USA - Geometry
Right triangles and trigonometry
Area of a triangle: A = ½ ab sin C
20 practice questions
2 video lessons
Theory + worked examples
Theory
When you know two sides and the angle between them, the area of a triangle is
\[\text{Area}=\dfrac12 ab\sin C,\]
where \(a\) and \(b\) are the two sides and \(C\) is the included angle.
This is base times height in disguise: \(b\sin C\) is the height dropped onto side \(a\).
\(\text{Area}=\dfrac12 ab\sin C\) with the included angle \(C\).
The SAS area formula.
The area formula:
\[\text{Area}=\dfrac12 ab\sin C\]
The angle must be included — between the two sides \(a\) and \(b\).
How to find the area
- Identify two sides and the angle between them.
- Substitute into \(\dfrac12 ab\sin C\).
- Evaluate, keeping square units.
- Rearrange to solve for a side if the area is given.
Example 1 — Basic area
Find the area with \(a=8\), \(b=5\), included angle \(C=30^\circ\).
Solution
Use \(\text{Area}=\dfrac12 ab\sin C\).
| \(\text{Area}\) | \(=\) | \(\dfrac12(8)(5)\sin 30^\circ\) |
| \(=\) | \(20(0.5)=10\) |
Example 2 — A larger angle
Find the area with \(a=12\), \(b=9\), \(C=105^\circ\).
Solution
Substitute into the formula.
| \(\text{Area}\) | \(=\) | \(\dfrac12(12)(9)\sin 105^\circ\) |
| \(\approx\) | \(52.2\) |
Example 3 — Solve for a side
A triangle has area \(30\), \(b=12\), and \(C=90^\circ\). Find \(a\).
Solution
Set up \(\dfrac12 ab\sin C=30\) with \(\sin 90^\circ=1\).
| \(\dfrac12\cdot a\cdot 12\cdot 1\) | \(=\) | \(30\) |
| \(6a\) | \(=\) | \(30\) |
| \(a\) | \(=\) | \(5\) |
Example 4 — Real-world area
A triangular plot has sides \(40\) ft and \(55\) ft meeting at \(105^\circ\). Find its area.
Solution
Two sides and the included angle.
| \(\text{Area}\) | \(=\) | \(\dfrac12(40)(55)\sin 105^\circ\) |
| \(\approx\) | \(1062\ \text{ft}^2\) |
Common pitfalls
The angle must be included between the two sides you use.
Area is in square units. Don't drop the \(^2\).
Half, not whole: the formula has a factor of \(\dfrac12\).
Frequently asked questions
How do you find a triangle's area from two sides and an angle?
Use \(\text{Area}=\dfrac12 ab\sin C\), where \(C\) is the angle between sides \(a\) and \(b\).
Does this formula work for any triangle?
Yes — any triangle where you know two sides and the included angle, not just right triangles.
Why does the formula work?
Because \(b\sin C\) is the height of the triangle on base \(a\), so \(\dfrac12 ab\sin C\) is one-half base times height.
What if the area is known and a side is missing?
Substitute the known values into \(\dfrac12 ab\sin C\) and solve for the unknown side.
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