Use two sides and their included angle for triangle area
Use (1/2)ab sinC with the included angle · Interpret perpendicular height for an obtuse angle · Avoid inferring congruence from equal areas
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
- Sine values
- Perpendicular triangle heights
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 (1/2)ab sinC with the included angle
- Interpret perpendicular height for an obtuse angle
- Avoid inferring congruence from equal areas
The included angle supplies a height
For positive sides a,b enclosing 0°<C<180°, area K=(1/2)ab sinC. The historical syllabus supplies this area formula. With a as the base, the perpendicular height is b sinC, even when an obtuse angle places the foot outside the base segment. An angle elsewhere in the triangle cannot simply replace the included C.
Equal sines can give equal areas
Supplementary angles C and 180°−C have equal sines, so the same two side lengths can enclose equal areas at different angles. Their cosine-rule third sides generally differ; equal area does not establish congruence. For fixed a,b, the maximum area is ab/2 at C=90°, because sine is at most 1.
Sources for this lesson
Pause and explain
Does a negative cosine for an obtuse angle make triangle area negative?
Worked example
Find area for sides 8 cm and 5 cm with included angle 30°.
Show the worked solution
- K=(1/2)×8×5×sin30°.
- Use sin30°=1/2.
Answer 10 cm².
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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 area for sides 8 cm and 5 cm enclosing 30°.
Give me a hint
Use the included-angle area formula.
Compare my reasoning
- K=(1/2)×8×5×(1/2).
- K=10 cm².
10 cm².
Look for these in your work
- Identify the included angle
- Use square units
Find area for sides 6 m and 9 m enclosing 120°, exactly and to 3 significant figures.
Give me a hint
Use the positive sine value √3/2.
Compare my reasoning
- K=27×√3/2 m².
- This is approximately 23.382686 m².
(27√3)/2 m² ≈ 23.4 m².
Look for these in your work
- Use sine rather than cosine
- Retain an exact form
Two triangles each have sides 5 cm and 8 cm, but their included angles are 40° and 140°. Compare their areas and decide whether equal area proves congruence.
Give me a hint
Supplementary angles have the same sine.
Compare my reasoning
- Both areas are 20sin40° cm²≈12.855752 cm².
- The third sides have different cosine-rule values, so equal area does not make these triangles congruent.
Both areas ≈12.9 cm²; the triangles are not congruent.
Look for these in your work
- Use the sine relationship
- Distinguish area from shape
Common mistakes
- Use the angle enclosed by the two side lengths.
- An obtuse angle can still give a positive height and area.
What can you explain now?
Does a negative cosine for an obtuse angle make triangle area negative?
Compare with the explanation
No; the area formula uses sine, which is positive for 0°<C<180°.
The positive perpendicular height is determined by sine, not cosine.
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Make it stick
Does a negative cosine for an obtuse angle make triangle area negative?
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