How this instrument works
Solving a triangle from side-side-angle, or SSA, behaves differently from the other combinations met in trigonometry. Hand a solver two sides and the included angle (SAS), two angles and a side (ASA), or all three sides (SSS), and exactly one triangle comes out every time. SSA breaks that pattern: because the known angle sits opposite one of the two known sides rather than between them, the unknown side can close the figure in more than one way, or sometimes cannot close it at all.
The deciding number is the height h = b·sin(A), the shortest possible reach of side a as it swings down from the far vertex toward the baseline. If a is shorter than that height, it never touches the base, and no triangle exists: zero solutions. If a matches h exactly, it lands at a single point, forming a right angle at that vertex: one solution, a boundary case. If a is longer than h yet still shorter than b, the swinging side crosses the base at two separate points, each closing a valid figure: two solutions, the genuinely ambiguous zone. Once a grows to at least the length of b, only one crossing stays possible, so the count drops back to one.
This page exists because SSA is the single exception among the standard triangle-solving cases, worth walking through on its own rather than folded into a general solver. Picture side b fixed at one end, angle A fixed at the far vertex, and side a hinged there like a swinging gate. Where the gate's free end can touch the base decides everything: too short and it never arrives, exactly right and it lands once, long enough and it can swing past and strike the base twice, once tipped toward an acute third angle and once tipped toward an obtuse one. That double crossing, and only that double crossing, is what makes SSA notorious among students first learning to solve triangles.
- Enter Side a, the length measured opposite the known angle.
- Enter Side b, the other known length in the setup.
- Set Known angle A, choosing degrees, radians, or turns.
- Check Number of valid triangles: it reports 0, 1, or 2.
- Nudge Side a up or down and watch the count flip between zero, one, and two.
Worked example — one setup, three different lengths for a
Fix b = 10 and angle A = 30°, so b·sin(A) = 10 × 0.5 = 5. With side a = 6, that places a strictly between 5 and 10, the ambiguous band, so the sheet reports numSolutions = 2. Geometrically, side a can swing down to meet the base at two separate points: one closes a triangle with an acute angle at the third vertex, the other closes one with an obtuse angle there, and both use the identical a, b, and A.
Shrink side a to 4, keeping b = 10 and A = 30°, and a now falls short of the 5-unit height, unable to reach the base at all: numSolutions = 0. Stretch it instead to a = 12, past b = 10 entirely, and only one crossing remains possible: numSolutions = 1. Three lengths for side a, the same b and A throughout, produce every possible outcome in turn.
Questions
Why can two different triangles share the same two sides and angle?
Because the known angle sits opposite side a rather than between the two known sides, side a can swing from the far vertex and cross the base at two separate points whenever a is longer than b·sin(A) but shorter than b. Each crossing closes a genuine figure with the same a, b, and A, but a different third angle, one acute and one obtuse.
What does the value b·sin(A) actually represent?
It is the height measured from the vertex opposite side b straight down to the line containing side b, in other words the shortest length side a could have and still reach that base at all. Compare a against this height first; it decides whether zero, one, or two triangles are even possible before any other check runs.
Why does a = 12 with b = 10 only produce one triangle?
Once side a reaches or passes the length of side b, the swinging arc can cross the base line at only one point on the valid side of the figure, so the ambiguity collapses to a single answer. This matches the third case on this page: a = 12, b = 10, A = 30° gives numSolutions = 1.
Does this same ambiguity turn up in SAS or ASA problems?
No. Side-angle-side and angle-side-angle both pin down a triangle uniquely, because the given angle sits between two known measurements rather than opposite one of them. SSA stands alone among the classic triangle-solving cases as the one arrangement where identical inputs can honestly describe more than one shape.
What happens right at the boundary, when a equals b·sin(A) exactly?
Exactly one triangle exists, and it carries a right angle at the vertex where side a meets the base. That is the single point where the swinging arc is tangent to the base rather than crossing it twice, so the count settles at one instead of two.
Which units can I use for the known angle A?
Degrees, radians, or turns; switch units on the angle field and the sheet converts automatically before comparing a against b·sin(A). Internally that comparison always happens in radians, so the reported count of triangles stays identical no matter which display unit gets selected.