How to Find the Area of a Triangle Using Trigonometry
The area of a triangle is normally half base times height, but what if you do not know the height? If you know two sides and the angle between them, the formula half ab sin C gives the area directly. On this page you will learn how to use it: find the area from two sides and the included angle, work backwards to find a missing side or angle when the area is given, and tackle the harder questions that ask for the perimeter or an upper bound. Worked examples show every step, and the practice room generates unlimited randomised questions across all of these types, auto-marked to three significant figures.
The area of a triangle using trigonometry
You already know the area of a triangle is half base times height. But what if you do not know the height? If you know two sides and the angle between them, you can find the area directly with a trigonometry formula, no height needed.
The angle must be the one between the two sides you are using, with one known side on either side of it. This is called the included angle. A quick check: if you stand at the angle, the two sides you know should be the two arms going out from it. Label the two sides \(a\) and \(b\), and the included angle \(C\) (the angle opposite the third side).
\[\text{Area} = \tfrac{1}{2}\,ab\sin C\]
In words: half the product of two sides, times the sine of the angle between them.Is this on the formula sheet? Yes. The area formula \(\tfrac{1}{2}ab\sin C\), along with the sine and cosine rules, is printed on the formula sheet at the front of the Edexcel IGCSE Maths A (4MA1) Higher tier paper, so you do not have to memorise it. The marks come from using it correctly: picking the two sides with the angle between them, and rearranging it when the area is given and a side or angle is missing.
When and how to use it
The method
- Identify two sides and check the angle is the one between them.
- Substitute into \(\text{Area} = \tfrac{1}{2}ab\sin C\).
- Work backwards if the area is given: rearrange for the missing side \(\left(a = \dfrac{2\,\text{Area}}{b\sin C}\right)\) or angle \(\left(\sin C = \dfrac{2\,\text{Area}}{ab}\right)\).
- Round to 3 significant figures only at the very end.
Worked examples
💡 Example 1: finding the area
In triangle ABC, a = 9 cm, b = 12 cm, and the included angle C = 58°. Find the area.
\[\begin{array}{rcl} \text{Area} &=& \tfrac{1}{2}ab\sin C \\ &=& \tfrac{1}{2}(9)(12)\sin 58^\circ \\ &=& 54\sin 58^\circ \\ &=& 45.8\ \text{cm}^2\ \text{(3 s.f.)} \end{array}\]
What's happening?
The 58° angle sits between the 9 cm and 12 cm sides, so it is the included angle and the formula applies directly.
Keep the full calculator value before rounding the final area.
💡 Example 2: an obtuse included angle
In triangle ABC, a = 7 cm, b = 10 cm, and the included angle C = 125°. Find the area.
\[\begin{array}{rcl} \text{Area} &=& \tfrac{1}{2}(7)(10)\sin 125^\circ \\ &=& 35\sin 125^\circ \\ &=& 28.7\ \text{cm}^2\ \text{(3 s.f.)} \end{array}\]
What's happening?
The formula works for obtuse angles too. The sine of an obtuse angle is still positive, so the area comes out positive as expected.
💡 Example 3: finding a side from the area
Triangle ABC has area 40 cm², side b = 11 cm, and included angle C = 47°. Find side a.
\[\begin{array}{rcl} \text{Area} &=& \tfrac{1}{2}ab\sin C \\ 40 &=& \tfrac{1}{2}(a)(11)\sin 47^\circ \\ a &=& \dfrac{2 \times 40}{11\sin 47^\circ} \\ &=& 9.94\ \text{cm}\ \text{(3 s.f.)} \end{array}\]
What's happening?
Rearrange the formula to make \(a\) the subject. Multiply the area by 2, then divide by \(b\sin C\).
💡 Example 4: finding the angle from the area
Triangle ABC has area 33 cm², with sides a = 9 cm and b = 10 cm. Find the included angle C (acute).
\[\begin{array}{rcl} 33 &=& \tfrac{1}{2}(9)(10)\sin C \\ \sin C &=& \dfrac{2 \times 33}{9 \times 10} \\ &=& 0.7333\ldots \\ C &=& \sin^{-1}(0.7333\ldots) \\ &=& 47.2^\circ\ \text{(3 s.f.)} \end{array}\]
What's happening?
Rearrange for \(\sin C\), then use inverse sine. We take the acute answer here; a question may specify if the obtuse angle is wanted.
💡 Example 5: finding the perimeter from the area
Triangle ABC has area 50 cm², side b = 14 cm, and included angle C = 60°. Find the perimeter.
First find side \(a\) from the area:
\[\begin{array}{rcl} a &=& \dfrac{2 \times 50}{14\sin 60^\circ} \\ &=& 8.25\ \text{cm} \end{array}\]
Now the third side \(c\) (opposite \(C\)) by the cosine rule:
\[\begin{array}{rcl} c^2 &=& a^2 + b^2 - 2ab\cos 60^\circ \\ c &=& 12.2\ \text{cm} \end{array}\]
Add the three sides:
\[\begin{array}{rcl} \text{Perimeter} &=& 8.25 + 14 + 12.2 \\ &=& 34.4\ \text{cm}\ \text{(3 s.f.)} \end{array}\]
What's happening?
This is a two-step problem. The area gives you the missing side, then the cosine rule gives the third side because you now know two sides and the angle between them.
Carry full accuracy through both steps; round only the final perimeter.
💡 Example 6: upper bound of the area
A triangular plot has two sides measured as 8 cm and 11 cm (each to the nearest cm) and an included angle of 40° (to the nearest degree). Find the upper bound for the area.
Use the upper bound of every measurement:
sides \(8.5\) cm and \(11.5\) cm, angle \(40.5^\circ\)
\[\begin{array}{rcl} \text{Area}_{\max} &=& \tfrac{1}{2}(8.5)(11.5)\sin 40.5^\circ \\ &=& 31.7\ \text{cm}^2\ \text{(3 s.f.)} \end{array}\]
What's happening?
Each measurement rounded to the nearest unit could be up to half a unit larger. Sine increases up to 90°, so using the larger angle here also increases the area. Using every upper bound gives the largest possible area.
💡 Example 7: an equilateral triangle
Find the area of an equilateral triangle of side 20 cm.
\[\begin{array}{rcl} \text{Area} &=& \tfrac{1}{2}(20)(20)\sin 60^\circ \\ &=& 200\sin 60^\circ \\ &=& 173\ \text{cm}^2\ \text{(3 s.f.)} \end{array}\]
What's happening?
An equilateral triangle has all three sides equal and every angle \(60^\circ\). So take any two of the equal sides (here \(20\) and \(20\)) with the \(60^\circ\) between them, and the formula does the rest — no height needed.
💡 Example 8: an isosceles triangle
An isosceles triangle has two equal sides of 16 cm and an area of 100 cm². Find the angle between the two equal sides.
\[\begin{array}{rcl} 100 &=& \tfrac{1}{2}(16)(16)\sin C \\ \sin C &=& \dfrac{2 \times 100}{16 \times 16} \\ &=& 0.78125 \\ C &=& \sin^{-1}(0.78125) \\ &=& 51.4^\circ\ \text{(3 s.f.)} \end{array}\]
What's happening?
The two equal sides are exactly the pair on either side of the angle you want, so \(a\) and \(b\) are both \(16\). From there it is the same rearrangement as finding any angle from the area.
🔑 Key points
- Area \(= \tfrac{1}{2}ab\sin C\): two sides and the angle between them.
- The angle must be the included angle (between the two sides).
- Given the area, rearrange to find a missing side or angle.
- Equilateral: all sides equal and every angle \(60^\circ\); isosceles: two equal sides.
- Perimeter problems need the cosine rule for the third side.
- Keep full accuracy; round only at the end (3 s.f.).
⚠ Common pitfalls
- Using an angle that is not between the two chosen sides.
- Forgetting the \(\tfrac{1}{2}\) in the formula.
- Forgetting to double the area when rearranging for a side or angle.
- For upper bounds, using the rounded values instead of the upper bounds.
- Calculator not in degree mode.