Distinguish zero, one and two triangles in SSA data
Use a perpendicular height to explain acute SSA cases · Handle given right or obtuse angles separately · Exclude zero-length and zero-angle degeneracy
Section Paper 2 · Historical Extended · non-calculator
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
- Ambiguous sine rule
- Triangle inequalities
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
- Use a perpendicular height to explain acute SSA cases
- Handle given right or obtuse angles separately
- Exclude zero-length and zero-angle degeneracy
For an acute given A
With A acute and known side b, the perpendicular distance from the endpoint of b to the other ray is h=b sinA. If a<h there is no intersection and no triangle; if a=h there is one right-angle contact; if h<a<b there are two triangles; if a≥b there is one non-degenerate triangle. At a=b, the other intersection is the starting point and gives a zero side, so it must be discarded.
Do not reuse that table for every A
For a given right or obtuse A, its opposite side a must be larger than b for a valid triangle; then one valid triangle results. If a≤b, none exists. This agrees with testing both candidate B values and requiring C>0. A ratio inside the inverse-sine range is only one necessary condition, not the complete existence test.
Sources for this lesson
Pause and explain
For a given obtuse A, can its opposite side a be no larger than b in a valid triangle?
Worked example
For A=30°, b=10 cm, compare a=4,5,7,10 cm.
Show the worked solution
- The perpendicular height h=10sin30°=5 cm.
- Compare each a with h and b; discard degenerate intersections.
Answer Respectively 0,1,2,1 non-degenerate triangles.
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.
For A=30°, b=10 cm, count non-degenerate triangles when a is 4,5,7,10 cm.
Give me a hint
Use h=5 and the acute-angle case conditions.
Compare my reasoning
- a=4 gives none; a=5 gives one tangent contact.
- a=7 gives two; a=10 gives one after discarding the zero-side case.
0,1,2,1 respectively.
Look for these in your work
- Use the height threshold
- Exclude the degenerate case
For A=120° and b=10 cm, count triangles when a is 12,10,8 cm.
Give me a hint
The side opposite the obtuse given angle must be larger.
Compare my reasoning
- a=12>b gives one valid triangle.
- a=10 or 8 is not larger than b, so each gives none.
1,0,0 respectively.
Look for these in your work
- Check the given-angle type
- Use the longest-side condition
A student accepts A=120°, a=10 cm, b=11 cm because the sine-rule ratio is less than 1. Identify the missing checks.
Give me a hint
Compare sides and check candidate angle sums.
Compare my reasoning
- a is smaller than b despite being opposite the obtuse angle.
- Both possible B values leave C≤0; a real inverse alone is insufficient.
No triangle: the side-angle condition and both angle sums fail.
Look for these in your work
- Distinguish necessary from sufficient conditions
- Verify the full triangle
Common mistakes
- The height-case table assumes the given angle is acute.
- A touching or repeated starting point can be degenerate.
What can you explain now?
For a given obtuse A, can its opposite side a be no larger than b in a valid triangle?
Compare with the explanation
No; the side opposite the obtuse angle must be the longest.
A≤b contradicts that necessary side-angle relationship.
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Make it stick
For a given obtuse A, can its opposite side a be no larger than b in a valid triangle?
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.