Find both SSA angle candidates and validate each triangle
Obtain sinB from a known opposite pair · Test the acute and supplementary candidates · Find a valid third angle and side for each accepted triangle
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Sine rule
- Supplementary sine values
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Obtain sinB from a known opposite pair
- Test the acute and supplementary candidates
- Find a valid third angle and side for each accepted triangle
An inverse sine is only the first candidate
Given A, its opposite side a, and another side b, sinB=(b sinA)/a. For a non-degenerate triangle require a,b>0 and 0°<A<180°. If the ratio exceeds 1, no real B exists. For a ratio between 0 and 1, inverse sine gives B₁ and the other possible angle is B₂=180°−B₁. If the ratio is exactly 1, both candidates are 90° and count only once.
Check the third angle for every candidate
Keep a candidate only if A+B<180°, making C=180°−A−B positive. Then c=a sinC/sinA. For A=30°, a=7, b=10, the two B values are approximately 45.5847° and 134.4153°. Both pass the sum test; their c values are approximately 13.5592 and 3.76127.
The second B is obtuse and still valid here. If the given A is obtuse, only a suitable acute B can fit; an apparently valid sine ratio alone does not guarantee even one triangle. Keep unrounded angles while checking the sum.
Sources for this lesson
Pause and explain
Must an obtuse supplementary B be rejected automatically?
Worked example
For A=30°, a=7 cm and b=10 cm, find both possible B angles and c lengths to 3 significant figures.
Show the worked solution
- sinB=5/7 gives B₁≈45.5847° and B₂≈134.4153°; both leave a positive C.
- Use each C in c=7sinC/sin30°.
Answer B≈45.6°, c≈13.6 cm; or B≈134°, c≈3.76 cm.
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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.
For A=30°, a=7 cm, b=10 cm, find the two B candidates to 1 decimal place and decide whether each is valid.
Give me a hint
Test the angle sum for both branches.
Compare my reasoning
- B₁≈45.5847° and B₂≈134.4153°.
- Adding A=30° leaves a positive third angle in both cases.
45.6° and 134.4°; both are valid.
Look for these in your work
- Generate both candidates
- Validate each angle sum
For those two triangles, find C to 1 decimal place and c to 3 significant figures, preserving branch order.
Give me a hint
Use the unrounded B values.
Compare my reasoning
- For the first: C≈104.4153°, c≈13.5592 cm.
- For the second: C≈15.5847°, c≈3.76127 cm.
C≈104.4°, c≈13.6 cm; or C≈15.6°, c≈3.76 cm.
Look for these in your work
- Keep each angle-side branch together
- Avoid premature rounding
Given A=120°, a=10 cm and b=11 cm, sinB is about 0.952628. Does its real inverse-sine value prove that a triangle exists?
Give me a hint
Test both resulting angle sums.
Compare my reasoning
- B₁≈72.2937° and B₂≈107.7063°.
- Either added to 120° exceeds 180°, so neither leaves a positive C.
No; both candidates fail the angle-sum condition.
Look for these in your work
- Check beyond the sine range
- Reject invalid third angles
Common mistakes
- Do not stop at the principal inverse-sine value.
- Reject a candidate with no positive third angle.
What can you explain now?
Must an obtuse supplementary B be rejected automatically?
Compare with the explanation
No; it is valid if A+B<180° and all side conditions hold.
An obtuse angle is allowed in a non-right triangle.
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Make it stick
Must an obtuse supplementary B be rejected automatically?
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