Triangle Calculator · Practical guide
When Does SSA Produce Two Triangles?
Two sides and a non-included angle can describe zero, one, or two triangles — here's how to tell which case you have.
At a glance
The Short Answer
Given sides a and b with angle A (not between them), two valid triangles exist when the side opposite the known angle is shorter than the other side but longer than the altitude: b × sin A < a < b. The SSA tab above detects this case and offers to show you the second solution.
Interactive tool
Triangle Calculator
Enter your own values, then compare the result with the examples below.
Triangle Solution
Worked Example: a = 8, b = 10, A = 30°
The altitude h = 10 × sin 30° = 5. Since 5 < 8 < 10, two triangles fit:
- Solution 1: B ≈ 38.68°, C ≈ 111.32°, c ≈ 14.91
- Solution 2: B ≈ 141.32°, C ≈ 8.68°, c ≈ 2.41 (the two B angles are supplementary: 180° − 38.68°)
All Four SSA Outcomes
| Condition | Number of Triangles |
|---|---|
| a < b × sin A | None — side a is too short to reach |
| a = b × sin A | One — a right triangle |
| b × sin A < a < b | Two — the ambiguous case |
| a ≥ b | One — only the acute solution closes |
How the Calculator Handles It
Enter the two sides and the non-included angle in the SSA tab. If two triangles are possible, the calculator displays the first solution and asks whether you'd like to see the second; if side a is too short, it alerts you that no triangle exists. This is exactly why SSA — unlike SSS, SAS, and ASA — is not a valid congruence rule in geometry proofs.
Frequently Asked Questions
Why is SSA called the ambiguous case?
Because the same three measurements can match zero, one, or two different triangles. The side opposite the known angle can swing to two positions and still satisfy the given values.
Is SSA a valid triangle congruence theorem?
No. SSS, SAS, ASA, and AAS each guarantee one unique triangle, but SSA can produce two — which is why it can't prove congruence.
How do I find the second SSA solution by hand?
Compute B₁ = arcsin(b × sin A ÷ a), then the second angle is B₂ = 180° − B₁. Finish each triangle with C = 180° − A − B and the Law of Sines.
What happens if side a is longer than side b?
Exactly one triangle exists. The supplementary second angle would push the angle sum past 180°, so only the acute solution is valid.