Triangle area from an included angle
Find non-right triangle area using two sides and their included angle.
Section Paper 4 · Extended · Calculator
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 sine to obtain a perpendicular component.
- Recognise an included triangle angle.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What height is needed for ½ × base × height?
What you will learn
- Use area = ½ab sin C.
- Explain the perpendicular height behind the formula.
Sine supplies the perpendicular height
For two sides a and b with included angle C, triangle area is ½ab sin C. Think of b as a base: the perpendicular height is a sin C, so the familiar half-base-times-height formula gives the result. The third side is not needed for this calculation.
For sides 8 and 10 cm with included angle 30°, area = ½ × 8 × 10 × 0.5 = 20 cm². Use square units. A side length along a slope is not a perpendicular height; substituting it directly into half-base-times-height would overestimate the area.
Obtuse angles still give positive areas
The formula also applies when C is obtuse. The perpendicular height can fall outside the triangle, but sin C remains positive for triangle angles between zero and 180 degrees. For sides 6 and 9 with included angle 120°, area = 27 sin 120° = 27√3/2 cm².
The angles C and 180° − C have equal sine, so the same two side lengths can give equal areas with different shapes. State which angle is included, keep exact surds if requested and round only the final numerical area. A zero or straight included angle gives a degenerate shape with no positive area.
Pause and explain
Sides 8 and 10 cm meet at 30°. What is the area?
Worked example
Sides 8 and 10 cm meet at 30°. Find the triangle area.
Show the worked solution
- The given angle lies between the two supplied sides.
- Area = ½ × 8 × 10 × sin 30°.
- The result is 20 cm², since sine thirty is one half.
Answer 20 cm².
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.
Sides 5 and 12 m meet at 90°. Find area.
Give me a hint
Sine ninety equals one.
Compare my reasoning
- Area = ½ × 5 × 12 × 1.
- The product is 30 m².
- At a right angle this is the usual half-base-times-height result.
30 m².
Look for these in your work
- Used the half factor.
- Reported square units.
Sides 6 and 9 cm meet at 120°. Find exact area.
Give me a hint
Use sine one hundred twenty = √3/2.
Compare my reasoning
- Area = 27 sin 120°.
- Substitute √3/2 to obtain 27√3/2 cm².
- The obtuse angle still has positive sine.
27√3/2 cm².
Look for these in your work
- Kept an exact surd.
- Used the included obtuse angle.
With sides 8 and 10 fixed, compare areas for included angles 30° and 150°.
Give me a hint
The two angles have equal sine.
Compare my reasoning
- Both sin 30° and sin 150° equal one half.
- Both areas are ½ × 8 × 10 × ½ = 20 cm².
- Equal areas do not imply equal third sides or identical triangles.
Both areas are 20 cm²; the triangles have different shapes.
Look for these in your work
- Recognised supplementary sine values.
- Separated equal area from congruence.
Common mistakes
- Using an angle opposite a supplied side.
- Omitting the half factor.
- Treating an obtuse triangle's area as negative.
What can you explain now?
Which angle is required in ½ab sin C?
Compare with the explanation
The angle between sides a and b.
That included angle determines the perpendicular component of one side relative to the other.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Tomorrow, derive the area formula by drawing a perpendicular height.
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.