Circular measure

CIE A-Level Mathematics· 15 min read

1. Radians: Unit of Angle★★☆☆☆⏱ 3 min

📘 Definition

Radian

rad

One radian is the angle subtended at the center of a circle by an arc that has a length equal to the radius of the circle. A full rotation is $2 π$ radians, equivalent to 360°.

Example:

Half a rotation = $π$ radians = 180°

To convert between degrees and radians, use these two conversion factors:

1^\circ = \frac{\pi}{180} \text{ rad}, \quad 1 \text{ rad} = \frac{180^\circ}{\pi}

📐 Worked Example

Convert (a) 120° to radians, (b) $\frac{3 π}{4}$ radians to degrees.

1
Part (a): Multiply the degree measure by $\frac{π}{180}$ :
2
$120^\circ \times \frac{\pi}{180} = \frac{120\pi}{180} = \frac{2\pi}{3} \text{ rad}$
3
Part (b): Multiply the radian measure by $\frac{180}{π}$ :
4
$\frac{3\pi}{4} \times \frac{180}{\pi} = \frac{3 \times 180}{4} = 135^\circ$

2. Arc Length Calculation★★☆☆☆⏱ 4 min

📘 Definition

Arc Length

s

The length of a portion of a circle's circumference between two radii. For a circle of radius $r$ and central angle $θ$ (in radians), arc length is given by:

Example:

Full circumference = $r \times 2 π = 2 π r$ , which matches the standard formula.

s = r\theta

📐 Worked Example

(a) Find the arc length of a sector with radius 5 cm and central angle 2.4 radians. (b) Find the radius of a circle with arc length 12 cm and central angle 1.5 radians.

1
Part (a): Substitute directly into the arc length formula:
2
$s = 5 \times 2.4 = 12 \text{ cm}$
3
Part (b): Rearrange the formula to solve for $r$ :
4
$r = \frac{s}{\theta} = \frac{12}{1.5} = 8 \text{ cm}$

✓ Quick check

Test your understanding:

What is the arc length for a 60° angle in a circle of radius 3 cm?
- $π$ cm
- $2 π$ cm
- $3 π$ cm
- $\frac{π}{2}$ cm
Reveal answer
$\pi$ cm —
First convert 60° to $\frac{π}{3}$ radians, then calculate $s = 3 \times \frac{π}{3} = π$ cm.

3. Area of Sectors and Segments★★★☆☆⏱ 4 min

The two most common area calculations in circular measure are for sectors (regions bounded by two radii and an arc) and segments (regions bounded by an arc and a chord).

📘 Definition

Sector Area

A

For a circle of radius $r$ and central angle $θ$ (in radians), sector area is:

Example:

Full circle area = $\frac{1}{2} r^{2} (2 π) = π r^{2}$ , which matches the standard formula.

A = \frac{1}{2} r^2 \theta

A segment is the part of the sector that lies beyond the chord connecting the two radii endpoints. Its area is calculated by subtracting the area of the triangle formed by the two radii from the sector area:

A_{\text{segment}} = \frac{1}{2}r^2(\theta - \sin\theta)

📐 Worked Example

Find the area of the sector and minor segment of a circle with radius 6 cm and central angle $\frac{π}{3}$ radians.

1
Calculate sector area first:
2
$A_{\text{sector}} = \frac{1}{2} (6)^2 \left(\frac{\pi}{3}\right) = 6\pi \approx 18.85 \text{ cm}^2$
3
Calculate area of the central triangle using $\frac{1}{2} ab sin θ$ :
4
$A_{\text{triangle}} = \frac{1}{2} (6)(6) \sin\left(\frac{\pi}{3}\right) = 9\sqrt{3} \approx 15.59 \text{ cm}^2$
5
Subtract triangle area from sector area to get segment area:
6
$A_{\text{segment}} = 6\pi - 9\sqrt{3} \approx 3.26 \text{ cm}^2$

4. Problem Solving with Circular Measure★★★★☆⏱ 4 min

📐 Worked Example

A square of side 10 cm has a quarter-circle cut from each corner, with each quarter-circle having radius 5 cm centered at the corner. Find the area of the remaining shape inside the square.

1
First calculate the total area of the square:
2
$A_{\text{square}} = 10 \times 10 = 100 \text{ cm}^2$
3
Four quarter circles have a total area equal to one full circle of radius 5 cm:
4
$A_{\text{cut out}} = 4 \times \left(\frac{1}{2} (5)^2 \left(\frac{\pi}{2}\right)\right) = 25\pi$
5
Subtract the cut out area from the square area:
6
$A_{\text{remaining}} = 100 - 25\pi \approx 21.46 \text{ cm}^2$

5. Common Pitfalls

Wrong move:

Using degrees instead of radians in $s = r θ$ and $A = \frac{1}{2} r^{2} θ$

Why:

These formulas are derived for radians, they give drastically incorrect results when using degree measure directly

Correct move:

Always convert all angles to radians before substituting into these standard circular measure formulas

Wrong move:

Adding the triangle area to the sector area to find segment area

Why:

The segment is the part of the sector outside the triangle, so the triangle area must be subtracted

Correct move:

Segment area = Sector area $-$ Area of the central triangle formed by the two radii

Wrong move:

Mixing up arc length and sector area formulas

Why:

Both formulas use $r$ and $θ$ , leading to common confusion about powers of $r$

Correct move:

Remember: length is one-dimensional (one power of $r$ ), area is two-dimensional (two powers of $r$ ): $s = r θ$ , $A = \frac{1}{2} r^{2} θ$

Wrong move:

Using the minor angle for major arc/sector area questions

Why:

Most problems default to minor segments unless stated otherwise, so you must actively check for major/mior labels

Correct move:

If asked for a major arc/sector, calculate the angle as $2 π - minor angle$ before substituting into formulas

6. Quick Reference Cheatsheet

Quantity	Formula (radians)	Formula (degrees)
Degree to radian	$θ_{rad} = \frac{π}{180} θ_{deg}$	$θ_{deg} = \frac{180}{π} θ_{rad}$
Arc length	$s = r θ$	$s = \frac{θ π r}{180}$
Sector area	$A = \frac{1}{2} r^{2} θ$	$A = \frac{θ}{360} π r^{2}$
Minor segment area	$A = \frac{1}{2} r^{2} (θ - sin θ)$	$A = \frac{θ π r ^{2}}{360} - \frac{1}{2} r^{2} sin θ$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Area of minor segment problem
2022 · 1
Composite shape perimeter problem
2021 · 1
Arc length calculation question

What's Next

Circular measure introduces radians, which are the standard unit for all advanced trigonometry and calculus of trigonometric functions. This topic regularly appears as a 5-7 mark question in CIE 9709 Paper 1, often combined with quadratic equations or trigonometric identities to create multi-step problems. Mastery of the core formulas saves significant exam time, avoiding mistakes from unnecessary unit conversions. It is a foundational topic for all further work in pure mathematics.