Area of a triangle: A = ½ ab sin C
Area of a Triangle with \(\tfrac12 ab\sin C\)
Area of a Triangle with \(\tfrac12 ab\sin C\) is a topic in Right Triangles & Trigonometry in the California Common Core State Standards. It is aligned to Standard G-SRT.9, which requires students to derive and use the formula for the area of a triangle as one half the product of two sides and the sine of the included angle.
The area of a triangle is \(\dfrac12 ab\sin C\) from two sides and the included angle.
Theory
When you know two sides and the angle between them, the area of a triangle is
where \(a\) and \(b\) are the two sides and \(C\) is the included angle.
The area formula:
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.
Use \(\text{Area}=\dfrac12 ab\sin C\).
| \(\text{Area}\) | \(=\) | \(\dfrac12(8)(5)\sin 30^\circ\) |
| \(=\) | \(20(0.5)=10\) |
Substitute into the formula.
| \(\text{Area}\) | \(=\) | \(\dfrac12(12)(9)\sin 105^\circ\) |
| \(\approx\) | \(52.2\) |
Set up \(\dfrac12 ab\sin C=30\) with \(\sin 90^\circ=1\).
| \(\dfrac12\cdot a\cdot 12\cdot 1\) | \(=\) | \(30\) |
| \(6a\) | \(=\) | \(30\) |
| \(a\) | \(=\) | \(5\) |
Two sides and the included angle.
| \(\text{Area}\) | \(=\) | \(\dfrac12(40)(55)\sin 105^\circ\) |
| \(\approx\) | \(1062\ \text{ft}^2\) |
Common pitfalls
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.