Two sides and the angle between them: When you know two sides and the angle between them (the included angle), the area is ½ × (one side) × (other side) × sin(included angle).
C must be the included angle: The angle in the formula must sit between the two sides you use — not just any angle of the triangle.
Substitute and evaluate: Identify the two sides and the angle between them, drop them into ½ab·sinC, and evaluate.
IB-style question — straight area
A triangle has sides 6 and 8 with an included angle of 30°. Find its area.
Step by step
- Substitute.
- Evaluate (sin 30° = ½).
Final answer
Area = 12.
Pick the right two sides: Use the two sides that enclose the given angle — if a different angle is given, use the sides next to it.
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Given the area, find a side or the angle: Set ½ab·sinC equal to the given area and solve for the unknown — a side (rearrange) or the included angle (then use sin⁻¹, watching for the obtuse possibility).
IB-style question — find the angle
A triangle with sides 10 and 12 has area 30. Find the included angle (acute).
Step by step
- Set up.
- Solve.
Final answer
C = 30° (the acute solution).
Two possible angles: sin C = 0.5 also gives C = 150°. Use the context (or 'acute/obtuse') to choose — both can give the same area.
Find a missing side/angle first, then the area: Many questions need a two-step approach: use the sine or cosine rule to find a missing side or the included angle, then apply ½ab·sinC.
IB-style question — cosine rule then area
A triangle has sides 5 and 7 with included angle 80°. Find its area, then the third side.
Step by step
- Area directly (included angle given).
- Third side by the cosine rule.
Final answer
Area ≈ 17.2; third side ≈ 7.85.
Read what you're given: If the included angle is missing, find it first (cosine rule from SSS, or sine rule); the area formula needs it.