The Ambiguous Case of the Law of Sines (SSA)

Given two sides and a non-included angle (SSA), you cannot assume one triangle. The ambiguous case law of sines can produce zero, one, or two valid triangles. The deciding factor is the height $h = b \sin A$. Compare side $a$ (the side opposite the known angle) to this height and to side $b$. That single comparison tells you how many triangles an SSA problem has. The decision table, the two-solution scenario, and a no-triangle case are explained with worked examples that show the full process.

In an SSA triangle, you know two side lengths and one angle that is not between them. Unlike ASA or AAS, where the known side is locked between the two known angles, the SSA configuration leaves the missing angle and the third side free to shift. The unknown angle opposite the second known side could be acute or obtuse, and the same sine value corresponds to two possible angles (the angle and its supplement). That is the core of the ambiguity: the law of sines, by itself, cannot distinguish between angle B and 180° − B.

The law of sines states a/sin(A) = b/sin(B) = c/sin(C). If you enter an SSA problem, you first solve for sin(B) = b sin(A) / a. If that value is > 1, no triangle exists. If it equals 1, B = 90°, exactly one triangle. If it is < 1, B could be either the acute angle that sin⁻¹ returns or its obtuse supplement. That is where the ambiguous case begins.

When the Ambiguity Disappears

An obtuse or right known angle (A ≥ 90°) removes the ambiguity. An obtuse angle can only oppose the longest side, so side a must be longer than side b. If a ≤ b, no triangle exists. If a > b, exactly one triangle exists. The supplement of an obtuse B would be acute and would not satisfy the triangle inequality, so only one solution is possible.

The height from vertex C to side b is the perpendicular distance h = b sin A. This height is the shortest possible length that side a can have while still reaching the line of side b. Compare side a to h and to side b:

• If a < h: side a is too short to reach the line of side b. No triangle.
• If a = h: side a just reaches the line at a right angle. Exactly one right triangle.
• If h < a < b: side a is longer than the height but shorter than side b. The arc of radius a crosses the line of side b at two points, giving two triangles.
• If a ≥ b: side a is at least as long as side b. The arc crosses at exactly one point, giving one triangle.

This height test is the geometric version of the algebraic condition sin(B) ≤ 1. The two approaches produce the same decision.

The following case table is the standard from Stewart/Redlin/Watson, Precalculus, 8th ed., Section 6.5, pages 498-499. It summarises every SSA scenario by the known angle A and the relationship between sides a and b.

When A is acute:

• a < b sin A → no triangle
• a = b sin A → one right triangle
• b sin A < a < b → two triangles
• a ≥ b → one triangle

When A is obtuse or right:

• a ≤ b → no triangle
• a > b → one triangle

To apply the table, first confirm that the sum of the known angles is less than 180°. Then identify which of the six cases matches your input. The hardest case to recognise is the third one (b sin A < a < b), because it requires the height calculation. Most SSA errors come from missing that second triangle.

The question ‘how many triangles SSA’ is answered by the height test and the case table above. For a quick check: compute h = b sin A. If h > a, zero. If h = a, one (right). If h < a and a < b, two. If a ≥ b, one. That sequence works for acute A. For obtuse A, the rule is a > b for one, a ≤ b for zero.

The two-triangle case is the one that catches most students. Both triangles share the same given sides a and b and the same given angle A. The first triangle uses B₁ = sin⁻¹(b sin A / a). The second uses B₂ = 180° − B₁. From each B, compute the third angle C = 180° − A − B, then the third side c = a sin C / sin A. Both sets satisfy the law of sines and the triangle sum theorem.

When the ambiguous case yields two solutions, finding the second triangle is a three‑step process:

1. Compute B₁ = sin⁻¹(b sin A / a). This is always an acute angle.
2. Compute B₂ = 180° − B₁. This is the obtuse supplement.
3. For each B, compute C = 180° − A − B. If C ≤ 0, that B is invalid and must be discarded.

The law of sines two solutions both share the same given sides. The difference is that one triangle has an acute B and the other has an obtuse B. The third side c will be longer in the obtuse‑B triangle because the opposite angle C is smaller, but side a and side b remain unchanged. Always check that C is positive before accepting a solution; a negative or zero C means the sum of A and B already exceeds 180°.

Given: A = 30°, a = 12, b = 20. Find all possible triangles.

Step 1, Compute h: h = b sin A = 20 × sin 30° = 20 × 0.5 = 10. Compare a to h: a = 12, h = 10, so a > h. Compare a to b: a = 12, b = 20, so a < b. Condition: h < a < b → two triangles.

Step 2, Find B₁: sin B₁ = b sin A / a = 10 / 12 ≈ 0.8333. B₁ = sin⁻¹(0.8333) ≈ 56.4°. This is acute.

Step 3, Find triangle 1: C₁ = 180° − A − B₁ = 180° − 30° − 56.4° = 93.6°. c₁ = a sin C₁ / sin A = 12 × sin 93.6° / 0.5 ≈ 12 × 0.998 / 0.5 ≈ 23.95.

Step 4, Find B₂: B₂ = 180° − B₁ = 180° − 56.4° = 123.6°. Check C₂: C₂ = 180° − A − B₂ = 180° − 30° − 123.6° = 26.4°. C₂ > 0, so this triangle is valid.

Step 5, Find triangle 2:5 ≈ 10.66.

Result: Triangle 1: B ≈ 56.4°, C ≈ 93.6°, c ≈ 23.95. Triangle 2: B ≈ 123.6°, C ≈ 26.4°, c ≈ 10.66. Both satisfy the law of sines. Use the context of the problem to pick the correct one; if no context exists, both are mathematically valid.

Given: A = 50°, a = 8, b = 15. Determine whether any triangle exists.

Step 1, Compute h: h = b sin A = 15 × sin 50° ≈ 15 × 0.766 = 11.49. Compare a to h: a = 8, h ≈ 11.49, so a < h.

Step 2, Apply the case table: A is acute (50°), and a < b sin A → no triangle. The side a is too short to reach the line of side b. No SSA triangle exists for these inputs.

Step 3, Confirm algebraically: sin B = b sin A / a ≈ 11.49 / 8 = 1.436. Since sin B > 1, no angle B exists in the real numbers. The law of sines cannot produce a valid angle, confirming the no‑triangle result.

What to do: Check the input values for a typo. If the numbers are correct, the problem has no solution. In surveying or navigation, this means the measured baseline or angle is inconsistent with the geometry of the site, and a re‑measurement is required.