For a sector of radius r with central angle θ in radians, the arc length is

The area of a sector of radius r with central angle θ in radians is

What is wrong formula for the triangle?

The triangle area is (1)/(2) r^2 sin θ (with sin θ, not θ). Confusing this with the sector formula gives a wrong segment.

What is calculator in the wrong mode?

Always check that your calculator is in radian mode when working from (π)/(6) etc.

NESA 2020 HSC Q11-style practice: A chord of a circle of radius 10 cm subtends an angle of (π)/(2) at the centre. Find the area of the minor segment cut off by the chord.

Segment area = sector area minus triangle area. Sector: (1)/(2) r^2 θ = (1)/(2) · 100 · (π)/(2) = 25 π. Triangle: (1)/(2) r^2 sin θ = (1)/(2) · 100 · sin(π)/(2) = 50. Segment: 25 π - 50 ≈ 78.54 - 50 = 28.54 cm^2. Markers expect the segment formula, the substitution into both terms, and a numerical answer with units.

NSWMaths AdvancedSyllabus dot point

How are radians defined, and how do we use them to find arc length and sector area?

Use radian measure to find arc length, the area of a sector, and the area of a segment of a circle

A focused answer to the HSC Maths Advanced dot point on radians and circular measure. Definition of radian built up stage by stage, conversion between radians and degrees, exact values, arc length $\ell = r \theta$ , sector area $A = \frac{1}{2} r^2 \theta$ , and area of a segment, with worked examples.

Generated by Claude Opus 4.814 min answerUpdated 2026-06-20

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to use the radian as the natural unit of angle, convert between radians and degrees, and apply the formulas for arc length and sector area, including segments cut off by a chord. Radians are the unit assumed by all calculus involving trig in Maths Advanced.

Why bother with a new unit at all? Because the radian is not an arbitrary choice like the degree (why $360$ ?), it is the angle measure that makes the geometry come out clean. Defining the angle as "arc length per radius" means the arc-length and sector-area formulas have no stray conversion factor, and, crucially for later, it is the only unit in which $\frac{d}{dx}(\sin x) = \cos x$ . Almost every error on this dot point traces back to one habit: substituting a degree value into a formula that is built for radians. Fix that habit and the topic is short.

The answer

One radian is the angle subtended at the centre of a circle by an arc of length equal to the radius. Equivalently, the radian measure of an angle is the ratio of arc length to radius:

\theta = \frac{\ell}{r}.

A full revolution is $2 \pi$ radians, because the full circumference $2 \pi r$ divided by the radius $r$ is $2 \pi$ .

From radius to sector area, stage by stage

The whole topic grows from one circle in four steps. Each formula is just the previous picture with one more piece added.

Stage 1, the radius. Start with a circle of radius $r$ and centre $O$ . Everything that follows is measured against this one length $r$ ; that is the point of radians.

Stage 2, define one radian. Swing a second radius round until the arc between the two radii is itself of length $r$ . The angle at the centre is then exactly $1$ radian (about $57.3^\circ$ ). This is the definition: the radian is the angle for which arc length equals radius.

Stage 3, arc length for any angle. Open the angle to a general size $\theta$ (in radians). Because $1$ radian spans an arc of $r$ , an angle of $\theta$ radians spans an arc of $\theta$ lots of $r$ :
$\ell = r\theta.$
This is just the definition $\theta = \frac{\ell}{r}$ rearranged, and it only works with $\theta$ in radians.

Stage 4, sector and segment area. Shade the sector. Its area is the fraction $\frac{\theta}{2\pi}$ of the whole circle $\pi r^2$ , which simplifies to $\frac{1}{2}r^2\theta$ . Join the two arc ends with a chord, and the slice between the chord and the arc is the segment, whose area is the sector minus the triangle, $\frac{1}{2}r^2(\theta - \sin\theta)$ .

Conversion

180^\circ = \pi \text{ radians}, \qquad 1^\circ = \frac{\pi}{180} \text{ rad}, \qquad 1 \text{ rad} = \frac{180^\circ}{\pi} \approx 57.30^\circ.

To convert: multiply degrees by $\frac{\pi}{180}$ to get radians; multiply radians by $\frac{180}{\pi}$ to get degrees.

Standard exact values:

Degrees	$0$	$30$	$45$	$60$	$90$	$180$	$270$	$360$
Radians	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\pi$	$\frac{3 \pi}{2}$	$2 \pi$

Arc length

For a sector of radius $r$ with central angle $\theta$ in radians, the arc length is

\ell = r \theta.

This is the formula behind the definition. Use radians, not degrees.

Sector area

The area of a sector of radius $r$ with central angle $\theta$ in radians is

A_{\text{sector}} = \frac{1}{2} r^2 \theta.

Derivation: the area is the fraction $\frac{\theta}{2 \pi}$ of the full circle area $\pi r^2$ , giving $\frac{\theta}{2 \pi} \cdot \pi r^2 = \frac{1}{2} r^2 \theta$ .

Triangle and segment

The triangle formed by the two radii and the chord has area

A_{\text{triangle}} = \frac{1}{2} r^2 \sin \theta.

The minor segment is the region between the chord and the arc. Its area is

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

The major segment (the larger region on the other side of the chord) has area $\pi r^2 - A_{\text{segment}}$ .

Key fact

For a circle of radius $r$ and central angle $\theta$ measured in radians:

\text{arc length } \ell = r\theta, \qquad \text{sector area } A = \tfrac{1}{2} r^2 \theta, \qquad \text{segment area} = \tfrac{1}{2} r^2 (\theta - \sin\theta).

The arc length and sector formulas use

\theta

alone; the segment subtracts the triangle

\tfrac{1}{2} r^2 \sin\theta

, so it mixes

\theta

(radians) with

\sin\theta

. Note the shared structure: both

\ell = r\theta

and

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

are the same fraction

\frac{\theta}{2\pi}

of the whole circle's circumference

2\pi r

and area

\pi r^2

respectively.

Chord length

By the cosine rule (or by splitting the isosceles triangle), the chord opposite the central angle $\theta$ has length

c = 2 r \sin\frac{\theta}{2}.

How exam questions ask about radians and circular measure

The wording tells you which formula to reach for:

"A sector subtends an angle of $\frac{\pi}{3}$ at the centre ... find the arc length / perimeter." Arc length is $\ell = r\theta$ . If they ask for the perimeter of the sector, add the two radii: $P = r\theta + 2r$ .
"... find the area of the sector." $A = \frac{1}{2} r^2 \theta$ , with $\theta$ in radians.
"Find the area of the segment / the area cut off by the chord / the shaded region." Sector minus triangle: $\frac{1}{2} r^2 (\theta - \sin\theta)$ . The word segment (or a shaded region between a chord and the arc) is the cue to subtract the triangle.
"A chord subtends an angle of $\theta$ at the centre." The angle named is the central angle; use it directly. The chord length, if needed, is $2r\sin\frac{\theta}{2}$ .
"The angle is $60^\circ$ ..." but the formula needs radians. Convert first: $60^\circ = \frac{\pi}{3}$ . A question can mix the two; the formula always wants radians.
"Find the angle, given the arc length / area." Rearrange: $\theta = \frac{\ell}{r}$ from arc length, or $\theta = \frac{2A}{r^2}$ from sector area.
"Express in radians / in degrees." A straight conversion: $\times \frac{\pi}{180}$ for degrees to radians, $\times \frac{180}{\pi}$ the other way.

Exam technique

The first move on every circular-measure question is to get $\theta$ into radians and write it down; substituting a degree value is the error that loses whole questions. Quote the formula you are using ( $\ell = r\theta$ , $A = \frac{1}{2}r^2\theta$ , segment $= \frac{1}{2}r^2(\theta - \sin\theta)$ ) before substituting, give an exact answer in terms of $\pi$ first and then a decimal if asked, and always attach units ( $\text{cm}$ for length, $\text{cm}^2$ for area). For a segment, set the working out in three labelled lines, sector, then triangle, then subtract, so a marker can follow it and award method marks even if the arithmetic slips. Keep the calculator in radian mode whenever a $\sin$ or $\cos$ of a radian value appears.

Worked example

Five calculations covering angle conversion, arc length, sector area, segment area, and chord length, each requiring $\theta$ to be in radians before substituting into a formula.

Convert between degrees and radians

To convert degrees to radians, multiply by $\frac{\pi}{180}$ ; to go the other way, multiply by $\frac{180}{\pi}$ .

120^\circ = 120 \times \frac{\pi}{180} = \frac{2\pi}{3} \text{ radians.}

\frac{5\pi}{4} \text{ rad} = \frac{5\pi}{4} \times \frac{180}{\pi} = 225^\circ.

Apply the arc length formula

A bicycle wheel of radius $35$ cm rotates through $4.5$ radians. The angle is already in radians, so substitute directly into $\ell = r\theta$ :

\ell = 35 \times 4.5 = 157.5 \text{ cm.}

The formula $\ell = r\theta$ only works with $\theta$ in radians because it comes from the definition of the radian as arc length divided by radius.

Convert the angle then compute sector area

A sector has radius $8$ cm and central angle $45^\circ$ . Convert the angle first, because the sector area formula requires radians:

45^\circ = \frac{\pi}{4} \text{ rad.}

A = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 64 \times \frac{\pi}{4} = 8\pi \approx 25.13 \text{ cm}^2.

Find segment area by subtracting the triangle from the sector

For a circle of radius $5$ cm with central angle $\frac{2\pi}{3}$ , compute sector and triangle areas separately, then subtract.

Sector area:

A_{\text{sector}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 25 \times \frac{2\pi}{3} = \frac{25\pi}{3} \approx 26.18 \text{ cm}^2.

Triangle area (the isosceles triangle formed by the two radii):

A_{\text{triangle}} = \frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 25 \times \sin\frac{2\pi}{3} = \frac{25}{2} \times \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{4} \approx 10.83 \text{ cm}^2.

Segment area:

A_{\text{segment}} = \frac{25\pi}{3} - \frac{25\sqrt{3}}{4} \approx 26.18 - 10.83 \approx 15.35 \text{ cm}^2.

Compute the chord length

For a circle with $r = 7$ and $\theta = \frac{\pi}{3}$ , the chord formula is $c = 2r\sin\frac{\theta}{2}$ . Halve the central angle first:

c = 2 \times 7 \times \sin\frac{\pi}{6} = 14 \times \frac{1}{2} = 7 \text{ cm.}

This result makes sense geometrically: a central angle of $\frac{\pi}{3}$ with equal radii of $7$ forms an equilateral triangle, so the chord equals the radius.

Final answers: $120^\circ = \frac{2\pi}{3}$ rad, $\frac{5\pi}{4}$ rad $= 225^\circ$ ; arc length $157.5$ cm; sector area $8\pi \approx 25.13$ cm $^2$ ; segment area $\approx 15.35$ cm $^2$ ; chord $= 7$ cm.

Common traps

Using degrees in radian formulas: $\ell = r \theta$ and $A = \frac{1}{2} r^2 \theta$ require $\theta$ in radians. Substituting $90$ instead of $\frac{\pi}{2}$ gives a wildly wrong answer.
Forgetting to subtract the triangle: A segment is not the sector; subtract the triangle.
Wrong formula for the triangle: The triangle area is $\frac{1}{2} r^2 \sin \theta$ (with $\sin \theta$ , not $\theta$ ). Confusing this with the sector formula gives a wrong segment.
Calculator in the wrong mode: Always check that your calculator is in radian mode when working from $\frac{\pi}{6}$ etc.
Confusing the minor and major segment: "Minor" is the smaller piece, between the chord and the shorter arc. Make sure $\theta$ refers to the central angle of that smaller piece (less than $\pi$ ).

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC Q103 marksA circle has radius

12

cm. A sector subtends an angle of

\frac{\pi}{3}

radians at the centre. Find the arc length and the area of the sector.

Show worked answer →

Arc length: $\ell = r \theta = 12 \cdot \frac{\pi}{3} = 4 \pi \approx 12.57$ cm.

Sector area: $A = \frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 144 \cdot \frac{\pi}{3} = 24 \pi \approx 75.40$ cm $^2$ .

Markers reward the correct formulas, the correct substitution with $\theta$ in radians, and exact then approximate answers with units.

2020 HSC Q113 marksA chord of a circle of radius

10

cm subtends an angle of

\frac{\pi}{2}

at the centre. Find the area of the minor segment cut off by the chord.

Show worked answer →

Segment area = sector area minus triangle area.

Sector: $\frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 100 \cdot \frac{\pi}{2} = 25 \pi$ .

Triangle: $\frac{1}{2} r^2 \sin \theta = \frac{1}{2} \cdot 100 \cdot \sin\frac{\pi}{2} = 50$ .

Segment: $25 \pi - 50 \approx 78.54 - 50 = 28.54$ cm $^2$ .

Markers expect the segment formula, the substitution into both terms, and a numerical answer with units.

What this dot point is asking

The answer

From radius to sector area, stage by stage

Conversion

Arc length

Sector area

Triangle and segment

Chord length

How exam questions ask about radians and circular measure

Exam-style practice questions

Related dot points