Check the area bound and find both possible included angles
Use sinC=2K/(ab) · Check whether the given area can occur · Distinguish two angles, one maximum-area angle and degenerate zero area
Section Paper 4 · Historical Extended · 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
- Triangle area formula
- Supplementary sine values and triangle feasibility
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 sinC=2K/(ab)
- Check whether the given area can occur
- Distinguish two angles, one maximum-area angle and degenerate zero area
Area alone may not determine the triangle
From K=(1/2)ab sinC, obtain sinC=2K/(ab). For 0<K<ab/2, both the acute inverse angle and its obtuse supplement can be possible included angles. For a=6,b=8,K=12, sine is 1/2, so C=30° or 150°. Both are non-degenerate, but their third sides differ.
Check the limits
At K=ab/2, sine is 1 and C=90° is the only included angle in (0°,180°). K>ab/2 is impossible for those sides. K=0 requires a straight or collapsed arrangement and is not a positive-area triangle with the stated positive sides. Additional angle or side information can select one branch when two are otherwise possible.
Sources for this lesson
Pause and explain
Can sides 6 cm and 8 cm enclose a triangle of area 30 cm²?
Worked example
For a=6 cm,b=8 cm and area 12 cm², find all possible included angles.
Show the worked solution
- sinC=24/48=1/2.
- Both 30° and 150° lie strictly between 0° and 180°.
Answer 30° or 150°.
Try a topic test
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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.
Find the possible included angles for sides 6 cm and 8 cm with area 12 cm².
Give me a hint
Calculate the sine ratio and both allowed branches.
Compare my reasoning
- sinC=1/2.
- The acute and supplementary candidates are 30° and 150°.
30° or 150°.
Look for these in your work
- Find both angles
- Check the triangle interval
For sides 6 cm and 8 cm with area 24 cm², find the included angle and explain whether there are two distinct branches.
Give me a hint
This is the maximum area.
Compare my reasoning
- sinC=48/48=1.
- The acute and supplementary constructions both give 90°, so count it once.
90° only.
Look for these in your work
- Check the area bound
- Do not duplicate an identical angle
For the same positive side lengths, classify requested areas 30 cm² and 0 cm² as possible non-degenerate triangles.
Give me a hint
Compare with the area range 0<K≤24.
Compare my reasoning
- 30 exceeds the maximum and is impossible.
- Zero area is degenerate rather than a non-degenerate triangle.
30 cm² is impossible; 0 cm² is degenerate; neither meets the request.
Look for these in your work
- Use the full area conditions
- Separate impossible from degenerate
Common mistakes
- The principal inverse may omit a valid obtuse angle.
- Check the maximum possible area before finding angles.
What can you explain now?
Can sides 6 cm and 8 cm enclose a triangle of area 30 cm²?
Compare with the explanation
No; their maximum area is 24 cm².
The area is at most (1/2)×6×8 because sine cannot exceed 1.
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Make it stick
Can sides 6 cm and 8 cm enclose a triangle of area 30 cm²?
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.