The Ambiguous Case of the Law of Sines (SSA): How to Solve It

You know two sides and an angle that is not between them (SSA), and the Law of Sines gives you an angle that could be two different things. That is the ambiguous case: the same measurements can describe no triangle, one triangle, or two triangles. This guide shows how to tell which, in under a minute, and how to find the second triangle when there is one.

Why SSA is ambiguous

The Law of Sines gives sin B = b · sin A / a. But a sine value matches two angles between 0° and 180°: an acute one and its supplement. For example, sin B = 0.8 is true for B = 53.13° and for B = 180° − 53.13° = 126.87°. Your calculator’s inverse sine only ever shows the acute one, so the second answer is easy to miss.

The picture makes it concrete. Fix angle A and side b. Side a is then a rod of fixed length hinged at C. Swing it down like a compass, and see where it lands on the base line:

Here A = 30°, b = 8 and a = 5. The height from C to the base is h = 8 × sin 30° = 4. Side a = 5 is longer than 4, so it reaches the base, and it reaches it in two places: B₁ (solid line, the larger triangle) and B₂ (dashed line, the smaller one). Both triangles have the same A, a and b.

That also explains each case of the test. A rod shorter than h can’t reach the base (none). A rod exactly h just touches it (one, a right triangle). A rod between h and b lands twice (two). A rod at least as long as b lands once on the correct side of A, because the other landing point would be behind A and doesn’t make a triangle (one).

The height test: how many triangles?

When A is obtuse (or 90°)

The height test is for an acute angle A. If A is 90° or more, the side opposite it must be the longest side. So there is one triangle when a > b, and none otherwise. There is never a second triangle.

How to solve an SSA triangle, step by step

1. Label the triangle. Side a is opposite angle A, side b is opposite angle B. Confirm the given angle is not between the two given sides. (If it is, that’s SAS, which is never ambiguous.)
2. Count the triangles with the height test above.
3. Find the first angle. sin B = b · sin A / a, so B₁ = sin⁻¹(b · sin A / a).
4. Find the second candidate: B₂ = 180° − B₁.
5. Check it fits. Is A + B₂ less than 180°? If yes, the second triangle exists. If not, there is only one.
6. Finish each triangle separately. C = 180° − A − B, then c = a · sin C / sin A.

Skip step 1 or step 5 and you will misreport the number of solutions. The calculator can do all of this at once: the Law of Sines Calculator in its two-sides-and-an-angle mode solves both triangles and tells you when no triangle exists.

Worked examples

Example 1: two triangles (A = 30°, a = 5, b = 8)

• h = 8 × sin 30° = 4. Since 4 < 5 < 8, expect two triangles.
• sin B = 8 × sin 30° / 5 = 0.8, so B₁ = 53.13° and B₂ = 180° − 53.13° = 126.87°.
• Check B₂: A + B₂ = 156.87°, which is under 180°, so it is valid.
• Triangle 1: C = 180° − 30° − 53.13° = 96.87°, so c = 5 × sin 96.87° / sin 30° ≈ 9.93.
• Triangle 2: C = 180° − 30° − 126.87° = 23.13°, so c = 5 × sin 23.13° / sin 30° ≈ 3.93.

Example 2: no triangle (A = 40°, a = 5, b = 10)

h = 10 × sin 40° = 6.43, and a = 5 is less than h. Check with the Law of Sines: sin B = 10 × sin 40° / 5 = 1.2856. A sine can never be more than 1, so no triangle exists. Side a is too short to reach the base.

Example 3: one triangle (A = 30°, a = 9, b = 8)

Here a > b, so there is one triangle. sin B = 8 × sin 30° / 9 = 0.4444, so B = 26.39°. The supplement, 153.61°, fails the check: 30° + 153.61° = 183.61°, which is over 180°. So C = 123.61° and c = 9 × sin 123.61° / sin 30° ≈ 14.99.

Example 4: exactly one right triangle (A = 30°, a = 4, b = 8)

h = 8 × sin 30° = 4, which equals a. Then sin B = 1, so B = 90°. There is no second angle, because 180° − 90° = 90° is the same one. C = 60.00° and c = 4 × sin 60.00° / sin 30° ≈ 6.93.

Example 5: obtuse angle A

A = 120°, a = 10, b = 6. Since A is obtuse and a > b, there is one triangle: sin B = 6 × sin 120° / 10 = 0.5196, so B = 31.31°, C = 28.69° and c ≈ 5.54.

A = 120°, a = 6, b = 10. Here a < b with an obtuse A, so there is none, and the Law of Sines agrees: sin B = 10 × sin 120° / 6 = 1.4434, which is bigger than 1. But some obtuse cases pass the sine test and still fail: with A = 100°, a = 9.9 and b = 10, sin B = 0.9948 gives B = 84.13°, but then A + B is 184.13°, which is more than 180°. That is why you always check that the angles add up to less than 180°.

Can I just use the Law of Cosines instead?

It does not remove the ambiguity. With SSA, the Law of Cosines turns into a quadratic equation for the missing side. For Example 1 it is c² − 16·cos 30°·c + 39 = 0, and its two solutions, c = 9.93 and c = 3.93, are exactly the two triangles. The ambiguity shows up either way, which is why SSA is the only case that needs the height test. For SAS and SSS, which are never ambiguous, see Law of Sines vs. Law of Cosines.

5 common mistakes

1. Reporting only the acute angle. Inverse sine gives B₁ only. Always test B₂ = 180° − B₁.
2. Assuming there are always two. Two triangles need h < a < b (for an acute A). Many SSA problems have one or none.
3. Checking the wrong thing. The second angle is valid when A + B₂ < 180°, not just when B₂ < 180°.
4. Using the height test with an obtuse A. For A of 90° or more, use the simpler rule: one triangle if a > b, otherwise none.
5. Confusing SSA with SAS. If the angle is between the two sides it is SAS, and there is exactly one triangle.

Practice: how many triangles?

1. A = 35°, a = 7, b = 9. How many triangles are there, and what are they?
2. A = 40°, a = 5, b = 10. How many triangles are there, and what are they?
3. A = 28°, a = 10, b = 7. How many triangles are there, and what are they?
4. A = 110°, a = 8, b = 12. How many triangles are there, and what are they?
5. A = 30°, a = 5, b = 10. How many triangles are there, and what are they?
6. A = 25°, a = 6, b = 9. How many triangles are there, and what are they?

Frequently Asked Questions

1. What is the ambiguous case of the Law of Sines?

   The ambiguous case is SSA: two sides and an angle that is not between them. Because the sine of an angle and the sine of its supplement are equal, the same measurements can describe no triangle, one triangle or two triangles.

2. How do you know if there are two triangles?

   For an acute angle A with opposite side a and the other known side b, compute h = b sin A. There are two triangles when h < a < b. There is one when a = h or a is at least b, and none when a < h.

3. How do you find the second triangle in the ambiguous case?

   Find the acute angle B1 from the Law of Sines, then take B2 = 180 degrees minus B1. If A + B2 is less than 180 degrees, the second triangle exists. Finish it by finding C = 180 - A - B2 and then c = a sin C / sin A.

4. Can the ambiguous case happen with an obtuse angle?

   No. If angle A is 90 degrees or more, the side opposite it must be the longest side. There is one triangle when a is greater than b and none otherwise, so there is never a second triangle.

5. Is there an ambiguous case for the Law of Cosines?

   Not for SAS or SSS, which always give exactly one triangle. If you apply the Law of Cosines to SSA, you get a quadratic equation for the missing side, and it has two solutions in exactly the cases where two triangles exist.

6. What if a equals b in an SSA problem?

   When a = b with an acute angle A, the triangle is isosceles and there is exactly one solution, with B = A. The supplement would make A + B2 equal to 180 degrees, which leaves no room for a third angle.

Keep going: solve any SSA triangle with the Law of Sines Calculator, choose the right rule with Law of Sines vs. Law of Cosines, or check values with the trig functions calculator.

Every worked example and answer on this page was computed and checked independently. Last reviewed October 2026.

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