle-congruence conditions: knowing all three side lengths determines a triangle's shape completely and uniquely, with no ambiguity about what the angles must be. This calculator solves for all three angles at once by applying the Law of Cosines three separate times, once with each side playing the role of the 'opposite' side being solved for, then finding the third angle as whatever remains once the first two are subtracted from 180°.

This is a genuinely different problem from the already-live SAS solver elsewhere on this site, which starts from two sides and the angle directly between them. Here, no angle is known at all going in — all three come out only after the three known sides are processed through the Law of Cosines in turn.

A useful cross-check built into this solver: since a triangle's three angles must always sum to exactly 180°, the third angle here is computed as 180° minus the other two rather than through a fourth independent Law of Cosines calculation — if the numbers didn't naturally sum correctly, that would signal an inconsistency in the input sides themselves (specifically, a violation of the Triangle Inequality).

A=arccos ⁣(b2+c2a22bc)A = \arccos\!\left(\frac{b^2+c^2-a^2}{2bc}\right)B=arccos ⁣(a2+c2b22ac)B = \arccos\!\left(\frac{a^2+c^2-b^2}{2ac}\right)C=180ABC = 180^{\circ}-A-B
a, b, c — the triangle's three side lengths; A, B, C — the three angles, each opposite its matching lowercase side.
  • Enter the triangle's three side lengths into the Side a, Side b, and Side c fields.
  • Read Angle A, Angle B, and Angle C: the sheet applies the Law of Cosines to recover the first two directly, then finds the third from the fact that all three must sum to 180°.
  • The three sides must satisfy the Triangle Inequality (any two sides summing to more than the third), or the sheet will flag the input.

Worked example — sides 5, 7, and 8

A triangle with sides 5, 7, and 8 solves completely to angle A ≈ 38.21°, angle B ≈ 60.00°, and angle C ≈ 81.79° — every angle recovered from the three sides alone, with the three values summing to exactly 180° as a built-in consistency check.

A 3-4-5 right triangle solves to angle A ≈ 36.87°, angle B ≈ 53.13°, and angle C = 90.00° exactly — the right angle correctly landing opposite the longest side, c, confirmed here directly from the side lengths with no right-angle assumption built in beforehand.

Questions

Can three side lengths always determine a unique triangle?

Yes, as long as they satisfy the Triangle Inequality (any two sides summing to more than the third) — this is exactly what the SSS congruence condition guarantees: three valid side lengths always determine one, and only one, possible triangle shape.

How does this differ from an SAS solver?

An SAS solver starts from two known sides and the angle directly between them. This SSS solver instead starts from all three sides with no angle known at all, recovering every angle purely from the side lengths via the Law of Cosines.

Why is the third angle found by subtraction instead of the Law of Cosines again?

Because a triangle's three angles always sum to exactly 180° — using that fact for the third angle is simpler than a third Law of Cosines calculation, and it doubles as a built-in consistency check on the two already computed.

What happens if the three sides don't form a valid triangle?

The Law of Cosines calculation would require taking the arccosine of a value outside the valid −1 to 1 range, which has no real solution — this calculator checks the Triangle Inequality directly and flags an invalid combination before that happens.

Does this work for an equilateral triangle?

Yes — three equal sides solve to three equal 60° angles exactly, confirmed independently through the Law of Cosines applied three times rather than assumed from the equal side lengths alone.

References