Maths · Further Trigonometry
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Area of a Triangle and Segments
Using half ab sin C to find areas, sides and angles, and finding the area of a segment of a circle.
Learning Objectives
- 1Use the formula \(\text{area} = \dfrac{1}{2}ab\sin C\) to find the area of a triangle.
- 2Find a missing side or angle from the area of a triangle.
- 3Find the area of a segment of a circle.
- 4Use exact values to give answers with surds, such as \(12\sqrt{3}\).
The area of any triangle
The usual formula, half the base times the perpendicular height, needs the height, which is often not given. If you know two sides and the angle between them, the height can be written using sine, and this gives a new formula for the area that does not need the height. The formula is also a quick way to find the area of a segment of a circle, which is the sector minus a triangle.
Area with a sine
The height \(h\) is found from the right-angled triangle that it makes with side \(b\), so \(h = b\sin C\). The area is \(\dfrac{1}{2} \times a \times h = \dfrac{1}{2}ab\sin C\).
Where the formula comes from
- Height \(h = b\sin C\).
- Area \(\dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times a \times b\sin C\).
- The angle It must be the angle between the two sides used.
- Other forms \(\dfrac{1}{2}bc\sin A\) and \(\dfrac{1}{2}ac\sin B\).
Using the formula
Choose two sides and the angle between them.
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The formula
\(\text{Area} = \dfrac{1}{2}ab\sin C\).
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Included angle
The angle \(C\) is between the sides \(a\) and \(b\).
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Exact values
Use \(\sin 30^\circ = \dfrac{1}{2}\), \(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\) and \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\).
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Units
The area is in square units.
The area of a triangle
In triangle \(ABC\), \(a = 6\) cm, \(b = 8\) cm and angle \(C = 60^\circ\). Work out the exact area of the triangle.
Show the solutionHide the solution
- 1 Choose the formula Two sides and the angle between them are known.
- 2 Substitute \(\text{Area} = \dfrac{1}{2} \times 6 \times 8 \times \sin 60^\circ\).
- 3 Exact value \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), so the area is \(24 \times \dfrac{\sqrt{3}}{2}\).
- 4 Simplify \(12\sqrt{3}\) cm\(^2\).
Answer\(12\sqrt{3}\) cm\(^2\)
Finding an angle or a side from the area
If the area is given, substitute it and solve.
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Write the equation
\(\text{Area} = \dfrac{1}{2}ab\sin C\), with the area on the left.
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Rearrange
\(\sin C = \dfrac{2 \times \text{Area}}{ab}\).
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Inverse sine
Use an exact value to find \(C\). If \(C\) could be obtuse, there are two possible angles, \(C\) and \(180^\circ - C\).
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Missing side
Rearrange for \(a\) or \(b\) in the same way.
Finding an angle from the area
A triangle has sides of 10 cm and 8 cm, and an area of \(20\) cm\(^2\). The angle between the sides is acute. Work out the size of this angle.
Show the solutionHide the solution
- 1 Write the equation \(20 = \dfrac{1}{2} \times 10 \times 8 \times \sin C\).
- 2 Simplify \(20 = 40\sin C\), so \(\sin C = \dfrac{1}{2}\).
- 3 Exact value \(\sin 30^\circ = \dfrac{1}{2}\), so \(C = 30^\circ\).
- 4 Check The other solution of \(\sin C = \dfrac{1}{2}\) is \(150^\circ\), which is obtuse, so it is not used.
Answer\(30^\circ\)
The area of a segment
A segment is the part of a circle between a chord and an arc.
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Sector
Area of the sector \(= \dfrac{\theta}{360} \times \pi r^2\).
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Triangle
The two radii and the chord make a triangle, with area \(\dfrac{1}{2}r^2\sin\theta\).
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Segment
Area of the segment \(=\) area of the sector \(-\) area of the triangle.
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Exact answers
Leave \(\pi\) and surds in the answer.
The area of a segment
\(O\) is the centre of a circle of radius 6 cm. \(A\) and \(B\) are points on the circle and angle \(AOB = 60^\circ\). Work out the exact area of the segment cut off by the chord \(AB\).
Show the solutionHide the solution
- 1 Sector \(\dfrac{60}{360} \times \pi \times 6^2 = \dfrac{1}{6} \times 36\pi = 6\pi\).
- 2 Triangle \(\dfrac{1}{2} \times 6 \times 6 \times \sin 60^\circ = 18 \times \dfrac{\sqrt{3}}{2} = 9\sqrt{3}\).
- 3 Subtract Segment \(= 6\pi - 9\sqrt{3}\).
- 4 Units The answer is in \(\text{cm}^2\).
Answer\((6\pi - 9\sqrt{3})\) cm\(^2\)
Test yourself
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1
What is the formula for the area of a triangle using a sine?
Show answerHide answer
\(\dfrac{1}{2}ab\sin C\).
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2
Which angle is used?
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The angle between the two sides.
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3
What is \(\sin 60^\circ\)?
Show answerHide answer
\(\dfrac{\sqrt{3}}{2}\).
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4
What is the area of a segment?
Show answerHide answer
Area of the sector minus area of the triangle.
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5
What is \(\sin 90^\circ\)?
Show answerHide answer
1.
Exam technique: area questions
Check the angle, then write each step.
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Angle between
Make sure the angle is between the two sides that you are using.
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Exact answers
If the question says exact, leave surds and \(\pi\).
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Two possible angles
If an angle is found from \(\sin\), check whether it could be obtuse.
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Show every step
The formula with the numbers in earns the first mark.
Summary and exam focus
- \(\text{Area} = \dfrac{1}{2}ab\sin C\) with \(C\) between \(a\) and \(b\).
- Rearrange the formula to find an angle or a side.
- A segment is a sector minus a triangle.
- Use exact values to give answers in surd form.
Exam focus
Triangle \(ABC\) has \(AB = 10\) cm, \(AC = 12\) cm and angle \(BAC = 30^\circ\). Work out the area of the triangle. (2 marks) (2 marks)
\(\text{Area} = \dfrac{1}{2} \times 10 \times 12 \times \sin 30^\circ = 60 \times \dfrac{1}{2} = 30\) cm\(^2\). Show the formula with the numbers substituted.
Key terms
The words this lesson expects you to use. Each one is linked from the first place it appears above.
- Area
- The amount of surface inside a shape.
- Included angle
- The angle between two given sides.
- Segment
- The part of a circle cut off by a chord.
- Sector
- The part of a circle between two radii.
- Exact value
- A value written with fractions, surds and \(\pi\).
- Chord
- A straight line joining two points on a circle.
- Perpendicular height
- The shortest distance from a vertex to the opposite side.
- Surd
- A root that cannot be written as a whole number or fraction.
- Radius
- The distance from the centre of a circle to its edge.
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