Area of a Triangle
Master the area of a triangle for NSW Year 12 Mathematics Standard 2. When you know two sides and the angle included between them, the area is \(A=\tfrac{1}{2}ab\sin C\) — a rule that works for acute and obtuse included angles alike.
You will learn to substitute two sides and the included angle to find the area, rearrange the rule to find an unknown side or the included angle, and extend it to parallelograms — core Standard 2 skills for the areas of blocks of land, garden beds, paddocks and sails.
Theory
The area of a triangle can be found from two sides and the included angle using \(A=\tfrac{1}{2}ab\sin C\). This Year 12 Standard 2 (NSW) guide shows how to apply the rule (including for obtuse included angles), rearrange it to find an unknown side or the included angle, and use it for parallelograms and real blocks of land.
The area of a triangle can be found whenever you know two sides and the angle included between them. The included angle is the angle formed where the two sides meet.
The rule is \(A=\tfrac{1}{2}ab\sin C\), where \(a\) and \(b\) are the two known sides and \(C\) is the included angle. It works for both acute and obtuse included angles, because \(\sin C\) stays positive for every angle from \(0^\circ\) to \(180^\circ\).
Rearranging the same rule lets you find an unknown side (\(a=\tfrac{2A}{b\sin C}\)) or the included angle (\(\sin C=\tfrac{2A}{ab}\), then \(\sin^{-1}\)). Because a diagonal splits a parallelogram into two equal triangles, a parallelogram of sides \(a\), \(b\) and included angle \(C\) has area \(ab\sin C\). This is a core Year 12 Standard 2 (NSW) skill for the areas of blocks of land, garden beds and sails.
For a triangle with two sides \(a\) and \(b\) and included angle \(C\):
Rearranged to find an unknown side or the included angle:
How to use the area rule
- Identify the two sides \(a\) and \(b\) and the angle \(C\) included between them.
- Substitute the values into \(A=\tfrac{1}{2}ab\sin C\) (calculator in degree mode).
- Evaluate to find the area, with the correct square units (\(\text{cm}^2\), \(\text{m}^2\)).
- To find a side, substitute the area and rearrange for the unknown side (\(a=\tfrac{2A}{b\sin C}\)).
- To find the angle, rearrange to \(\sin C=\tfrac{2A}{ab}\), then apply \(\sin^{-1}\) and round as asked.
Substitute the two sides and the included angle into \(A=\tfrac{1}{2}ab\sin C\).
| \(A\) | \(=\) | \(\dfrac{1}{2}ab\sin C\) |
| \(\) | \(=\) | \(\dfrac{1}{2}\times 9 \times 14 \times \sin 38^\circ\) |
| \(\) | \(=\) | \(63\sin 38^\circ\) |
| \(\therefore\ A\) | \(\approx\) | \(38.79\text{ cm}^2\) |
The rule works for the obtuse angle too — use \(\sin 125^\circ\).
| \(A\) | \(=\) | \(\dfrac{1}{2}ab\sin C\) |
| \(\) | \(=\) | \(\dfrac{1}{2}\times 11 \times 16 \times \sin 125^\circ\) |
| \(\) | \(=\) | \(88\sin 125^\circ\) |
| \(\therefore\ A\) | \(\approx\) | \(72.09\text{ m}^2\) |
Substitute the known values, then rearrange for \(a\).
| \(45\) | \(=\) | \(\dfrac{1}{2}\times a \times 15 \times \sin 40^\circ\) |
| \(45\) | \(=\) | \(4.8209\ldots\, a\) |
| \(a\) | \(=\) | \(\dfrac{45}{4.8209\ldots}\) |
| \(\therefore\ a\) | \(\approx\) | \(9.33\text{ cm}\) |
Make \(\sin C\) the subject, then apply the inverse.
| \(60\) | \(=\) | \(\dfrac{1}{2}\times 12 \times 17 \times \sin C\) |
| \(60\) | \(=\) | \(102\sin C\) |
| \(\sin C\) | \(=\) | \(\dfrac{60}{102}=0.5882\ldots\) |
| \(C\) | \(=\) | \(\sin^{-1}(0.5882\ldots)=36.03\ldots^\circ\) |
| \(\therefore\ C\) | \(\approx\) | \(36^\circ\) |
Common pitfalls
Frequently asked questions
When can I use A = 1/2 ab sin C?
Use it whenever you know two sides of a triangle and the angle included between those two sides. If you instead know three sides, or a side and a non-included angle, you first find the included angle another way (for example with the cosine rule or the angle sum) before applying the area rule.
What is the included angle?
The included angle is the angle formed where the two known sides meet. In the formula A equals a half a b sin C, the sides a and b are the two sides and C is the angle between them, not any other angle in the triangle.
Does the area rule work for an obtuse triangle?
Yes. The sine of any angle between 0 and 180 degrees is positive, so sin of an obtuse angle such as 125 degrees is a positive number. You substitute it in exactly the same way, so the rule works for acute, right-angled and obtuse included angles.
How do I find a missing side or the included angle from the area?
Substitute what you know into A equals a half a b sin C. To find a side, rearrange to a equals 2A over b sin C. To find the included angle, rearrange to sin C equals 2A over a b, then take the inverse sine and round as asked.
How is the area of a parallelogram related to this rule?
A diagonal splits a parallelogram into two congruent triangles. Each triangle has area a half a b sin C, so the whole parallelogram has area a b sin C, which is exactly twice the triangle. That is why forgetting the one half gives the parallelogram area instead.