The ambiguous sine-rule case
Check the acute and supplementary angles when two sides and a non-included angle are given.
Extended only. This is the separate 9–1 qualification 0980. Cambridge intends Core for grades 4–1 and Extended for grades 9–5; the eligible award bands are grades 5–1 for the Core papers and grades 9–3 for the Extended papers.
Before you begin
- Use the sine rule and inverse sine.
- Know sin θ = sin(180° − θ).
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which other angle has the same sine as 40° in a triangle?
What you will learn
- Identify when inverse sine can produce two triangle shapes.
- Reject supplementary candidates that violate the angle sum.
Two candidate angles may share the same sine
With A = 30°, a = 8 and b = 12 cm, the sine rule gives sin B = 12 sin 30°/8 = 0.75. Inverse sine gives B ≈ 48.6°, but its supplement 131.4° has the same sine. Both can form a triangle here because each leaves a positive third angle.
For the acute candidate, C ≈ 101.4°; for the obtuse candidate, C ≈ 18.6°. Use unrounded angles when calculating another side. The data are side–side–angle rather than two sides with an included angle, so one sketch alone should not silently select a unique solution.
Test every candidate against the geometry
If the sine-rule calculation gives a value greater than one, no real triangle is possible. For A = 30°, a = 5 and b = 12, sin B = 1.2, so the data are impossible. If the sine equals one, B is 90° and its supplement is the same candidate, not a second triangle.
A supplementary angle is allowed only when A + B < 180°. For A = 40°, a = 10 and b = 7, inverse sine gives about 26.7°, whereas the supplement is about 153.3°; the latter plus forty exceeds 180°. Report only the valid candidate. Check any diagram constraints as well as the angle sum.
Pause and explain
A = 30°, a = 8, b = 12. How many valid triangle possibilities are there?
Worked example
A = 30°, a = 8 cm, b = 12 cm. Find all possible B values to one decimal place.
Show the worked solution
- sin B = 12 sin 30°/8 = 0.75.
- B₁ ≈ 48.6°, B₂ = 180° − B₁ ≈ 131.4°.
- Both leave a positive C after subtracting A and B from 180°.
Answer 48.6° or 131.4°.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A = 30°, a = 5, b = 12. Decide whether a triangle exists.
Give me a hint
Calculate sin B before taking an inverse.
Compare my reasoning
- sin B = 12 × 0.5/5 = 1.2.
- No real angle has sine above one.
- Therefore no triangle satisfies these data.
No triangle.
Look for these in your work
- Checked the permissible sine range.
- Explained the impossibility.
A = 40°, a = 10, b = 7. Find valid B values to one decimal place.
Give me a hint
Test the supplement against A + B < 180°.
Compare my reasoning
- sin B = 7 sin 40°/10, giving B₁ ≈ 26.7°.
- The supplementary B₂ ≈ 153.3° makes A + B₂ exceed 180°.
- Only B = 26.7° is possible.
26.7° only.
Look for these in your work
- Considered both sine candidates.
- Rejected the invalid angle sum.
A = 30°, a = 6, b = 12. How many triangles are possible, and what is B?
Give me a hint
A computed sine of one gives a unique right angle.
Compare my reasoning
- sin B = 12 × 0.5/6 = 1.
- B = 90°; the supplement is also ninety, not a distinct candidate.
- C = 60°, so exactly one triangle is possible.
One triangle; B = 90°, C = 60°.
Look for these in your work
- Avoided double-counting identical candidates.
- Checked the remaining angle.
Common mistakes
- Assuming every side–side–angle case has two solutions.
- Accepting a candidate with zero or negative remaining angle.
- Rounding before finding a second side.
What can you explain now?
What extra check follows inverse sine for a triangle angle?
Compare with the explanation
Test the supplementary angle and the triangle angle sum.
Inverse sine returns a principal angle; another triangle may satisfy the same side–angle data.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Tomorrow, describe examples with zero, one and two possible triangles.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.