Finding the area of a triangle
The area rule depends on the data that is given. With two sides and their included angle, use one half times the side product and the sine of that angle
\[A=\frac{1}{2}ab\sin C\]
- Identify the included angle For two known sides, use the angle at their common vertex \(\displaystyle \angle BAC=70^{\circ}\text{ is between }AB\text{ and }AC\)
- Use the sine area formula Multiply one half by the two known sides and the sine of their included angle \(\displaystyle A=\frac{1}{2}(13)(13)\sin70^{\circ}\approx79.4\ \mathrm{cm}^{2}\)
- Keep the one-half factor The product of the two sides and the sine value is twice the triangle area \(\displaystyle \frac{1}{2}(158.8)=79.4\)
- Use the angle value correctly An obtuse included angle is allowed, and its sine is still positive \(\displaystyle \sin(180^{\circ}-\theta)=\sin\theta\)
Worked example
Triangle XYZ has sides XY = 12 m and XZ = 18 m. The included angle YXZ is 65°. Find the area to 2 decimal places
\[A=\frac{1}{2}(12)(18)\sin65^{\circ}\]
- \(\displaystyle \angle YXZ=65^{\circ}\)The angle at X lies between the two known sides, so it is the included angle
- \(\displaystyle A=\frac{1}{2}(12)(18)\sin65^{\circ}\)Use the sine area formula with the two sides and the included angle
- \(\displaystyle 12\times18\sin65^{\circ}\approx195.76248\)The product before halving is approximately 195.76248 square metres
- \(\displaystyle A=\frac{1}{2}(12)(18)\sin65^{\circ}\approx97.88124\ \mathrm{m}^{2}\approx97.88\ \mathrm{m}^{2}\)Halve the product and round the final area to 2 decimal places
Common mistakes
Using a non-included angle
✕\(\displaystyle \frac{1}{2}(13)(13)\sin55^{\circ}\approx69.2\ \mathrm{cm}^{2}\)✓\(\displaystyle \frac{1}{2}(13)(13)\sin70^{\circ}\approx79.4\ \mathrm{cm}^{2}\)
The sine area formula needs the angle between the two given sides, so angle 55° is not the included angleDropping the one-half factor
✕\(\displaystyle A=13\times13\sin70^{\circ}\approx158.8\ \mathrm{cm}^{2}\)✓\(\displaystyle A=\frac{1}{2}(13)(13)\sin70^{\circ}\approx79.4\ \mathrm{cm}^{2}\)
The sine area formula includes one half, so the side product times the sine must be divided by 2.Using the angle as a multiplier
✕\(\displaystyle A=\frac{1}{2}(13)(13)(70)=5915\ \mathrm{cm}^{2}\)✓\(\displaystyle A=\frac{1}{2}(13)(13)\sin70^{\circ}\approx79.4\ \mathrm{cm}^{2}\)
The formula uses sin of the included angle, not the angle's degree number as a multiplierDifficulty levels
introUse the sine area or base-height formula for one triangle with simple data
standardApply the appropriate area rule to a contextual triangle, including obtuse included angles and three-side data
challengeDerive an included angle or compare a sine-formula area with a Heron area
examChoose the correctly calculated sine-formula area from options containing half-factor and angle errors
Questions about this topic
Why is the included angle needed in the sine area formula?+
The angle between the two known sides determines the perpendicular component of one side. That component is the side length multiplied by the sine of the included angle
Can the included angle be obtuse?+
Yes. The sine of an obtuse angle between 90° and 180° is positive, so the calculated triangle area is positive
When is Heron's rule useful?+
Use Heron's rule when all three side lengths are known and no perpendicular height or included angle is supplied. First find the semi-perimeter, s, then use the four factors in Heron's formula
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, General Mathematics, chapter 11: Trigonometry
In this chapter
Name sides and state trigonometric ratios11A Trigonometric ratios: Evaluate a trigonometric ratio, and find an angle from its valueChoose a trigonometric ratioFind an unknown side of a right-angled triangle11C Finding unknown angles: Evaluate a trigonometric ratio, and find an angle from its valueFind an angle of a right-angled triangle from two sidesSolve a practical right-angled-triangle problemAngles of elevation and depressionRead and state three-figure bearingsSolve a navigation problem with bearingsUse the sine rule in a non-right-angled triangleThe ambiguous case of the sine ruleUse the cosine rule in a non-right-angled triangle11H Cosine rule: Choose the rule a triangle problem needs11H Cosine rule: Solve a practical triangle problem end to end11I Triangle area: Choose the rule a triangle problem needsFind the area of a triangle11I Triangle area: Solve a practical triangle problem end to end