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Area of a Triangle using 1/2ab sin C

HigherHigher tier onlyAQAEdexcelOCR

Practise area of a triangle using 1/2ab sin c with this free Higher GCSE Maths worksheet from Virtus Academy. You'll work through finding the area of a triangle using ½ab sin C, building confidence for your exam, and every question comes with worked solutions in the mark scheme. Suitable for AQA, Edexcel and OCR. Area = ½ab sin C uses two sides and the angle between them.

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These worksheets and mark schemes are original, written for Virtus Academy and checked against the current AQA, Edexcel and OCR specifications. Every worksheet comes with a full mark scheme.

This is a Higher tier only topic, so there's no Foundation paper — only the Higher worksheet and mark scheme below.

Higher

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Topic overview

The area of a triangle can be found without knowing its perpendicular height, using \(\text{Area} = \frac{1}{2}ab\sin C\).

Here \(a\) and \(b\) are two sides and \(C\) is the angle between them. The angle must be the included angle — the one formed where those two sides meet.

The formula is useful precisely when the standard half-base-times-height cannot be applied, because the height is not given and would be awkward to find. It also connects neatly with the sine rule and cosine rule, which use the same labelling.

Revision notes

The formula

Area is \(\frac{1}{2}ab\sin C\), where \(C\) is the angle between sides \(a\) and \(b\).

For sides \(6\)cm and \(10\)cm with an included angle of \(30^\circ\): \(\frac{1}{2} \times 6 \times 10 \times 0.5 = 15\)cm².

Identifying the included angle

The angle must sit between the two sides you are using.

If the given angle is elsewhere in the triangle, find the included angle first using the angle sum, or use a different method.

Working backwards

If the area and two sides are known, substitute and solve for the angle.

With area \(20\)cm² and sides \(8\) and \(10\): \(20 = \frac{1}{2}(8)(10)\sin C\), so \(\sin C = 0.5\) and \(C = 30^\circ\).

Key points

  • Area is \(\frac{1}{2}ab\sin C\).
  • \(C\) must be the angle between sides \(a\) and \(b\).
  • No perpendicular height is needed.
  • Useful when the height is not given.
  • Area is in squared units.
  • Rearrange to find an angle if the area is known.

Worked examples

Example 1

Find the area of a triangle with sides \(8\)cm and \(12\)cm and included angle \(30^\circ\).

Working

\[\frac{1}{2} \times 8 \times 12 \times \sin 30\]substitute into the formula
\[48 \times 0.5\]evaluate, using sin 30 = 0.5
\[= 24\text{cm}^2\]state the area

Example 2

Find the area of a triangle with sides \(5\)cm and \(9\)cm and included angle \(90^\circ\).

Working

\[\frac{1}{2} \times 5 \times 9 \times \sin 90\]substitute into the formula
\[22.5 \times 1\]sin 90 equals 1
\[= 22.5\text{cm}^2\]this matches the standard half-base-times-height

Example 3

A triangle has area \(30\)cm² with sides \(10\)cm and \(12\)cm. Find the included angle.

Working

\[30 = \frac{1}{2}(10)(12)\sin C\]substitute the known values
\[\sin C = 0.5\]rearrange for sin C
\[C = 30^\circ\]apply the inverse sine

Common mistakes

  • Using an angle that is not between the two sides.

    The formula needs the included angle, formed where the two sides meet.

  • Forgetting the half.

    The formula begins with ½, exactly like the standard triangle area.

  • Using the perpendicular height as one of the sides.

    The formula uses two actual sides, not a height.

  • Giving the answer in ordinary units.

    Area needs squared units.

Exam tips

  • Check the angle sits between the two sides you are using.
  • Write the formula out before substituting.
  • Keep full accuracy and round only at the end.
  • Include squared units in the answer.

Key terms

Included angle
The angle between the two sides being used.
Sine
The trigonometric ratio used in the area formula.
Area
The space inside the triangle.
Rearrange
To make a different quantity the subject.

Related topics

Written and reviewed against the current AQA, Edexcel and OCR specifications. Spotted an error? Let us know.