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This lesson covers radian measure, the formulae for arc length and sector area, and calculations involving segments of circles. Radian measure is essential throughout A-Level Mathematics — it simplifies many formulae in calculus, trigonometry, and mechanics, and is used exclusively in the AQA specification from this point onward.
A radian is a unit for measuring angles. One radian is defined as the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle.
Since the circumference of a circle is 2πr, a full turn (360°) corresponds to an arc length of 2πr, which gives:
2π radians = 360°
Therefore:
π radians = 180°
This single relationship is the key to all conversions between degrees and radians.
Multiply the angle in degrees by π/180:
θ (radians) = θ (degrees) × π/180
Example 1: Convert 45° to radians.
45° × π/180 = π/4 radians
Example 2: Convert 120° to radians.
120° × π/180 = 2π/3 radians
Multiply the angle in radians by 180/π:
θ (degrees) = θ (radians) × 180/π
Example 3: Convert 5π/6 radians to degrees.
5π/6 × 180/π = 150°
Example 4: Convert 1.2 radians to degrees.
1.2 × 180/π = 68.8° (3 s.f.)
| Degrees | Radians |
|---|---|
| 0° | 0 |
| 30° | π/6 |
| 45° | π/4 |
| 60° | π/3 |
| 90° | π/2 |
| 120° | 2π/3 |
| 135° | 3π/4 |
| 150° | 5π/6 |
| 180° | π |
| 270° | 3π/2 |
| 360° | 2π |
You must know these by heart. In AQA examinations, angles will frequently be given in radians, and you are expected to be fluent in working with them.
An arc is a section of the circumference of a circle. The length of an arc is proportional to the angle it subtends at the centre.
If the radius is r and the angle at the centre is θ (in radians):
s = rθ
where s is the arc length.
This formula is strikingly simple — and that simplicity is one of the main reasons we use radians in advanced mathematics.
The full circumference is 2πr, corresponding to a full angle of 2π radians. The fraction of the circle used is θ/(2π). Therefore:
s = 2πr × θ/(2π) = rθ
Example 5: A sector has radius 8 cm and angle 1.5 radians. Find the arc length.
s = rθ = 8 × 1.5 = 12 cm
Example 6: A sector has radius 6 cm and arc length 4π cm. Find the angle in radians.
s = rθ
4π = 6θ
θ = 4π/6 = 2π/3 radians
Example 7: The arc length of a sector is 15 cm and the angle is 2.5 radians. Find the radius.
s = rθ
15 = r × 2.5
r = 15/2.5 = 6 cm
If the angle is given in degrees, the arc length formula is:
s = (θ/360) × 2πr = πrθ/180
However, at A-Level you should almost always work in radians.
A sector is the region enclosed by two radii and the arc between them (think of a slice of pizza).
A = ½r²θ
where r is the radius and θ is the angle in radians.
The area of the full circle is πr², and the fraction used is θ/(2π):
A = πr² × θ/(2π) = ½r²θ
Example 8: Find the area of a sector with radius 10 cm and angle π/4 radians.
A = ½r²θ = ½ × 10² × π/4 = ½ × 100 × π/4 = 25π/2 cm²
This is approximately 39.3 cm².
Example 9: A sector has area 27 cm² and radius 6 cm. Find the angle in radians.
A = ½r²θ
27 = ½ × 36 × θ
27 = 18θ
θ = 27/18 = 3/2 = 1.5 radians
Example 10: A sector has area 50 cm² and angle 0.8 radians. Find the radius.
A = ½r²θ
50 = ½ × r² × 0.8
50 = 0.4r²
r² = 125
r = √125 = 5√5 cm ≈ 11.2 cm
The perimeter of a sector consists of two radii and the arc:
Perimeter = 2r + rθ = r(2 + θ)
Example 11: Find the perimeter of a sector with radius 7 cm and angle 1.2 radians.
Perimeter = 7(2 + 1.2) = 7 × 3.2 = 22.4 cm
A segment is the region between a chord and the arc it cuts off. There are two types:
The area of a segment is found by subtracting the area of the triangle from the area of the sector:
Area of segment = Area of sector − Area of triangle
= ½r²θ − ½r²sinθ
= ½r²(θ − sinθ)
This uses the triangle area formula ½r²sinθ (since the triangle has two sides of length r and the included angle θ).
Example 12: Find the area of the minor segment of a circle with radius 12 cm and central angle π/3 radians.
Area = ½r²(θ − sinθ)
= ½ × 144 × (π/3 − sin(π/3))
= 72 × (π/3 − √3/2)
= 72π/3 − 72√3/2
= 24π − 36√3
≈ 75.4 − 62.4
= 13.0 cm² (3 s.f.)
Example 13: A chord AB subtends an angle of 2 radians at the centre of a circle with radius 5 cm. Find the area of the minor segment.
Area = ½ × 25 × (2 − sin 2)
= 12.5 × (2 − 0.9093...)
= 12.5 × 1.0907...
= 13.6 cm² (3 s.f.)
Problem: The diagram shows a sector OAB of a circle, centre O, radius 8 cm. The angle AOB is 0.9 radians. Find:
(a) the length of arc AB
(b) the area of sector OAB
(c) the area of the shaded segment
Solution:
(a) Arc length:
s = rθ = 8 × 0.9 = 7.2 cm
(b) Sector area:
A = ½r²θ = ½ × 64 × 0.9 = 28.8 cm²
(c) Segment area:
A = ½r²(θ − sinθ) = ½ × 64 × (0.9 − sin 0.9)
= 32 × (0.9 − 0.7833...)
= 32 × 0.1167...
= 3.73 cm² (3 s.f.)
Exam Tip: The formulae s = rθ and A = ½r²θ are in the AQA formula booklet, but you should know them by heart. Remember that these formulae only work when θ is in radians — if the question gives an angle in degrees, convert it first. Always include units in your final answer and give answers in exact form (involving π or surds) where appropriate.