The ambiguous case appears when you are given two sides and an angle not between them. The same data can make no triangle, one triangle or two triangles.
This lesson is part of solution of triangles. It assumes you can already choose the sine rule for a side-angle pair.
Why can there be two answers?
Sine gives the same value for an angle and its supplement, so sin 30° = sin 150° = 0.5. When you use the sine rule to find an angle B, the calculator returns only the smaller one, yet the larger one, 180° − B, may also form a triangle.
Both answers are valid if the larger angle plus the given angle is still less than 180°.
Worked example: two triangles
In triangle ABC, angle A = 30°, a = 5 cm and b = 8 cm. Find angle B and the side c.
Step 1. Sine rule: sin B = b sin A ÷ a = 8 × 0.5 ÷ 5 = 0.8.
Step 2. B = 53.13° or B = 180° − 53.13° = 126.87°.
Step 3. Test each: 30° + 53.13° = 83.13° and 30° + 126.87° = 156.87°. Both are less than 180°, so both triangles exist.
| Triangle 1 | Triangle 2 | |
|---|---|---|
| B | 53.13° | 126.87° |
| C = 180° − 30° − B | 96.87° | 23.13° |
| c = a sin C ÷ sin A | 10 sin 96.87° ≈ 9.93 cm | 10 sin 23.13° ≈ 3.93 cm |
A short check with sides: the largest angle faces the longest side. In triangle 1, C = 96.87° is largest and c = 9.93 cm is longest. In triangle 2, B = 126.87° is largest and b = 8 cm is longest.
When is there no triangle, one triangle or two?
For A acute, compare a with b and with b sin A.
| Condition | Number of triangles |
|---|---|
| a < b sin A | None (sin B would exceed 1) |
| a = b sin A | One (a right angle at B) |
| b sin A < a < b | Two |
| a ≥ b | One |
Here are two quick cases. With A = 40°, a = 9 and b = 6: sin B = 6 sin 40° ÷ 9 ≈ 0.4285, so B ≈ 25.4°. The other angle, 154.6°, plus 40° is more than 180°, so only one triangle fits.
With A = 30°, a = 3 and b = 8: sin B = 8 × 0.5 ÷ 3 = 1.33, which is impossible, so no triangle exists.
The mistake that costs marks
The common slip is to stop after the first angle. The working looks complete and the angle is correct, yet half the solution is missing.
| Step | Wrong | Right |
|---|---|---|
| Find B | B = 53.13° | B = 53.13° or 126.87° |
| Test | (skipped) | Does 30° + 126.87° < 180°? Yes |
| Answer | One triangle | Two triangles |
Make it a rule: after every sin⁻¹ in a sine-rule question, write the supplement and test it.
Check yourself
In triangle ABC, angle A = 35°, a = 7 cm and b = 10 cm. How many triangles are possible? Find the possible values of angle B.
Answer
sin B = 10 sin 35° ÷ 7 = 5.736 ÷ 7 ≈ 0.8194, so B ≈ 55.0° or B ≈ 180° − 55.0° = 125.0°.
Test: 35° + 55.0° = 90.0° < 180° and 35° + 125.0° = 160.0° < 180°. Both fit, so there are two triangles, with B = 55.0° or 125.0°.
Condition check: b sin A = 5.74 < a = 7 < b = 10, which matches the two-triangle row.
What to study next
Next, see how area works when the given angle is not between the known sides, in calculating triangle area with a non-included height. Then try the solution of triangles practice set.
If you want a teacher to work through ambiguous-case questions with you, see online one-to-one Additional Mathematics tuition.