Radians, Arcs and Sectors

Measure angles in radians, convert to and from degrees, and use s = rθ and A = ½r²θ for arcs, sectors, segments and compound shapes.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 31 real past-paper questions below

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What is a radian?

Why is a full turn 360°? It is a choice people made long ago. Nothing about the circle forces it. Mathematicians prefer a unit that comes from the circle itself: the radian.

Rule

Take the radius of a circle and bend it round the edge. The angle at the centre that this arc makes is 1 radian. So in 1 radian, the arc length equals the radius.

How many radii fit round the whole circle? The circumference is $2 π r$ , which is $2 π$ lots of r. So a full turn is $2 π$ radians, and half a turn is $π$ radians.

Formula

π radians = 18 0^{\circ}

$π$ — half a turn measured in radians, about 3.14.
$18 0^{\circ}$ — the same half turn measured in degrees.

Example

$1 radian = \frac{18 0 ^{\circ}}{π} \approx 57. 3^{\circ}$ . A radian is a big angle: just under 60°.

Radians are often written with no unit at all. If an angle has no ° sign, such as $θ = 1.2$ or $θ = \frac{π}{4}$ , it is in radians. Some books write rad or c.

Key idea

The word “radian” first appeared in print in 1873, in exam questions set by James Thomson at Queen's College, Belfast. The idea is older, but the name stuck.

Key idea

A radian measures an angle by the circle itself: how many radii of arc the angle cuts off.

Degrees and radians

Since $π$ radians is 180°, you convert by multiplying by a fraction that equals 1.

Formula

radians = degrees \times \frac{π}{180}

$π$ — on top, because you want radians.
$180$ — underneath, to cancel the degrees.

Formula

degrees = radians \times \frac{180}{π}

$180$ — on top, because you want degrees.
$π$ — underneath, to cancel the radians.

Degrees	30°	45°	60°	90°	180°	360°
Radians	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$π$	$2 π$

Worked example

1
Change 150° to radians, and 2.5 radians to degrees.
2
$150 \times \frac{π}{180} = \frac{5 π}{6}$ — Multiply by π over 180 and cancel: 150 ÷ 180 = 5 ÷ 6.
3
$2.5 \times \frac{180}{π} = 143. 2^{\circ}$ — Multiply by 180 over π. 143.239..., to 1 decimal place.

Rule

No calculator on 0606 and 4037 Paper 1: write angles exactly as fractions of π, such as $\frac{5 π}{6}$ . Learn the table above.

Common mistake

Leaving your calculator in degree mode. Before any radian question, switch it to RAD, or every sin and cos will be wrong.

Key idea

To radians: times π over 180. To degrees: times 180 over π.

Arc length

An arc is part of the circumference. If the angle is $θ$ radians, the arc is the fraction $\frac{θ}{2 π}$ of the whole circle.

$s = \frac{θ}{2 π} \times 2 π r$ — the arc is this fraction of the circumference.

$s = r θ$ — the $2 π$ cancels.

Formula

s = r θ

$s$ — the arc length, in the same units as r.
$r$ — the radius.
$θ$ — the angle at the centre, in radians.

Rule

Learn this formula. 0606 and 4037 do not give it, and 4PM1 expects you to know it.

Worked example

1
A sector has radius 6 cm and angle $\frac{2 π}{3}$ . Find its arc length and its perimeter.
2
$s = 6 \times \frac{2 π}{3} = 4 π$ — Arc length $s = r θ$ . Leave it exact: 4π cm.
3
$P = 6 + 6 + 4 π = 12 + 4 π$ — The perimeter of a sector is two radii plus the arc.

Worked example

1
An arc of length 9 cm is on a circle of radius 6 cm. Find the angle at the centre.
2
$9 = 6 θ$ — Put the values into $s = r θ$ .
3
$θ = 1.5$ — Divide by 6. The angle is 1.5 radians.

Common mistake

Using degrees in $s = r θ$ , such as $6 \times 120$ . The formula only works with $θ$ in radians.

Key idea

In radians, arc length is simply radius times angle. No 360 needed.

Sector area

A sector is a slice of the circle. With angle $θ$ , it is the fraction $\frac{θ}{2 π}$ of the whole area $π r^{2}$ .

$A = \frac{θ}{2 π} \times π r^{2}$ — the sector is this fraction of the circle.

$A = \frac{1}{2} r^{2} θ$ — the π cancels, leaving a half.

Formula

A = \frac{1}{2} r^{2} θ

$A$ — the area of the sector, in square units.
$r$ — the radius. Only r is squared.
$θ$ — the angle at the centre, in radians.

Rule

Learn this formula too. No board prints it.

Worked example

1
Find the area of a sector with radius 8 cm and angle 0.75 radians.
2
$A = \frac{1}{2} \times 8^{2} \times 0.75$ — Put the values into the formula.
3
$A = 24$ — Half of 64 is 32, and 32 × 0.75 = 24. The area is 24 cm².

Worked example

1
A sector of radius 5 cm has area 30 cm². Find its angle.
2
$30 = \frac{1}{2} \times 25 \times θ$ — Put the values into $A = \frac{1}{2} r^{2} θ$ .
3
$θ = \frac{60}{25} = 2.4$ — Multiply by 2, then divide by 25. The angle is 2.4 radians.

Common mistake

Forgetting the half, or squaring $θ$ instead of r. It is half, times r squared, times $θ$ .

Key idea

Arc: $r θ$ . Area: $\frac{1}{2} r^{2} θ$ . Both need $θ$ in radians.

Finding r and θ together

Some questions give the area and the perimeter of a sector, and you must find both r and $θ$ . That is two equations with two unknowns, so substitute one into the other.

Worked example

1
A sector has area 5 cm² and perimeter 12 cm. Find the radius and the angle.
2
$\frac{1}{2} r^{2} θ = 5 \Rightarrow θ = \frac{10}{r ^{2}}$ — Make $θ$ the subject of the area equation.
3
$r θ + 2 r = 12$ — Perimeter: the arc plus two radii.
4
$\frac{10}{r} + 2 r = 12$ — Substitute: $r \times \frac{10}{r ^{2}} = \frac{10}{r}$ .
5
$2 r^{2} - 12 r + 10 = 0$ — Multiply every term by r, then move everything to one side.
6
$(r - 1) (r - 5) = 0$ — Divide by 2 and factorise.
7
$r = 1 \Rightarrow θ = 10$ — Reject: 10 radians is more than a full turn, $2 π \approx 6.28$ .
8
$r = 5 \Rightarrow θ = 0.4$ — $θ = \frac{10}{25} = 0.4$ . So r = 5 cm and θ = 0.4 radians.

Rule

A sector angle must be between 0 and $2 π$ . Check every answer and reject any that is not.

Common mistake

Keeping both pairs of answers. Always test each one against the shape.

Key idea

Two facts about one sector means two equations. Make θ the subject of one, then substitute.

Segments

A segment is the part of a sector cut off by a chord. Its area is the sector minus the triangle. The triangle has two sides r with the angle $θ$ between them, so its area is $\frac{1}{2} r^{2} sin θ$ .

Formula

Segment = \frac{1}{2} r^{2} (θ - sin θ)

$r$ — the radius.
$θ$ — the angle at the centre, in radians. Your calculator must be in radian mode for sin θ.

Worked example

1
A chord cuts off an angle of $\frac{π}{3}$ at the centre of a circle of radius 10 cm. Find the exact area and the perimeter of the minor segment.
2
$Sector = \frac{1}{2} \times 100 \times \frac{π}{3} = \frac{50 π}{3}$ — $A = \frac{1}{2} r^{2} θ$ .
3
$Triangle = \frac{1}{2} \times 100 \times sin \frac{π}{3} = 253$ — $sin \frac{π}{3} = \frac{3}{2}$ .
4
$Segment = \frac{50 π}{3} - 253$ — Sector minus triangle. About 9.06 cm².
5
$Chord = 10$ — The triangle has two sides of 10 and a 60° angle, so it is equilateral.
6
$Perimeter = 10 + \frac{10 π}{3}$ — Chord plus arc. Arc $= 10 \times \frac{π}{3}$ .

Common mistake

Working out $sin \frac{π}{3}$ with the calculator in degree mode. That gives sin of about 1.05°, which is wrong.

Key idea

Segment = sector − triangle. Its perimeter is the arc plus the chord.

Compound shapes

Harder questions join sectors, triangles and tangents. First plan which pieces add and which subtract. Then work out each piece.

Recall: a tangent meets the radius at a right angle. So right-angled trigonometry can give you the angle exactly.

Worked example

1
A circle has centre O and radius 6 cm. The tangent at A meets a line from O at T, where OT = 12 cm. The line OT crosses the circle at B. Find the exact area and perimeter of the region between AT, BT and the arc AB.
2
$cos A O T = \frac{6}{12} = \frac{1}{2} \Rightarrow A O T = \frac{π}{3}$ — Angle OAT is a right angle, so use cos. Give the angle in radians.
3
$A T = 1 2^{2} - 6^{2} = 63$ — Pythagoras in triangle OAT.
4
$Triangle O A T = \frac{1}{2} \times 6 \times 63 = 183$ — Half base times height: OA and AT are at right angles.
5
$Sector O A B = \frac{1}{2} \times 36 \times \frac{π}{3} = 6 π$ — $A = \frac{1}{2} r^{2} θ$ .
6
$Area = 183 - 6 π$ — Triangle minus sector. About 12.3 cm².
7
$B T = 12 - 6 = 6$ — OT minus the radius OB.
8
$Perimeter = 63 + 6 + 2 π$ — AT + BT + arc AB. Arc $= 6 \times \frac{π}{3} = 2 π$ .

Rule

Mark schemes give method marks for each piece, so write what each line is: sector, triangle, arc. Keep exact values until the end.

Common mistake

Adding the radius OB into the perimeter. Only count the edges that are on the edge of the region.

Key idea

Plan the pieces first: what to add, what to take away. Then each piece is a formula you already know.

Which method? Read the question

Before you start, read what the question asks for. The wording tells you which formula to use.

You see	You do
An angle in degrees	Multiply by $\frac{π}{180}$
“Arc length”	$s = r θ$
“Perimeter of the sector”	$r θ + 2 r$
“Area of the sector”	$\frac{1}{2} r^{2} θ$
“Area of the segment”	Sector − triangle: $\frac{1}{2} r^{2} (θ - sin θ)$
Area and perimeter both given	Make θ the subject, substitute, solve; reject bad roots
A tangent in the diagram	Right angle at the radius: use trig or Pythagoras
“Exact” or no calculator	Keep π and surds in the answer

Common mistake

Mixing units. Once you are in radians, stay in radians, and keep the calculator in RAD mode.

Key idea

Every circle question in radians uses just two formulas, $r θ$ and $\frac{1}{2} r^{2} θ$ , plus the triangle pieces around them.

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