The arc-length formula falls straight out of the definition. Since θ = Lr, multiplying both sides by r gives

A sector is the "pizza slice" region bounded by the two radii and the arc. Its area is the fraction θ2π of the whole circle (the angle as a share of a full turn), and the whole circle has area π r^2:

NSWMaths AdvancedSyllabus dot point

What is a radian, why does measuring an angle by arc length over radius make pi rad = 180 deg, what are the exact radian values of the common angles, and how do the clean formulas arc length L = r theta, sector area A = (1/2) r^2 theta and segment area (sector minus triangle) follow once an angle is measured in radians?

Define a radian as the angle whose arc length equals the radius, convert between degrees and radians using pi rad = 180 deg, know the exact radian values of the common angles, and use the radian formulas for arc length L = r theta, sector area A = (1/2) r^2 theta and the area of a segment as the sector minus the triangle

The Year 11 Maths Advanced answer on radian measure, arcs and sectors: a radian defined by arc length equal to radius, converting with pi rad = 180 deg, exact radian values of common angles, arc length L = r theta, sector area half r squared theta, and segment area as sector minus triangle, with code-checked numbers and diagrams.

Generated by Claude Opus 4.818 min answerUpdated 2026-06-21

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

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Quick answer

A radian measures an angle by arc length over radius, $\theta = \dfrac{L}{r}$ , so one radian is the angle whose arc equals the radius (about $57.3^\circ$ ). A full turn is $2\pi$ and a straight angle is $\pi$ , giving the single conversion fact $\pi \text{ rad} = 180^\circ$ : to go from degrees to radians multiply by $\dfrac{\pi}{180}$ , and to go back multiply by $\dfrac{180}{\pi}$ , cancelling the fraction to keep the exact form (for example $150^\circ = \dfrac{5\pi}{6}$ and $\dfrac{3\pi}{4} = 135^\circ$ ). Learn the common values by sight: $30^\circ = \dfrac{\pi}{6}$ , $45^\circ = \dfrac{\pi}{4}$ , $60^\circ = \dfrac{\pi}{3}$ , $90^\circ = \dfrac{\pi}{2}$ , $120^\circ = \dfrac{2\pi}{3}$ , $180^\circ = \pi$ , $360^\circ = 2\pi$ . With $\theta$ in radians the circle formulas are clean: the arc length is $L = r\theta$ , the sector area is $A = \dfrac{1}{2} r^2 \theta$ , and the perimeter of a sector is $r\theta + 2r$ . The area of a segment (between a chord and its arc) is the sector minus the isosceles triangle, $\dfrac{1}{2} r^2 \theta - \dfrac{1}{2} r^2 \sin\theta = \dfrac{1}{2} r^2 (\theta - \sin\theta)$ . For example a Ferris wheel cabin of radius $30$ m through $40^\circ = \dfrac{2\pi}{9}$ travels $\dfrac{20\pi}{3} \approx 20.94$ m; one eighth of a $14$ cm pizza ( $\theta = \dfrac{\pi}{4}$ ) has area $\dfrac{49\pi}{2} \approx 76.97 \text{ cm}^2$ ; and a chord at $\dfrac{\pi}{3}$ in a $10$ cm circle cuts off a segment of $\dfrac{50\pi}{3} - 25\sqrt{3} \approx 9.06 \text{ cm}^2$ . Always convert degrees to radians before using $L = r\theta$ or $A = \tfrac{1}{2} r^2 \theta$ , keep $\pi$ in exact answers, and add both radii for a sector perimeter. The calculus of radians comes later.

What this dot point is asking

Every angle you have measured so far has been in degrees, where a full turn is $360^\circ$ . This dot point introduces a second, more natural unit, the radian, in which a full turn is $2\pi$ . A radian is defined purely from a circle: it is the angle for which the arc length equals the radius. NESA expects you to know that definition, to convert between degrees and radians using the single fact $\pi \text{ rad} = 180^\circ$ , to recognise the exact radian values of the common angles ( $\dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2}, \pi, 2\pi$ and friends), and then to use radians in three clean formulas: the arc length $L = r\theta$ , the sector area $A = \tfrac{1}{2} r^2 \theta$ , and the area of a segment, which is a sector with its triangle removed. The pay-off is that these formulas are short and exact only because the angle is in radians; the same results in degrees carry clumsy factors of $\dfrac{\theta}{360}$ . Real circular shapes, a fan blade sweeping an arc, a slice of pizza, the arc a Ferris wheel cabin travels, are all sector and arc problems, and radians make them quick.

This is the introductory radian page. It stays with the definition and the basic arc, sector and segment. The fuller treatment, including circular-measure problems that combine these formulas, the small-angle results and radians inside calculus, is the Year-12 page radian measure, arc length and sector area; this page deliberately does not differentiate any trig function or use $\sin x \approx x$ . Here radians are a unit and the circle formulas that follow from it.

The answer

What a radian is: arc length over radius

Take a circle of radius $r$ and a central angle. The two arms of the angle cut off an arc along the circle. The size of the angle in radians is defined as the ratio

\theta = \frac{\text{arc length}}{\text{radius}} = \frac{L}{r}.

So one radian is the angle whose arc is exactly as long as the radius: walk a distance $r$ around the rim, and the angle you have swept from the centre is $1$ radian. Because both the arc and the radius scale together when you change the size of the circle, this ratio gives the same angle whatever the radius, which is exactly what an angle measure must do.

This definition makes a radian a pure number (a length divided by a length), with no units of its own. That is why "an angle of $1.2$ " with no degree symbol means $1.2$ radians, and why your calculator has a separate RAD mode. From now on, an angle written without a degree symbol is in radians.

To pin down the size, look at a whole revolution. The arc for a full turn is the entire circumference $2\pi r$ , so

\text{one revolution} = \frac{2\pi r}{r} = 2\pi \text{ radians}.

A full turn is also $360^\circ$ , so $360^\circ = 2\pi$ radians, and halving gives the single conversion fact you actually use.

Converting between degrees and radians

Everything flows from $\pi \text{ rad} = 180^\circ$ . Dividing both sides two ways gives the two multipliers:

Degrees to radians: multiply by $\dfrac{\pi}{180}$ .
Radians to degrees: multiply by $\dfrac{180}{\pi}$ .

For example, $150^\circ = 150 \times \dfrac{\pi}{180} = \dfrac{150\pi}{180} = \dfrac{5\pi}{6}$ , and going back, $\dfrac{5\pi}{6} \times \dfrac{180}{\pi} = 150^\circ$ . The trick that keeps the work tidy is to cancel the fraction rather than reach for a decimal: $\dfrac{150}{180}$ reduces to $\dfrac{5}{6}$ , so the answer is $\dfrac{5\pi}{6}$ in one step. Keeping $\pi$ in the answer is what "exact value" wants.

A single radian is $1 \times \dfrac{180}{\pi} = \dfrac{180}{\pi} \approx 57.30^\circ$ , which is why a sector with arc equal to its radius looks like a slightly squashed equilateral triangle (a $60^\circ$ slice). Going the other way, $1^\circ = \dfrac{\pi}{180} \approx 0.0175$ radians.

The exact radian values worth memorising

A handful of angles come up constantly, so learn their radian forms by sight rather than converting each time. They are just $\dfrac{\pi}{180}$ times the degree value, cancelled:

Degrees	Radians	Degrees	Radians
$30^\circ$	$\dfrac{\pi}{6}$	$180^\circ$	$\pi$
$45^\circ$	$\dfrac{\pi}{4}$	$210^\circ$	$\dfrac{7\pi}{6}$
$60^\circ$	$\dfrac{\pi}{3}$	$225^\circ$	$\dfrac{5\pi}{4}$
$90^\circ$	$\dfrac{\pi}{2}$	$270^\circ$	$\dfrac{3\pi}{2}$
$120^\circ$	$\dfrac{2\pi}{3}$	$330^\circ$	$\dfrac{11\pi}{6}$
$135^\circ$	$\dfrac{3\pi}{4}$	$360^\circ$	$2\pi$

The pattern is easy: the three "special" acute angles are $\dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}$ (that is $30, 45, 60$ ), a right angle is $\dfrac{\pi}{2}$ , a straight angle is $\pi$ , and the rest are built from these (for instance $120^\circ = 2 \times 60^\circ = \dfrac{2\pi}{3}$ ). Recognising, say, $\dfrac{3\pi}{4}$ as $135^\circ$ on sight is what lets you evaluate trig functions and read sector problems quickly.

Arc length: L = r theta

The arc-length formula falls straight out of the definition. Since $\theta = \dfrac{L}{r}$ , multiplying both sides by $r$ gives

L = r\theta.

That is the whole rule: arc length is radius times the angle in radians. There is no $360$ , no $\pi$ buried in it, which is the entire reason radians are worth the trouble. The diagram shows the set-up: the two radii $r$ enclose the angle $\theta$ at the centre $O$ , and the arc $L$ they cut off is highlighted.

A worked instance: a Ferris wheel of radius $30$ m carries a cabin through a central angle of $40^\circ$ . First convert, $40^\circ = 40 \times \dfrac{\pi}{180} = \dfrac{2\pi}{9}$ , then $L = 30 \times \dfrac{2\pi}{9} = \dfrac{20\pi}{3} \approx 20.94$ m of travel. The conversion to radians is not optional: $L = r\theta$ is false if $\theta$ is left in degrees.

Sector area: A = half r squared theta

A sector is the "pizza slice" region bounded by the two radii and the arc. Its area is the fraction $\dfrac{\theta}{2\pi}$ of the whole circle (the angle as a share of a full turn), and the whole circle has area $\pi r^2$ :

A = \frac{\theta}{2\pi} \times \pi r^2 = \frac{1}{2} r^2 \theta.

The $\pi$ and the $2\pi$ cancel, leaving the compact $A = \tfrac{1}{2} r^2 \theta$ , again only because $\theta$ is in radians. For example a single slice of a $14$ cm pizza cut into eight has central angle $\dfrac{\pi}{4}$ , so its area is $\tfrac{1}{2} \times 14^2 \times \dfrac{\pi}{4} = \dfrac{49\pi}{2} \approx 76.97 \text{ cm}^2$ . The perimeter of a sector, when asked, is the arc plus the two straight radii, $P = L + 2r = r\theta + 2r$ , not $2\pi r$ .

Segment area: the sector minus the triangle

A chord joins the two ends of an arc and slices off a segment, the region between the chord and the arc. To find its area, draw the two radii to the ends of the chord. They split the figure into the sector and the isosceles triangle $AOB$ , and the minor segment is what is left when the triangle is taken out of the sector:

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

The triangle's area uses the two-sides-and-included-angle rule $\tfrac{1}{2} ab \sin C$ with both sides equal to $r$ , giving $\tfrac{1}{2} r^2 \sin\theta$ . Notice the two terms differ only by $\theta$ versus $\sin\theta$ , so the segment area is $\tfrac{1}{2} r^2 (\theta - \sin\theta)$ . The diagram shades the segment and shows the triangle beneath it.

The one thing to watch is the mode: the sector term $\tfrac{1}{2} r^2 \theta$ needs $\theta$ in radians, while the triangle term $\tfrac{1}{2} r^2 \sin\theta$ is the same number whether you read $\theta$ in degrees or radians as long as the calculator is set to match. The safest habit is to convert to radians once and stay there.

How exam questions ask about radians, arcs and sectors

The command words map onto specific actions:

"Express $150^\circ$ in radians" or "convert to radian measure" wants $\times \dfrac{\pi}{180}$ with the fraction cancelled, left in terms of $\pi$ unless a decimal is requested.
"Find the exact value" is the signal to keep $\pi$ (and any $\sqrt{3}$ from a triangle term) in the answer, stopping at $\dfrac{49\pi}{2}$ or $48\pi - 36\sqrt{3}$ rather than rounding.
"Find the length of the arc" means $L = r\theta$ ; if the angle is in degrees, convert first and say so.
"Find the area of the sector" means $A = \tfrac{1}{2} r^2 \theta$ ; "perimeter of the sector" means $r\theta + 2r$ , a common place to forget the two radii.
"Find the area of the (minor) segment" means sector $-$ triangle: write both areas, then subtract. A "major" segment, if ever asked, is the rest of the circle, found by adding the triangle to the major sector or subtracting the minor segment from $\pi r^2$ .
A real context (a fan, a slice, a Ferris wheel arc, a windscreen wiper, a goat tethered in a paddock) is a sector or segment problem in disguise: identify $r$ and $\theta$ , convert $\theta$ to radians, and apply the matching formula with units.

This page builds on the earlier exp/log work only in style; for the circle formulas the prerequisite is right-triangle trig and the area rule $\tfrac{1}{2} ab \sin C$ . The next step up, combining these formulas in multi-part circular-measure problems and meeting radians inside calculus, is the Year-12 page radian measure, arc length and sector area.

Worked example

Seven examples covering the degree-radian conversions both ways, the exact values, an arc length in a real context, a sector area and perimeter, and a segment as sector minus triangle.

Convert 150 degrees to radians

Apply the degrees-to-radians multiplier and cancel.

Write the conversion. To convert degrees to radians, multiply by $\dfrac{\pi}{180}$ :

150^\circ = 150 \times \frac{\pi}{180}.

Cancel the fraction. Divide $150$ and $180$ by their common factor $30$ :

\frac{150\pi}{180} = \frac{5\pi}{6}.

State the answer. So $150^\circ = \dfrac{5\pi}{6}$ radians, exactly. As a decimal, $\dfrac{5\pi}{6} = 2.617993\ldots \approx 2.618$ .

Check. Convert back: $\dfrac{5\pi}{6} \times \dfrac{180}{\pi} = \dfrac{5 \times 180}{6} = \dfrac{900}{6} = 150^\circ$ , the original angle.

Convert 3 pi over 4 radians to degrees

Apply the radians-to-degrees multiplier.

Write the conversion. To convert radians to degrees, multiply by $\dfrac{180}{\pi}$ :

\frac{3\pi}{4} = \frac{3\pi}{4} \times \frac{180}{\pi}.

Cancel the $\pi$ and simplify. The $\pi$ cancels, leaving

\frac{3 \times 180}{4} = \frac{540}{4} = 135^\circ.

State the answer. So $\dfrac{3\pi}{4} = 135^\circ$ .

Check. $135^\circ$ is between $90^\circ$ and $180^\circ$ , and $\dfrac{3\pi}{4}$ is between $\dfrac{\pi}{2}$ and $\pi$ , so the size is in the right place.

Recognise the exact radian value of a common angle

Read a common angle straight off without a full conversion.

Spot the building block. The angle $120^\circ$ is $2 \times 60^\circ$ , and $60^\circ = \dfrac{\pi}{3}$ is one of the three special acute values.

Combine. Doubling the radian value,

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

Confirm by the formula. Directly, $120 \times \dfrac{\pi}{180} = \dfrac{120\pi}{180} = \dfrac{2\pi}{3}$ , the same.

State the answer. So $120^\circ = \dfrac{2\pi}{3} \approx 2.0944$ , worth knowing on sight.

Arc length of a Ferris wheel cabin

A Ferris wheel of radius $30$ m carries a cabin through a central angle of $40^\circ$ . Find the distance the cabin travels along its circular path, to two decimal places.

Convert the angle to radians. Since $L = r\theta$ needs radians,

40^\circ = 40 \times \frac{\pi}{180} = \frac{40\pi}{180} = \frac{2\pi}{9}.

Apply $L = r\theta$ . With $r = 30$ ,

L = 30 \times \frac{2\pi}{9} = \frac{60\pi}{9} = \frac{20\pi}{3} \text{ m}.

Convert to a decimal.

\frac{20\pi}{3} = 20.943951\ldots \approx 20.94 \text{ m}.

Check. A full revolution would be $2\pi \times 30 = 60\pi \approx 188.5$ m; $40^\circ$ is $\dfrac{40}{360} = \dfrac{1}{9}$ of a turn, and $\dfrac{1}{9} \times 60\pi = \dfrac{20\pi}{3}$ , agreeing with the radian calculation. The cabin travels about $20.94$ m.

Area and perimeter of a fan blade sector

A ceiling-fan blade of length $24$ cm sweeps out a sector of central angle $\dfrac{2\pi}{3}$ in one position. Find the exact area swept and the exact perimeter of that sector.

Apply the sector-area formula $A = \tfrac{1}{2} r^2 \theta$ . With $r = 24$ and $\theta = \dfrac{2\pi}{3}$ (already in radians),

A = \frac{1}{2} \times 24^2 \times \frac{2\pi}{3} = \frac{1}{2} \times 576 \times \frac{2\pi}{3} = \frac{576\pi}{3} = 192\pi \text{ cm}^2.

Find the arc for the perimeter. Using $L = r\theta$ ,

L = 24 \times \frac{2\pi}{3} = \frac{48\pi}{3} = 16\pi \text{ cm}.

Add the two radii. The perimeter of a sector is the arc plus two straight edges:

P = L + 2r = 16\pi + 2 \times 24 = 16\pi + 48 \text{ cm}.

State the decimals. $A = 192\pi \approx 603.19 \text{ cm}^2$ and $P = 16\pi + 48 \approx 98.27 \text{ cm}$ .

Check. The angle $\dfrac{2\pi}{3}$ is a third of a revolution, so the area should be $\dfrac{1}{3}$ of the full circle $\pi \times 24^2 = 576\pi$ ; indeed $\dfrac{1}{3} \times 576\pi = 192\pi$ , as found.

Area of a pizza slice

A pizza of radius $14$ cm is cut into eight equal slices. Find the area of one slice, in exact form and to two decimal places.

Find the central angle. Eight equal slices split a full turn $2\pi$ into eighths:

\theta = \frac{2\pi}{8} = \frac{\pi}{4}.

Apply $A = \tfrac{1}{2} r^2 \theta$ . With $r = 14$ ,

A = \frac{1}{2} \times 14^2 \times \frac{\pi}{4} = \frac{1}{2} \times 196 \times \frac{\pi}{4} = \frac{196\pi}{8} = \frac{49\pi}{2} \text{ cm}^2.

Convert to a decimal.

\frac{49\pi}{2} = 76.969020\ldots \approx 76.97 \text{ cm}^2.

Check. Eight slices rebuild the whole pizza: $8 \times \dfrac{49\pi}{2} = 196\pi = \pi \times 14^2$ , the full circle's area, so one slice is right.

Area of a segment, sector minus triangle

In a circle of radius $10$ cm, a chord subtends an angle of $\dfrac{\pi}{3}$ at the centre. Find the exact area of the minor segment cut off by the chord, then a decimal to two places. (Use $\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$ .)

Set up segment $=$ sector $-$ triangle. Draw the two radii to the chord's ends to form the sector of angle $\dfrac{\pi}{3}$ and the isosceles triangle $AOB$ .

Find the sector area. With $A = \tfrac{1}{2} r^2 \theta$ ,

\text{sector} = \frac{1}{2} \times 10^2 \times \frac{\pi}{3} = \frac{1}{2} \times 100 \times \frac{\pi}{3} = \frac{50\pi}{3} \text{ cm}^2.

Find the triangle area. Two sides are $r = 10$ with included angle $\dfrac{\pi}{3}$ , so $\tfrac{1}{2} r^2 \sin\theta$ gives

\text{triangle} = \frac{1}{2} \times 10^2 \times \sin\frac{\pi}{3} = \frac{1}{2} \times 100 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} \text{ cm}^2.

Subtract.

\text{segment} = \frac{50\pi}{3} - 25\sqrt{3} \text{ cm}^2.

Convert to a decimal. Since $\dfrac{50\pi}{3} = 52.359877\ldots$ and $25\sqrt{3} = 43.301270\ldots$ ,

\text{segment} = 52.359877\ldots - 43.301270\ldots = 9.058607\ldots \approx 9.06 \text{ cm}^2.

Check. The triangle ( $\approx 43.30$ ) is smaller than the sector ( $\approx 52.36$ ), so the segment area is positive and is the small sliver between chord and arc, about $9.06 \text{ cm}^2$ .

Final answers: $150^\circ = \dfrac{5\pi}{6}$ and $\dfrac{3\pi}{4} = 135^\circ$ ; $120^\circ = \dfrac{2\pi}{3}$ on sight; the Ferris wheel cabin travels $\dfrac{20\pi}{3} \approx 20.94$ m; the fan sector has area $192\pi \approx 603.19 \text{ cm}^2$ and perimeter $16\pi + 48 \approx 98.27$ cm; a pizza slice is $\dfrac{49\pi}{2} \approx 76.97 \text{ cm}^2$ ; and the segment is $\dfrac{50\pi}{3} - 25\sqrt{3} \approx 9.06 \text{ cm}^2$ .

Common traps

Using a degree angle in $L = r\theta$ or $A = \tfrac{1}{2} r^2 \theta$: Both formulas require radians. For a $40^\circ$ arc you must first write $40^\circ = \dfrac{2\pi}{9}$ ; substituting $40$ gives nonsense.
Leaving the calculator in the wrong mode: When you do need a decimal for $\sin\theta$ or to evaluate an angle, set the calculator to RAD for a radian angle and DEG for a degree angle. A radian answer computed in degree mode is silently wrong.
Forgetting the two radii in a sector perimeter: The perimeter of a sector is $r\theta + 2r$ , the arc plus the two straight edges, not just the arc and not the whole circumference $2\pi r$ .
Rounding $\pi$ too early or dropping the exact form: "Exact value" means keep $\pi$ (and any $\sqrt{3}$ ), so stop at $\dfrac{49\pi}{2}$ or $48\pi - 36\sqrt{3}$ . Converting to a decimal mid-working both loses the exact-value mark and accumulates error.
Mixing up sector and segment: A sector is the pizza slice bounded by two radii and the arc; a segment is the region between a chord and the arc. The segment is the sector with the triangle removed, so it is always the smaller of the two.
Computing the triangle area wrongly: The triangle $AOB$ is isosceles with both sides equal to $r$ , so its area is $\tfrac{1}{2} r^2 \sin\theta$ , not $\tfrac{1}{2} r^2 \theta$ and not $\tfrac{1}{2} \times \text{base} \times r$ unless you have actually found the base.
Treating "the arc" as unambiguous: A chord or two radii define a minor arc and a major arc. Unless told otherwise, the marked (smaller) angle gives the minor arc and minor sector; a major arc or sector uses the reflex angle $2\pi - \theta$ .

Exam technique

Markers reward the conversion to radians, the right formula, exact answers where asked, and units. To secure the marks:

Convert degree angles to radians first and show it. Write $40^\circ = \dfrac{2\pi}{9}$ on its own line before $L = r\theta$ ; that line is often a method mark.
Cancel fractions to keep the exact form. $150 \times \dfrac{\pi}{180}$ should become $\dfrac{5\pi}{6}$ , not a decimal, unless a decimal is requested.
Quote the formula you are using. A line such as " $A = \tfrac{1}{2} r^2 \theta$ " before substituting shows the examiner the method even if the arithmetic slips.
For a sector perimeter, add the two radii. Write $P = r\theta + 2r$ ; losing the $+2r$ is the most common perimeter error.
For a segment, write sector and triangle separately, then subtract. Show $\tfrac{1}{2} r^2 \theta$ and $\tfrac{1}{2} r^2 \sin\theta$ on their own lines so the subtraction is clear.
Leave $\pi$ and surds in an exact answer; round only when asked. $\dfrac{49\pi}{2}$ and $48\pi - 36\sqrt{3}$ are exact; convert at the very end if "to two decimal places" is stated, and keep the units ( $\text{cm}$ , $\text{cm}^2$ , m).
Set the calculator mode to match the angle whenever you evaluate a trig value, and state your decimal to the requested accuracy.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksConvert to radians in terms of

\pi

: (a)

60^\circ

, (b)

135^\circ

, and (c)

240^\circ

Show worked solution →

Use the conversion rule. To go from degrees to radians, multiply by $\dfrac{\pi}{180}$ , then simplify the fraction.

(a) Convert $60^\circ$ .

60 \times \frac{\pi}{180} = \frac{60\pi}{180} = \frac{\pi}{3}.

(b) Convert $135^\circ$ .

135 \times \frac{\pi}{180} = \frac{135\pi}{180} = \frac{3\pi}{4}.

(c) Convert $240^\circ$ .

240 \times \frac{\pi}{180} = \frac{240\pi}{180} = \frac{4\pi}{3}.

Check. Reversing one: $\dfrac{\pi}{3} \times \dfrac{180}{\pi} = 60^\circ$ , so part (a) is right. As decimals these are $\dfrac{\pi}{3} \approx 1.0472$ , $\dfrac{3\pi}{4} \approx 2.3562$ and $\dfrac{4\pi}{3} \approx 4.1888$ .

foundation2 marksAn arc of a circle of radius

8\text{ cm}

subtends an angle of

\dfrac{\pi}{4}

at the centre. Find the exact arc length, then give it to two decimal places.

Show worked solution →

State the arc-length formula. With the angle in radians, $L = r\theta$ .

Substitute $r = 8$ and $\theta = \dfrac{\pi}{4}$ .

L = 8 \times \frac{\pi}{4} = \frac{8\pi}{4} = 2\pi \text{ cm}.

Convert to a decimal.

2\pi = 6.283185\ldots \approx 6.28 \text{ cm}.

Check. A quarter-circle of this radius would have arc $\dfrac{1}{4}(2\pi \times 8) = 4\pi$ ; here $\theta = \dfrac{\pi}{4}$ is one eighth of a revolution, giving half of that, $2\pi$ , as found.

core3 marksA sector of a circle of radius

10\text{ cm}

has a central angle of

1.2

radians. Find (a) the arc length and (b) the area of the sector, each to two decimal places.

Show worked solution →

(a) Apply $L = r\theta$ . The angle is already in radians, so substitute directly:

L = 10 \times 1.2 = 12 \text{ cm}.

(b) Apply $A = \tfrac{1}{2} r^2 \theta$ .

A = \frac{1}{2} \times 10^2 \times 1.2 = \frac{1}{2} \times 100 \times 1.2 = 60 \text{ cm}^2.

Check. The two answers are consistent: $A = \tfrac{1}{2} r L = \tfrac{1}{2} \times 10 \times 12 = 60 \text{ cm}^2$ , the same area. Both come out exact here because $\theta = 1.2$ is a plain number, not a multiple of $\pi$ .

core3 marksA sector has area

24\text{ cm}^2

and radius

6\text{ cm}

. Find the central angle, in radians as an exact fraction and then in degrees to two decimal places.

Show worked solution →

Substitute into $A = \tfrac{1}{2} r^2 \theta$ . Put $A = 24$ and $r = 6$ :

24 = \frac{1}{2} \times 6^2 \times \theta = \frac{1}{2} \times 36 \times \theta = 18\theta.

Solve for $\theta$ .

\theta = \frac{24}{18} = \frac{4}{3} \text{ radians}.

Convert to degrees. Multiply by $\dfrac{180}{\pi}$ :

\frac{4}{3} \times \frac{180}{\pi} = \frac{240}{\pi} = 76.394372\ldots \approx 76.39^\circ.

Check. Substituting back, $\tfrac{1}{2} \times 36 \times \tfrac{4}{3} = 18 \times \tfrac{4}{3} = 24 \text{ cm}^2$ , the given area.

exam5 marksA pizza of radius

14\text{ cm}

is cut into eight equal slices, so each slice is a sector with central angle

\dfrac{\pi}{4}

. For one slice find, leaving each answer in exact form and then to two decimal places: (a) the length of the curved crust, (b) the perimeter of the slice, and (c) the area of the slice.

Show worked solution →

(a) The curved crust is the arc, $L = r\theta$ .

L = 14 \times \frac{\pi}{4} = \frac{14\pi}{4} = \frac{7\pi}{2} \text{ cm} = 10.995574\ldots \approx 11.00 \text{ cm}.

(b) The perimeter is the arc plus the two straight edges. Each straight edge is a radius, so

P = L + 2r = \frac{7\pi}{2} + 2 \times 14 = \frac{7\pi}{2} + 28 \text{ cm} = 38.995574\ldots \approx 39.00 \text{ cm}.

(c) The area is the sector area, $A = \tfrac{1}{2} r^2 \theta$ .

A = \frac{1}{2} \times 14^2 \times \frac{\pi}{4} = \frac{1}{2} \times 196 \times \frac{\pi}{4} = \frac{49\pi}{2} \text{ cm}^2 = 76.969020\ldots \approx 76.97 \text{ cm}^2.

Check. Eight slices should rebuild the whole pizza: $8 \times \dfrac{49\pi}{2} = 196\pi = \pi \times 14^2$ , the area of the full circle, so the slice area is right.

exam5 marksA circular stained-glass window has radius

12\text{ cm}

. A chord cuts off a minor segment whose two bounding radii meet at the centre at an angle of

\dfrac{2\pi}{3}

. Find the area of the minor segment, in exact form and then to two decimal places. (Use

\sin\dfrac{2\pi}{3} = \dfrac{\sqrt{3}}{2}

Show worked solution →

Set up: segment $=$ sector $-$ triangle. The minor segment is the sector of angle $\dfrac{2\pi}{3}$ with the isosceles triangle $AOB$ removed.

Find the sector area with $A = \tfrac{1}{2} r^2 \theta$ .

\text{sector} = \frac{1}{2} \times 12^2 \times \frac{2\pi}{3} = \frac{1}{2} \times 144 \times \frac{2\pi}{3} = 48\pi \text{ cm}^2.

Find the triangle area with $\tfrac{1}{2} r^2 \sin\theta$ .

\text{triangle} = \frac{1}{2} \times 12^2 \times \sin\frac{2\pi}{3} = \frac{1}{2} \times 144 \times \frac{\sqrt{3}}{2} = 36\sqrt{3} \text{ cm}^2.

Subtract.

\text{segment} = 48\pi - 36\sqrt{3} \text{ cm}^2 = 150.796447\ldots - 62.353829\ldots = 88.442618\ldots \approx 88.44 \text{ cm}^2.

Check. The triangle ( $36\sqrt{3} \approx 62.35$ ) is smaller than the sector ( $48\pi \approx 150.80$ ), so the segment is positive, as a region must be; and the segment ( $\approx 88.44$ ) is less than the sector, exactly as "sector minus triangle" requires.

What this dot point is asking

The answer

What a radian is: arc length over radius

Converting between degrees and radians

The exact radian values worth memorising

Arc length: L = r theta

Sector area: A = half r squared theta

Segment area: the sector minus the triangle

How exam questions ask about radians, arcs and sectors

Practice questions

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