Last updated: 2026-05-18

Solving a Triangle from Three Sides

The side-side-side (SSS) case is the configuration in which all three side lengths of a triangle are known. Three fixed side lengths determine a triangle completely (up to reflection), so every angle, the area, and the radii of the inscribed and circumscribed circles follow from the sides alone. The angles are recovered through the Law of Cosines.

The Law of Cosines

For a triangle with sides $a$, $b$, $c$ opposite to angles $A$, $B$, $C$ respectively:

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$

Rearranging: $A = \arccos\!\left(\dfrac{b^2 + c^2 - a^2}{2bc}\right)$

The formula is applied twice (for $A$ and $B$), and the third angle follows from $C = 180° - A - B$.

Worked example — the 3-4-5 triangle:

$A = \arccos\!\left(\frac{16 + 25 - 9}{40}\right) = \arccos(0.8) \approx 36.87°$

$B = \arccos\!\left(\frac{9 + 25 - 16}{30}\right) = \arccos(0.6) \approx 53.13°$

$C = 180° - 36.87° - 53.13° = 90°$

The 3-4-5 triangle is a right triangle, as expected.

The Triangle Inequality

Not every set of three lengths forms a valid triangle. The triangle inequality requires:

$a + b > c, \quad b + c > a, \quad a + c > b$

If any one of these fails, the triangle is degenerate (zero area) or impossible. For example, sides 1, 2, and 10 cannot form a triangle because $1 + 2 = 3 < 10$.

Area from Heron's Formula

Once the sides are validated, the area follows without needing any angle. First compute the semi-perimeter:

$s = \frac{a + b + c}{2}$

Then apply Heron's formula:

$S = \sqrt{s(s-a)(s-b)(s-c)}$

For the 3-4-5 triangle: $s = 6$, so $S = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6$.

Circumradius and Inradius

Two circles are naturally associated with every triangle.

Circumradius $R$ — the radius of the circle passing through all three vertices (circumscribed circle):

$R = \frac{abc}{4S}$

Inradius $r$ — the radius of the circle tangent to all three sides (inscribed circle):

$r = \frac{S}{s}$

For the 3-4-5 triangle: $R = \dfrac{3 \times 4 \times 5}{4 \times 6} = 2.5$ and $r = \dfrac{6}{6} = 1$.

Note that for any right triangle, the circumradius equals half the hypotenuse: $R = c/2 = 5/2 = 2.5$ — a convenient check.

Connection to the Pythagorean Theorem

When one angle is exactly $90°$, $\cos(90°) = 0$ and the Law of Cosines simplifies to:

$c^2 = a^2 + b^2 - 2ab \cdot 0 = a^2 + b^2$

This is the Pythagorean theorem — confirming that it is a special case of the more general law.

Common Pythagorean Triples

These are integer-sided right triangles. Scaling by any positive integer gives another valid triple.