Enter side a (opposite angle A), side b (adjacent), and angle A to determine whether the SSA configuration produces 0, 1, or 2 triangles. Includes the height test, a full decision tree, both solutions when applicable, and an arc diagram showing every possible configuration.
SSA Inputs
Must be between 0° and 180° (exclusive).
Examples
Results
| Triangle 1 | Triangle 2 |
|---|
The Height Test — Why h = b·sin A Is Critical
h = b · sin A (altitude from B perpendicular to base)
In an SSA triangle, fix angle A at the left vertex and draw side b extending to point B. The missing vertex C must lie on an arc of radius a centered at B. The shortest possible distance from B down to the base line AC is the altitude h = b · sin A.
If a < h, the arc falls entirely above the base line and never intersects it — 0 triangles. If a = h exactly, the arc just grazes the base line at a right angle — 1 triangle (right triangle). If h < a < b, the arc crosses the base line in two places — both give valid triangles. If a ≥ b (and A is acute), only the crossing on the same side as A is valid — 1 triangle.
Why SSA Alone Is Ambiguous — SSS / SAS / AAS / ASA Are Not
SSS, SAS, AAS, ASA → exactly 1 triangle (or 0 if invalid)
SAS locks in two sides and the angle between them — the third vertex is uniquely forced by the Law of Cosines. ASA / AAS fix two angles (so the third is known) plus at least one side; only one triangle satisfies all three angle constraints.
SSS is unambiguous because the Law of Cosines uniquely determines every angle once all three sides are given (assuming the triangle inequality holds).
SSA is different: angle A is not between the two known sides, so the opposite side a can "swing" to two different positions. The second solution uses B₂ = 180° − B₁ and produces a genuinely different triangle — same a, b, A, but different B, C, c, and area.
Our tutors walk you through SSA, the Law of Sines, and every triangle congruence case — step by step, at your pace.