Area of a Triangle Using Sine
What happens when you need to find the area of a triangle but don’t know its perpendicular height? This guide will show you how to use a powerful trigonometric formula to find the area of any triangle when you know two sides and the angle between them — a key skill for higher-tier GCSE Maths.
The Area Formula Using Sine
This formula is used to find the area of a non-right-angled triangle when you know two sides and the included angle (the angle sandwiched between those two sides).
The Formula
The area of a triangle can be found using:
Here, $a$ and $b$ are the lengths of two sides, and $C$ is the angle between them.
Worked Examples
Example 1: Finding the Area
Find the area of a triangle with sides 8 cm and 11 cm, and an included angle of 40°.
- Area $= \frac{1}{2}ab\sin C$.
- Area $= \frac{1}{2} \times 8 \times 11 \times \sin(40°)$.
- $= 44 \times \sin(40°) \approx 28.3$.
Answer: 28.3 cm² (to 3 s.f.)
Example 2: Finding a Missing Side
The area of a triangle is 50 m². Two sides are 12 m and $x$ m, with an included angle of 65°. Find $x$.
- $50 = \frac{1}{2} \times 12 \times x \times \sin(65°)$.
- $50 = 6x\sin(65°)$.
- $x = \frac{50}{6\sin(65°)} \approx 9.20$.
Answer: 9.20 m (to 3 s.f.)
Example 3: Finding a Missing Angle
The area of a triangle is 25 cm². Two sides are 7 cm and 9 cm. Find the included angle $\theta$.
- $25 = \frac{1}{2} \times 7 \times 9 \times \sin\theta$.
- $25 = 31.5\sin\theta$.
- $\sin\theta = \frac{25}{31.5}$.
- $\theta = \sin^{-1}\!\left(\frac{25}{31.5}\right) \approx 52.5°$.
Answer: 52.5° (to 1 d.p.)
Tutor Insights
🤔 Common Misunderstandings
- Using the wrong angle. The formula only works if you use the angle that is between the two sides you know.
- Forgetting the $\frac{1}{2}$. A very common slip that will double your answer!
- Calculator in the wrong mode. Ensure it’s set to Degrees (DEG), not Radians (RAD).
📝 Common Exam Mistakes
- Not memorising the formula, as it’s often not provided on the formula sheet.
- Rounding too early. Keep the full calculator value until the very final step.
- Algebraic errors when rearranging to find a missing side or angle.
Practice Questions
- Calculate the area of a triangle with sides 10 cm and 12 cm, and an included angle of 75°. Give your answer to 3 significant figures.
- The area of a triangle is 40 cm². Two of its sides are 8 cm and 15 cm. Find the size of the included angle to 1 decimal place.
- The area of triangle ABC is 72 cm². Side $AB = 16$ cm and side $AC = 10$ cm. Find the two possible values for angle $BAC$ to 1 decimal place.
Show Answers
- Working: Area $= \frac{1}{2} \times 10 \times 12 \times \sin(75°) \approx 58.0$.
Answer: 58.0 cm². - Working: $40 = \frac{1}{2} \times 8 \times 15 \times \sin\theta \implies 40 = 60\sin\theta \implies \theta = \sin^{-1}\!\left(\frac{40}{60}\right) \approx 41.8°$.
Answer: 41.8°. - Working: $72 = \frac{1}{2} \times 16 \times 10 \times \sin A \implies 72 = 80\sin A \implies A = \sin^{-1}\!\left(\frac{72}{80}\right) \approx 64.2°$. The obtuse solution: $180° – 64.2° = 115.8°$.
Answers: 64.2° and 115.8°.
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