Circular Measure

Syllabus coverage

Understand the definition of a radian, and use the relationship between radians and degrees
Use the formulae $s = r θ$ and $A = \frac{1}{2} r^{2} θ$ in solving problems concerning the arc length and sector area of a circle
Calculate lengths and angles in the triangles formed by two radii and a chord, including the chord itself
Calculate the area of a triangle with two sides of length $r$ and the area of the segment cut off by a chord
Choose the unit the question needs, work in radians without converting back to degrees, and give exact answers in terms of $π$ where the question asks for them

Part of the AS & A Level Maths syllabus 9709.

A sector: the shaded region is bounded by the two radii and the arc, and

θ

is the angle at the centre. The dotted line through the centre is a diameter.

What a radian is

A radian measures an angle by the arc it cuts off rather than by a fraction of a full turn. One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius: lay the radius along the circumference and the angle you have marked out is 1 radian.

Because the circumference is $2 π r$ , a full turn is $2 π$ radians. Half a turn is $π$ radians, a right angle is $\frac{π}{2}$ radians, and in degrees one radian is $\frac{180}{π} = {57.3}^{\circ}$ to 3 significant figures - a little under $60^{\circ}$ , which is a useful check on any conversion you do.

Radians are not a different kind of angle, only a different unit for the same thing. What they buy you is that arc length and angle are then related by multiplication rather than by a fraction of $360^{\circ}$ , which is why every formula in this topic is written in radians and why the rest of the course uses them too.

Put your calculator into radian mode the moment a question mentions radians. Most wrong answers in this topic are the right method worked out with the calculator still in degree mode.

Converting between degrees and radians

One fact does all the conversions: $π$ radians $= 180^{\circ}$ . To go from degrees to radians, multiply by $\frac{π}{180}$ . To go from radians to degrees, multiply by $\frac{180}{π}$ . Keep the answer in terms of $π$ when the question asks for it, because the exact form $\frac{2 π}{5}$ is wanted more often than its decimal $1.2566$ .

So $72^{\circ} = 72 \times \frac{π}{180} = \frac{2 π}{5}$ radians, and $\frac{5 π}{6}$ radians $= \frac{5 π}{6} \times \frac{180}{π} = 150^{\circ}$ . The $π$ cancels when you go the other way, which is the sign that you used the right fraction. Getting the fraction upside down gives a nonsense answer such as $72 \times \frac{180}{π} = 4125^{\circ}$ , so the size of the result is its own warning.

Degrees	Radians	Degrees	Radians
$15^{\circ}$	$\frac{π}{12}$	$120^{\circ}$	$\frac{2 π}{3}$
$30^{\circ}$	$\frac{π}{6}$	$135^{\circ}$	$\frac{3 π}{4}$
$45^{\circ}$	$\frac{π}{4}$	$150^{\circ}$	$\frac{5 π}{6}$
$60^{\circ}$	$\frac{π}{3}$	$180^{\circ}$	$π$
$72^{\circ}$	$\frac{2 π}{5}$	$270^{\circ}$	$\frac{3 π}{2}$
$90^{\circ}$	$\frac{π}{2}$	$360^{\circ}$	$2 π$

0.8 radians is

{45.8}^{\circ}

and 2.4 radians is

{137.5}^{\circ}

. Knowing roughly what a radian value looks like in degrees tells you at once whether a sector angle is sensible.

Arc length: $s = r θ$

The definition of the radian hands you the arc length formula. Since the angle in radians is arc divided by radius, $θ = \frac{s}{r}$ , multiplying both sides by $r$ gives $s = r θ$ . The angle must be in radians: a question that gives degrees is really testing the conversion.

For a sector of radius 8 cm and angle 1.5 radians the arc is $s = 8 \times 1.5 = 12$ cm, which is one line of working. In degrees the same calculation would need $1.5 \times \frac{180}{π} = {85.9}^{\circ}$ and then a fraction of the circumference, so radians are doing real work rather than just being notation.

The formula scales the way you would expect. Doubling the radius doubles the arc, doubling the angle doubles the arc, and a full turn $θ = 2 π$ gives $s = 2 π r$ , which is the circumference you have known since IGCSE. The special case $θ = 1$ gives $s = r$ , which is the definition of the radian written as a formula.

$s = r θ$ with $θ$ in radians - always convert first
$θ = 2 π$ gives $s = 2 π r$ , the whole circumference
$θ = π$ gives $s = π r$ , half the circumference
$θ = 1$ gives $s = r$ , which is what a radian means

Sector area: $A = \frac{1}{2} r^{2} θ$

A sector is the fraction $\frac{θ}{2 π}$ of the whole circle, so its area is that fraction of $π r^{2}$ : $A = \frac{θ}{2 π} \times π r^{2} = \frac{1}{2} r^{2} θ$ . Being able to derive it in one line is worth more than memorising it, because that line is often the method mark.

The formula also has a second shape worth knowing: $A = \frac{1}{2} r^{2} θ = \frac{1}{2} r (r θ) = \frac{1}{2} r s$ . The sector area is half the base times the height, with the arc playing the part of the base - which is exactly why it looks like the triangle formula.

For the sector of radius 8 cm and angle 1.5 radians, $A = \frac{1}{2} \times 8^{2} \times 1.5 = 48$ cm2. The angle contributes as a pure number, with no unit, and the answer is in square units because it is an area.

Check the formula against the whole circle:

θ = 2 π

gives

A = π r^{2}

and

θ = π

gives

A = \frac{1}{2} π r^{2}

. If your formula fails those two tests, the wrong quantity has been squared.

The perimeter of a sector, and working backwards

The perimeter of a sector is the arc plus the two radii: $P = r θ + 2 r$ , which is $r (θ + 2)$ factorised. Students lose this mark by writing down the arc and stopping. The two radii are part of the boundary, so they are part of the perimeter, and for the sector of radius 8 cm and angle 1.5 radians the answer is $12 + 16 = 28$ cm.

Working backwards is a favourite question. Given the arc length and the area, both formulae contain the product $r θ$ , so dividing one by the other removes it. From $s = r θ = 15$ and $A = \frac{1}{2} r^{2} θ = 45$ : write the area as $A = \frac{1}{2} r (r θ) = \frac{1}{2} \times 15 r$ , so $7.5 r = 45$ , $r = 6$ cm, and then $θ = \frac{15}{6} = 2.5$ radians.

The same idea gives a general shortcut: from $A = \frac{1}{2} r s$ , the radius is $\frac{2 A}{s}$ and then the angle follows from $θ = \frac{s}{r}$ . Two quantities given, two unknowns, two formulae - the whole skill is choosing which to use first.

Perimeter of a sector $= r θ + 2 r$ : arc plus two radii
From the arc and the area: $r = \frac{2 A}{s}$ , then $θ = \frac{s}{r}$
From the arc and the angle: $r = \frac{s}{θ}$
From the area and the angle: $r = \sqrt{\frac{2 A}{θ}}$

A sector of radius 5 cm and angle 2.4 radians has arc

12

cm, area

30

cm2 and perimeter

22

cm. Work all three from the same two numbers until it is automatic.

Chords, triangles and the area of a segment

The chord joining the ends of an arc is the third side of the isosceles triangle $A O B$ , where $O$ is the centre. Splitting that triangle down the middle gives two right-angled triangles, each with hypotenuse $r$ and angle $\frac{θ}{2}$ at the centre, so the half-chord is $r \sin \frac{θ}{2}$ and the chord is $A B = 2 r \sin \frac{θ}{2}$ . For a radius of 10 cm and an angle of 0.9 radians, $A B = 20 \sin 0.45 = 8.70$ cm to 3 significant figures.

The area of that triangle is $A = \frac{1}{2} r^{2} \sin θ$ , the familiar half $a b \sin C$ with both sides equal to $r$ . The same triangle can be done as two right-angled triangles, but the formula is faster and it is the one the mark scheme has in mind.

The segment between the chord and the arc is the sector minus the triangle: $A = \frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} \sin θ = \frac{1}{2} r^{2} (θ - \sin θ)$ . For a radius of 6 cm and an angle of $\frac{π}{3}$ radians the sector is $\frac{1}{2} \times 36 \times \frac{π}{3} = 6 π$ , the triangle is $\frac{1}{2} \times 36 \times \sin \frac{π}{3} = 9 \sqrt{3}$ , and the segment is $6 π - 9 \sqrt{3} = 3.26$ cm2 to 3 significant figures.

Region	Formula	For $r = 6$ cm and $θ = \frac{π}{3}$
Sector	$\frac{1}{2} r^{2} θ$	$6 π = 18.85$ cm2
Triangle $A O B$	$\frac{1}{2} r^{2} \sin θ$	$9 \sqrt{3} = 15.59$ cm2
Segment	$\frac{1}{2} r^{2} (θ - \sin θ)$	$6 π - 9 \sqrt{3} = 3.26$ cm2
Chord $A B$	$2 r \sin \frac{θ}{2}$	$12 \sin \frac{π}{6} = 6$ cm

A segment is always smaller than the sector it comes from, and the difference is the triangle. If your segment area is bigger than the sector area, the triangle has been added instead of subtracted.

Which unit, and when

The arc and sector formulae only work in radians, so a question that gives degrees is really testing the conversion. Convert once, at the start, and write the conversion on the page: in most of these questions it carries a mark of its own, and it is also the step that catches the calculator left in the wrong mode.

Degrees are still the unit to use for angles you can picture - a $60^{\circ}$ slice, a right angle, a bearing - and they are what a context question usually supplies. Radians are the unit of the rest of the course: the derivative of $\sin x$ is $\cos x$ only when the angle is measured in radians, so this topic is where the pure maths course settles into its own unit.

Exact answers in terms of $π$ are expected whenever the question says 'in terms of $π$ ', and a decimal answer there loses the accuracy mark. When the question asks for a decimal instead, work in exact form to the end and round once, to the accuracy named. Rounding the angle first and then using it is how a 3 significant figure question ends up wrong in the third figure.

If the question gives an angle in degrees and asks for a length, the conversion is a mark. If it gives an angle in radians and asks for an area, the conversion is not needed - but neither is a conversion back to degrees.

Modelling with sectors

Most circular measure questions are dressed up as something physical: a windscreen wiper, a bend on a running track, a slice of pizza, a security camera sweeping a car park, the minute hand of a clock. The mathematics never changes. Find the radius, find the angle in radians, then decide whether the question wants an arc, a sector area, a chord or a perimeter.

Two decisions catch students out. The first is which length is the radius: a windscreen wiper's radius is the length of the arm rather than the width of the blade, and a running track's radius belongs to the bend rather than to the straight. The second is whether the angle is the part swept out or the part left behind, since a question about what remains of a lawn or a clock face wants $2 π - θ$ .

One context worked through: a windscreen wiper arm 45 cm long sweeps through $100^{\circ}$ . In radians that is $100 \times \frac{π}{180} = \frac{5 π}{9} = 1.7453$ . The tip of the arm travels $45 \times 1.7453 = 78.5$ cm, and the area swept out is $\frac{1}{2} \times 45^{2} \times 1.7453 = 1770$ cm2 to 3 significant figures.

Windscreen wiper: the radius is the arm length and the angle is the sweep
Running track: each bend is an arc $r θ$ ; the straights are added separately
Clock hand: the radius is the hand's length, and 20 minutes is $120^{\circ} = \frac{2 π}{3}$ radians
Slice of pizza: the crust is an arc, the topping is a sector, the cut edge is a chord

Key terms you must be able to define

Radian: The angle subtended at the centre of a circle by an arc equal in length to the radius. One radian is $\frac{180}{π} = {57.3}^{\circ}$ .

Degree: One 360th of a full turn. The bridge between the two units is $π$ radians $= 180^{\circ}$ .

Arc: A part of the circumference. Its length is $s = r θ$ when $θ$ is in radians.

Minor arc: The shorter of the two arcs between two points on a circle, with angle less than $π$ radians.

Major arc: The longer of the two arcs between two points, with angle greater than $π$ radians. The two arcs together make the circumference.

Sector: The region bounded by two radii and an arc. Its area is $A = \frac{1}{2} r^{2} θ$ .

Segment: The region bounded by a chord and an arc. It is the sector minus the triangle formed with the centre.

Chord: The straight line joining two points on a circle. For a sector of angle $θ$ , the chord is $2 r \sin \frac{θ}{2}$ .

Subtend: The angle that an arc or a chord makes at a point, usually the centre of the circle.

Perimeter of a sector: The arc plus the two radii: $r θ + 2 r$ , or $r (θ + 2)$ factorised.

Triangle in a sector: The isosceles triangle formed by two radii and a chord. Its area is $\frac{1}{2} r^{2} \sin θ$ .

Area of a segment: Sector minus triangle: $\frac{1}{2} r^{2} (θ - \sin θ)$ with $θ$ in radians.

Conversion factor: Multiply degrees by $\frac{π}{180}$ to get radians; multiply radians by $\frac{180}{π}$ to get degrees.

Full turn: $2 π$ radians, or $360^{\circ}$ . Half a turn is $π$ radians and a right angle is $\frac{π}{2}$ radians.

Applied

At AS & A Level the marks sit in the setup: choosing the model, stating the variable and writing the equation before any working. These show the shape of a full-mark answer, then you do the algebra yourself.

A windscreen wiper

A wiper arm 45 cm long sweeps through $100^{\circ}$ . In radians the sweep is $100 \times \frac{π}{180} = \frac{5 π}{9} = 1.7453$ . The tip travels $s = 45 \times 1.7453 = 78.5$ cm, and the area cleared is $A = \frac{1}{2} \times 45^{2} \times 1.7453 = 1770$ cm2 to 3 significant figures.

A bend on a running track

Each bend of a track is a semicircle of radius 36.5 m, so its angle is $π$ radians. The arc of one bend is $36.5 π = 114.7$ m, which is the length a runner covers round the bend, and the area of the semicircular section is $\frac{1}{2} \times {36.5}^{2} \times π = 2090$ m2 to 3 significant figures.

A slice of pizza

A slice has an angle of $40^{\circ}$ at the point and a radius of 18 cm. The angle in radians is $40 \times \frac{π}{180} = \frac{2 π}{9} = 0.6981$ . The crust is the arc, $18 \times 0.6981 = 12.6$ cm, and the topping covers $\frac{1}{2} \times 18^{2} \times 0.6981 = 113$ cm2.

A security camera

A camera turns through 1.4 radians, and the nearest ground it sees is 15 m from the lens. The arc of ground along that nearest edge is $15 \times 1.4 = 21$ m, and the area of ground inside the sweep is $\frac{1}{2} \times 15^{2} \times 1.4 = 158$ m2 to 3 significant figures. The angle is already in radians, so no conversion is needed.

A clock hand

The minute hand of a clock is 12 cm long. In 20 minutes it turns through a third of a full turn, which is $\frac{1}{3} \times 2 π = \frac{2 π}{3} = 2.0944$ radians. The tip travels $12 \times 2.0944 = 25.1$ cm and the hand sweeps an area of $\frac{1}{2} \times 12^{2} \times 2.0944 = 150.8$ cm2.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What is a radian?

The angle subtended at the centre of a circle by an arc equal in length to the radius.

Q2How many radians are in a full turn?

2 π

. Half a turn is

π

radians and a right angle is

\frac{π}{2}

radians.

Q3How do you convert degrees to radians?

Multiply by

\frac{π}{180}

. For example

72^{\circ} = \frac{2 π}{5}

Q4How do you convert radians to degrees?

Multiply by

\frac{180}{π}

. For example

\frac{5 π}{6} = 150^{\circ}

Q5What is 1 radian in degrees?

\frac{180}{π} = {57.3}^{\circ}

to 3 significant figures, a little under

60^{\circ}

Q6What is

60^{\circ}

in radians?

\frac{π}{3}

. Halving

120^{\circ} = \frac{2 π}{3}

gives the same answer.

Q7State the arc length formula.

s = r θ

, with the angle

θ

in radians.

Q8State the sector area formula.

A = \frac{1}{2} r^{2} θ

, which is also

A = \frac{1}{2} r s

because

s = r θ

Q9What is the perimeter of a sector?

r θ + 2 r

: the arc plus the two radii.

Q10How do you find the chord of a sector of angle

θ

A B = 2 r \sin \frac{θ}{2}

, from splitting the isosceles triangle

A O B

down the middle.

Q11What is the area of the triangle formed by two radii and a chord?

\frac{1}{2} r^{2} \sin θ

, which is half

a b \sin C

with both sides equal to

r

Q12How do you find the area of a segment?

Sector minus triangle:

\frac{1}{2} r^{2} (θ - \sin θ)

Q13A sector has radius 5 cm and angle 2.4 radians. What is its area?

\frac{1}{2} \times 25 \times 2.4 = 30

cm2.

Q14A sector has arc length 15 cm and area 45 cm2. What is its radius?

r = \frac{2 A}{s} = \frac{90}{15} = 6

cm, and then

θ = \frac{15}{6} = 2.5

radians.

Q15How many radians is

270^{\circ}

\frac{3 π}{2}

radians, three quarters of a full turn of

2 π

0 of 15 marked as known

Exam technique for this topic

Convert degrees to radians before you do anything else, and write the conversion down. In this topic it is usually worth a mark on its own, and a calculator left in degree mode is the commonest single cause of a wrong answer.
Decide what the question wants before you calculate. An arc, a sector area, a chord and a perimeter all start from the same radius and angle, and the mark scheme rewards choosing the right one.
Keep the working exact, in terms of $π$ , until the final line and then round once. Rounding the angle early and squaring the rounded value is how the third significant figure goes wrong.
The perimeter of a sector is the arc plus two radii, so $P = r θ + 2 r$ . Writing down the arc alone is the classic one-mark loss here.
When a question gives two of $r$ , $θ$ , $s$ and $A$ , write both formulae down before choosing. The equation that removes the product $r θ$ is almost always the quickest way in.

Why students lose marks here

A student wroteA sector of radius 6 cm and angle

60^{\circ}

has arc length

6 \times 60 = 360

cm.

Why it loses marksThe formula

s = r θ

needs the angle in radians.

60^{\circ}

is not 60 radians, and the answer is larger than the whole circle, which is the giveaway.

The answer that scores

60^{\circ} = \frac{π}{3}

radians, so

s = 6 \times \frac{π}{3} = 2 π = 6.28

cm.

A student wroteThe perimeter of a sector of radius 8 cm and angle 1.5 radians is 12 cm.

Why it loses marksThat is the arc on its own. The boundary of a sector is the arc together with the two radii that close it.

The answer that scores

P = r θ + 2 r = 8 \times 1.5 + 16 = 28

cm.

A student wroteThe area of a sector of radius 4 cm and angle 3 radians is

\frac{1}{2} \times 4 \times 3 = 6

cm2.

Why it loses marksThe radius is squared in the area formula. Using

r

instead of

r^{2}

gives an answer four times too small here.

The answer that scores

A = \frac{1}{2} \times 4^{2} \times 3 = 24

cm2.

A student wroteThe area of the segment cut off by a chord of a circle of radius 6 cm and angle

\frac{π}{3}

6 π = 18.8

cm2.

Why it loses marksThat is the area of the sector. The segment stops at the chord, so the triangle formed by the two radii and the chord has to come off.

The answer that scores

6 π - 9 \sqrt{3} = 18.85 - 15.59 = 3.26

cm2 to 3 significant figures.

A student wrote1 radian is

180^{\circ}

Why it loses marks

π

radians is

180^{\circ}

. The conversion factor has been confused with the unit itself, which makes every answer about 3 times too large.

The answer that scores1 radian is

\frac{180}{π} = {57.3}^{\circ}

to 3 significant figures.

Syllabus coverage

What a radian is

Converting between degrees and radians

Arc length: $s = r θ$

Sector area: $A = \frac{1}{2} r^{2} θ$

The perimeter of a sector, and working backwards

Chords, triangles and the area of a segment

Which unit, and when

Modelling with sectors

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here

Circular Measure

Syllabus coverage

What a radian is

Converting between degrees and radians

Arc length: s=rθ

Sector area: A=12r2θ

The perimeter of a sector, and working backwards

Chords, triangles and the area of a segment

Which unit, and when

Modelling with sectors

Key terms you must be able to define

Applied

Flashcards

Exam-style questions

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Mark scheme — tick each point you actually made

Exam technique for this topic

Why students lose marks here

Arc length: $s = r θ$

Sector area: $A = \frac{1}{2} r^{2} θ$