Add Maths · 9 Circular measure

9 Circular measure

Before you start

The circumference $2 π r$ and the area $π r^{2}$ of a circle.
Exact values of $\sin$ and $\cos$ for $30^{\circ}$ , $45^{\circ}$ , $60^{\circ}$ and $90^{\circ}$ from Extended Mathematics, for example $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $\cos 60^{\circ} = \frac{1}{2}$ .
Right-angled triangle trigonometry, Pythagoras' theorem, and the fact that a tangent is perpendicular to the radius at its point of contact.
The triangle area $\frac{1}{2} a b \sin C$ and the cosine rule. Both are on the List of formulas.
Solving a quadratic by factorising, and a pair of simultaneous equations by substitution.

Syllabus objectives covered (1)

Solve problems involving the arc length and sector area of a circle, including knowledge and use of radian measure.

Part 1 of 7

What a radian is

Degrees split a full turn into $360$ equal parts. The number $360$ is a convention. Radians measure an angle with the circle itself, and that is why the arc and sector formulas become so short in radians.
Draw a circle with centre $O$ and radius $r$ . Mark an arc whose length is also $r$ . The angle at $O$ between the radii to the two ends of that arc is one radian.
graph diagram — not drawn yet
A circle with centre O; two radii OA and OB, each labelled r; the minor arc AB drawn thick and labelled 'arc length r'; the angle AOB at the centre marked with a small arc and labelled '1 rad'.
When the arc is exactly one radius long, the angle at the centre is $1$ radian (about ${57.3}^{\circ}$ ).
So an angle in radians counts how many radius-lengths of arc it cuts off. An arc of length $3 r$ makes an angle of $3$ radians at the centre.
The whole circumference is $2 π r$ , which is $2 π$ radius-lengths. So a full turn is $2 π$ radians, and half a turn is $π$ radians.
$2 π rad = 360^{\circ} π rad = 180^{\circ}$
So $1$ radian is $\frac{180^{\circ}}{π}$ , about ${57.3}^{\circ}$ . That is a little less than $60^{\circ}$ . The reason: the arc is curved, so the straight chord across it is shorter than $r$ , and a triangle with sides $r$ , $r$ and a side shorter than $r$ has an angle smaller than the $60^{\circ}$ of an equilateral triangle.
Notation: an angle with no degree sign is in radians. $θ = 1.2$ means $1.2$ radians.

Part 2 of 7

Converting between degrees and radians

Every conversion comes from $π$ radians $= 180^{\circ}$ . To change degrees to radians, multiply by $\frac{π}{180}$ . To change radians to degrees, multiply by $\frac{180}{π}$ .
Degrees to radians: $150^{\circ} = 150 \times \frac{π}{180} = \frac{5 π}{6}$ . Simplify the fraction $\frac{150}{180}$ to $\frac{5}{6}$ and keep $π$ as a symbol. That is the exact form, and it is the form to use on Paper 1.
Radians to degrees: $2.5$ radians $= 2.5 \times \frac{180}{π} = {143.2}^{\circ}$ to 1 decimal place.
Learn these standard angles so that you do not have to convert them each time.
$30^{\circ} = \frac{π}{6}, 45^{\circ} = \frac{π}{4}, 60^{\circ} = \frac{π}{3}, 90^{\circ} = \frac{π}{2}, 120^{\circ} = \frac{2 π}{3}, 135^{\circ} = \frac{3 π}{4}, 150^{\circ} = \frac{5 π}{6}, 180^{\circ} = π, 270^{\circ} = \frac{3 π}{2}$
Your exact trigonometric values carry straight across, because it is the same angle with a different name: $\sin \frac{π}{6} = \frac{1}{2}$ , $\cos \frac{π}{3} = \frac{1}{2}$ , $\sin \frac{π}{4} = \cos \frac{π}{4} = \frac{\sqrt{2}}{2}$ and $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$ .
On the calculator paper, set your calculator to radian mode before any trigonometry in this topic. In degree mode, $\sin 1.2$ gives $0.0209$ , which is the sine of ${1.2}^{\circ}$ . In radian mode it gives $0.932$ , the sine of $1.2$ radians. The wrong mode still gives a number, so nothing tells you it is wrong.

Part 3 of 7

Arc length: $s = r θ$

An arc is part of the circumference. A sector with angle $θ$ at the centre is the fraction $\frac{θ}{2 π}$ of the whole circle, because a full turn is $2 π$ .
So the arc is that fraction of the circumference. The $2 π$ cancels:
$s = \frac{θ}{2 π} \times 2 π r = r θ$
This agrees with the definition: $θ$ radians means $θ$ radius-lengths of arc.
The formula is true only when $θ$ is in radians. For a $60^{\circ}$ sector of radius $5$ , ' $5 \times 60 = 300$ ' is nonsense: that arc would go round the circle more than nine times. Convert first: $60^{\circ} = \frac{π}{3}$ , so $s = \frac{5 π}{3}$ .
Example: radius $5$ cm and angle $1.2$ radians give $s = 5 \times 1.2 = 6$ cm. No $π$ appears, because the angle was not given as a multiple of $π$ .
The perimeter of a sector is the arc plus the two straight radii. Read the question: 'arc length' means the curve only; 'perimeter' means the whole boundary.
$P = 2 r + r θ$
Arc length is not on the List of formulas, so you must recall it. If you forget it, rebuild it from the fraction $\frac{θ}{2 π}$ , as above.

Part 4 of 7

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

The sector's area is the same fraction $\frac{θ}{2 π}$ of the circle's area $π r^{2}$ . This formula is not on the List of formulas either.
$A = \frac{θ}{2 π} \times π r^{2} = \frac{1}{2} r^{2} θ$
Example: radius $6$ cm and angle $\frac{π}{3}$ give $A = \frac{1}{2} \times 36 \times \frac{π}{3} = 6 π$ cm². On Paper 1, $6 π$ is the finished exact answer. On Paper 2 you may give $18.8$ cm² to 3 significant figures.
To keep the two formulas apart, think about units. An arc is a length, so it has one factor of $r$ . An area needs $r^{2}$ . The angle in radians has no units, so it does not change this.
Two radii split a circle into two sectors. The smaller one is the minor sector, with angle $θ$ . The larger one is the major sector, with angle $2 π - θ$ . Use $2 π - θ$ in both formulas for the major sector.
Working backwards uses the same formula, rearranged. A sector of radius $4$ cm with area $10$ cm² has $\frac{1}{2} \times 16 \times θ = 10$ , so $8 θ = 10$ and $θ = 1.25$ radians.

Part 5 of 7

When both $r$ and $θ$ are unknown

If you know two facts about a sector, such as its perimeter and its area, write each fact as an equation. That gives two simultaneous equations in $r$ and $θ$ .
Example: a sector has perimeter $20$ cm and area $25$ cm².
$2 r + r θ = 20 \frac{1}{2} r^{2} θ = 25$
Notice that $r θ$ is in the first equation, and $\frac{1}{2} r^{2} θ = \frac{1}{2} r \times r θ$ . From the first equation, $r θ = 20 - 2 r$ . Substitute that whole block into the second equation:
$\frac{1}{2} r (20 - 2 r) = 25 \Rightarrow 10 r - r^{2} = 25 \Rightarrow r^{2} - 10 r + 25 = 0$
${(r - 5)}^{2} = 0$ , so $r = 5$ . Then $5 θ = 20 - 10$ , so $θ = 2$ radians.
Check that each answer is possible: $r$ must be positive, and a sector angle must be more than $0$ and less than $2 π$ (about $6.28$ ). Reject any root that breaks this. Here $θ = 2$ is fine.

Part 6 of 7

Chords, triangles and segments

A chord is the straight line that joins the ends of an arc. The two radii and the chord make an isosceles triangle, with two sides $r$ and the angle $θ$ between them.
The triangle's area comes from $\frac{1}{2} a b \sin C$ , which is on the List of formulas. Put $a = b = r$ and $C = θ$ :
$triangle = \frac{1}{2} r^{2} \sin θ$
The segment is the region between the chord and the arc. It is the sector with the triangle removed:
$segment = \frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} \sin θ = \frac{1}{2} r^{2} (θ - \sin θ)$
graph diagram — not drawn yet
A circle with centre O; radii OA and OB, each labelled r, with the angle AOB labelled θ; the chord AB drawn; the triangle OAB left unshaded and the minor segment between the chord AB and the minor arc AB shaded.
Sector $O A B$ = triangle $O A B$ + shaded segment, so segment = $\frac{1}{2} r^{2} θ - \frac{1}{2} r^{2} \sin θ$ .
Look at the two terms. In the sector term, $θ$ is a plain number of radians. In the triangle term, $θ$ is inside $\sin$ . They are different numbers, so $θ - \sin θ$ does not simplify.
To find the chord length, draw the line from $O$ to the midpoint $M$ of the chord. In an isosceles triangle this line is perpendicular to the chord and halves the angle at $O$ . In each right-angled half, half the chord is opposite the angle $\frac{θ}{2}$ and the hypotenuse is $r$ .
$chord = 2 r \sin \frac{θ}{2}$
graph diagram — not drawn yet
An isosceles triangle OAB with OA and OB each labelled r and the angle at O labelled θ; a dashed line from O to the midpoint M of AB, with a right-angle mark at M; the two half-angles at O each labelled θ/2 and AM labelled r sin(θ/2).
Halving the isosceles triangle gives two right-angled triangles, each with hypotenuse $r$ and angle $\frac{θ}{2}$ at $O$ .
The cosine rule gives the same length: ${chord}^{2} = r^{2} + r^{2} - 2 r^{2} \cos θ$ . The cosine rule is on the List of formulas; the half-angle method needs only right-angled trigonometry. Use whichever is quicker for you.
The perimeter of the segment is its arc plus its chord: $r θ + 2 r \sin \frac{θ}{2}$ .

Part 7 of 7

Compound shapes: name every piece first

A compound region is made from sectors, triangles, segments and other simple shapes. Before you calculate, make two separate lists: the pieces of its boundary and the pieces of its area.
For a perimeter, go round the outside edge and add each piece: arcs ( $r θ$ ) and straight edges (from trigonometry or Pythagoras). Lines inside the region are not part of its perimeter.
For an area, decide what to add and what to take away. A region inside a triangle but outside a sector is triangle minus sector. A region inside a sector but outside a triangle is a segment.
Look for right angles. A tangent is perpendicular to the radius at its point of contact, so the centre, the point of contact and any other point on the tangent make a right-angled triangle. That is where $\tan$ , $\cos$ and Pythagoras come in.
graph diagram — not drawn yet
A circle with centre O and radius 5; an external point P to the right with OP = 13 drawn dashed; tangents PA (upper) and PB (lower) touching the circle at A and B, each labelled 12; radii OA and OB drawn with right-angle marks at A and B; the region bounded by PA, PB and the minor arc AB shaded.
The shaded region is the kite $O A P B$ minus the sector $O A B$ . Its perimeter is $P A + P B +$ arc $A B$ ; the radii are inside the kite but not on the region's edge.
On the calculator paper, keep full accuracy in every intermediate value, and round only the final answer to 3 significant figures. An angle rounded early can change the third figure of the answer.
On Paper 1, or when a question says 'exact', leave $π$ and surds in the answer. $12 π - 9 \sqrt{3}$ is exact; $22.1$ is not.

Worked examples

Example 1. A sector of a circle has radius $8$ cm and perimeter $28$ cm. Find the angle of the sector in radians and the area of the sector.

Show the working

The perimeter is two radii plus the arc. Write that as an equation, with $θ$ the angle in radians.
$2 (8) + 8 θ = 28$
So $8 θ = 12$ and $θ = 1.5$ radians. That is less than $2 π$ , so it is a possible sector angle.
Now find the area with the angle you have just found.
$A = \frac{1}{2} \times 8^{2} \times 1.5 = 32 \times 1.5 = 48$

Answer: $θ = 1.5$ radians; area $= 48$ cm².

Example 2. A circle has centre $O$ and radius $6$ cm. The points $A$ and $B$ lie on the circle and angle $A O B = \frac{2 π}{3}$ radians. Find the exact perimeter and the exact area of the minor segment cut off by the chord $A B$ .

Show the working

The minor segment is bounded by the minor arc $A B$ and the chord $A B$ . Its perimeter is arc plus chord. Its area is sector minus triangle.
Arc: $s = r θ = 6 \times \frac{2 π}{3} = 4 π$ .
Chord: draw the line from $O$ to the midpoint $M$ of $A B$ . Angle $A O M = \frac{π}{3}$ , and $A M$ is opposite it in a right-angled triangle with hypotenuse $6$ .
$A M = 6 \sin \frac{π}{3} = 6 \times \frac{\sqrt{3}}{2} = 3 \sqrt{3}, A B = 6 \sqrt{3}$
Perimeter $= 4 π + 6 \sqrt{3}$ cm. Leave it in this form: it is exact, and the two terms cannot be combined.
Sector: $\frac{1}{2} \times 36 \times \frac{2 π}{3} = 12 π$ .
Triangle: its base is $A B = 6 \sqrt{3}$ and its height is $O M = 6 \cos \frac{π}{3} = 3$ , so its area is $\frac{1}{2} \times 6 \sqrt{3} \times 3 = 9 \sqrt{3}$ . The formula $\frac{1}{2} r^{2} \sin θ$ agrees: $\frac{1}{2} \times 36 \times \sin \frac{2 π}{3} = 18 \times \frac{\sqrt{3}}{2} = 9 \sqrt{3}$ , because $\sin 120^{\circ} = \sin 60^{\circ}$ .
Segment $=$ sector $-$ triangle $= 12 π - 9 \sqrt{3}$ cm².

Answer: Perimeter $4 π + 6 \sqrt{3}$ cm; area $12 π - 9 \sqrt{3}$ cm².

Example 3. A circle has centre $O$ and radius $5$ cm. A chord $A B$ has length $8$ cm. (a) Show that angle $A O B = 1.855$ radians, correct to 3 decimal places. (b) Find the area of the minor segment cut off by $A B$ .

Show the working

(a) Let $M$ be the midpoint of $A B$ . Triangle $O A M$ is right-angled at $M$ , with hypotenuse $O A = 5$ and $A M = 4$ .
Angle $A O M$ is half of angle $A O B$ , and $A M$ is opposite it, so use sine. Calculator in radian mode.
$\sin (∠ A O M) = \frac{4}{5} \Rightarrow ∠ A O M = 0.92729 \dots$
Double it: angle $A O B = 1.85459 \dots = 1.855$ to 3 decimal places. Write the longer value before the rounded one. It shows that you worked the angle out, and did not copy the printed $1.855$ .
(b) Use the unrounded angle from your calculator's memory, not $1.855$ .
$segment = \frac{1}{2} \times 5^{2} \times (1.85459 - \sin 1.85459) = 12.5 \times (1.85459 - 0.96)$
That is $12.5 \times 0.89459 = 11.18$ , so $11.2$ cm² to 3 significant figures.
Check another way. The triangle has base $8$ and height $O M = \sqrt{25 - 16} = 3$ , so its area is $12$ . The sector is $12.5 \times 1.85459 = 23.18$ . Then $23.18 - 12 = 11.18$ , the same.

Answer: (a) Angle $A O B = 2 \sin^{- 1} 0.8 = 1.85459 \dots = 1.855$ radians. (b) $11.2$ cm².

Example 4. A circle has centre $O$ and radius $5$ cm. A point $P$ is $13$ cm from $O$ . The tangents from $P$ touch the circle at $A$ and $B$ . The region $R$ is bounded by $P A$ , $P B$ and the minor arc $A B$ . Find the perimeter and the area of $R$ .

Show the working

List the pieces. Boundary of $R$ : the tangents $P A$ and $P B$ and the minor arc $A B$ . Area of $R$ : the kite $O A P B$ minus the minor sector $O A B$ .
A tangent is perpendicular to the radius at its point of contact, so triangle $O A P$ is right-angled at $A$ . By symmetry $P B = P A$ .
$P A = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12$
Angle $A O P$ : $\cos (∠ A O P) = \frac{5}{13}$ , so $∠ A O P = 1.17600 \dots$ radians. Angle $A O B$ is twice this: $2.35201 \dots$ radians.
Arc $A B = 5 \times 2.35201 = 11.760$ . Perimeter $= 12 + 12 + 11.760 = 35.760$ , so $35.8$ cm.
Kite: two right-angled triangles, each $\frac{1}{2} \times 5 \times 12 = 30$ , so $60$ in total. Sector: $\frac{1}{2} \times 25 \times 2.35201 = 29.400$ .
Area of $R = 60 - 29.400 = 30.600$ , so $30.6$ cm².
See what early rounding does. With angle $A O P$ rounded to $1.18$ , angle $A O B = 2.36$ , the sector is $29.5$ and the area of $R$ comes out as $30.5$ , which is wrong in the third figure.

Answer: Perimeter $35.8$ cm; area $30.6$ cm².

Where people go wrong

Mistake: Using an angle in degrees in $s = r θ$ or $A = \frac{1}{2} r^{2} θ$ , for example arc $= 5 \times 60 = 300$ for a $60^{\circ}$ sector of radius $5$ .

Instead: Convert to radians first: $60^{\circ} = \frac{π}{3}$ , so the arc is $\frac{5 π}{3}$ .

Both formulas come from the fraction $\frac{θ}{2 π}$ of a circle, which is correct only when $θ$ is in radians.

Mistake: Leaving the calculator in degree mode when finding $\sin θ$ for a triangle or segment.

Instead: Switch to radian mode. In radians, $\sin 1.2 = 0.932$ ; degree mode gives $0.0209$ .

An angle with no degree sign is in radians. The wrong mode still gives a number, so the error is easy to miss.

Mistake: Giving only the arc when the question asks for the perimeter of a sector, or adding the radii when it asks for the arc length.

Instead: Perimeter of a sector $= 2 r + r θ$ ; arc length $= r θ$ only.

A perimeter is the whole boundary, and a sector's boundary includes both straight radii.

Mistake: Mixing up the formulas, for example writing the area as $\frac{1}{2} r θ^{2}$ or the arc as $r^{2} θ$ .

Instead: Arc $= r θ$ (one factor of $r$ ); area $= \frac{1}{2} r^{2} θ$ (two factors of $r$ ).

An angle in radians has no units, so the power of $r$ decides the units: a length needs one, an area needs two.

Mistake: Writing the segment area as $\frac{1}{2} r^{2} θ - \frac{1}{2} r^{2}$ , or cancelling $θ$ against $\sin θ$ .

Instead: Segment $= \frac{1}{2} r^{2} (θ - \sin θ)$ , with $\sin θ$ worked out on its own.

The triangle's area is $\frac{1}{2} r^{2} \sin θ$ from $\frac{1}{2} a b \sin C$ . It equals $\frac{1}{2} r^{2}$ only when $θ = \frac{π}{2}$ , where $\sin θ = 1$ .

Check yourself

Convert $135^{\circ}$ to radians. Give an exact answer.
Answer
$135 \times \frac{π}{180} = \frac{3 π}{4}$ .
Convert $2.5$ radians to degrees, correct to 1 decimal place.
Answer
$2.5 \times \frac{180}{π} = {143.2}^{\circ}$ .
Find the arc length and the perimeter of a sector with radius $7$ cm and angle $0.8$ radians.
Answer
Arc $= 7 \times 0.8 = 5.6$ cm. Perimeter $= 14 + 5.6 = 19.6$ cm.
Find the exact area of a sector with radius $10$ cm and angle $\frac{π}{5}$ .
Answer
$\frac{1}{2} \times 100 \times \frac{π}{5} = 10 π$ cm².
A sector has radius $6$ cm and area $30$ cm². Find its angle and its arc length.
Answer
$\frac{1}{2} \times 36 \times θ = 30$ , so $θ = \frac{5}{3}$ radians. Arc $= 6 \times \frac{5}{3} = 10$ cm.
Two radii of a circle of radius $4$ cm make an angle of $\frac{π}{3}$ . Find the exact area of the major sector.
Answer
Major angle $= 2 π - \frac{π}{3} = \frac{5 π}{3}$ . Area $= \frac{1}{2} \times 16 \times \frac{5 π}{3} = \frac{40 π}{3}$ cm².
A chord of a circle of radius $10$ cm makes an angle of $1.2$ radians at the centre. Find the area of the minor segment, correct to 3 significant figures.
Answer
$\frac{1}{2} \times 100 \times (1.2 - \sin 1.2) = 50 \times (1.2 - 0.93204) = 13.4$ cm².
Explain why a full turn is $2 π$ radians.
Answer
One radian cuts off an arc one radius long. The circumference is $2 π r$ , which is $2 π$ radius-lengths, so a full turn is $2 π$ radians.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The points $A$ and $B$ lie on a circle with centre $O$ and radius $12$ cm. Angle $A O B = \frac{π}{2}$ radians. The minor segment is the region between the chord $A B$ and the minor arc $A B$ .

(a) Find the exact perimeter of the minor segment. 3 marks No calculator

Mark scheme

B1 minor arc $A B = 12 \times \frac{π}{2} = 6 π$
M1 a correct method for the chord, e.g. $\sqrt{12^{2} + 12^{2}}$ or $2 \times 12 \sin \frac{π}{4}$
A1 $6 π + 12 \sqrt{2}$ cm oe

(b) Find the exact area of the minor segment. 3 marks No calculator

Mark scheme

M1 sector area $\frac{1}{2} \times 12^{2} \times \frac{π}{2}$ $[= 36 π]$
M1 subtracts the triangle area $\frac{1}{2} \times 12 \times 12$ $[= 72]$ from their sector area, or uses $\frac{1}{2} r^{2} (θ - \sin θ)$ with $r = 12$ , $θ = \frac{π}{2}$
A1 $36 π - 72$ cm² oe, e.g. $36 (π - 2)$

Question 2. $O A B$ is a sector of a circle with centre $O$ , radius $10$ cm and angle $A O B = 0.9$ radians. The tangent to the circle at $A$ meets the line $O B$ extended at the point $T$ , so that $B$ lies between $O$ and $T$ . The shaded region is bounded by the straight lines $A T$ and $T B$ and the arc $A B$ .

(a) Show that $A T = 12.60$ cm, correct to 2 decimal places. 2 marks

Mark scheme

M1 uses the right angle at $A$ : $A T = 10 \tan 0.9$ , or a correct equivalent method, e.g. $\sqrt{O T^{2} - 10^{2}}$ with $O T = \frac{10}{\cos 0.9}$
A1 $12.6015 \dots$ seen, leading to $12.60$

(b) Find the perimeter of the shaded region. 4 marks

Mark scheme

M1 $O T = \frac{10}{\cos 0.9}$ or $O T = \sqrt{10^{2} + {12.60}^{2}}$
A1 $T B = O T - 10 = 6.087 \dots$ , awrt $6.09$
B1 arc $A B = 10 \times 0.9 = 9$
A1 perimeter $= 27.7$ cm, awrt $27.7$

(c) Find the area of the shaded region. 2 marks

Mark scheme

M1 triangle $O A T$ minus sector $O A B$ : $\frac{1}{2} \times 10 \times 12.60 - \frac{1}{2} \times 10^{2} \times 0.9$ oe, e.g. using $10 \tan 0.9$ for $A T$
A1 $18.0$ cm², awrt $18.0$

Question 3. A sector of a circle has radius $r$ cm and angle $θ$ radians. The perimeter of the sector is $24$ cm and its area is $32$ cm².

(a) Show that $r^{2} - 12 r + 32 = 0$ . 3 marks No calculator

Mark scheme

B1 both equations: $2 r + r θ = 24$ and $\frac{1}{2} r^{2} θ = 32$
M1 eliminates $θ$ , e.g. substitutes $r θ = 24 - 2 r$ into $\frac{1}{2} r (r θ) = 32$
A1 $12 r - r^{2} = 32$ rearranged to the given equation with no errors; nfww

(b) Hence find the two possible values of $r$ and the corresponding values of $θ$ . 3 marks No calculator

Mark scheme

M1 solves the given quadratic, e.g. $(r - 4) (r - 8) = 0$
A1 $r = 4$ , $θ = 4$
A1 $r = 8$ , $θ = 1$ [both values of $θ$ are less than $2 π$ , so both are valid]

How this is examined

Finding the perimeter and the area of a compound region made from sectors, triangles and segments: a multi-part Find on a described or printed diagram, with the shaded region's perimeter and its area as separate parts Each piece found with a correct method: arcs as $r θ$ , straight edges from trigonometry or Pythagoras, sectors as $\frac{1}{2} r^{2} θ$ , triangles as $\frac{1}{2} a b \sin C$ or $\frac{1}{2} \times$ base $\times$ height, with $θ$ in radians. Method marks go to each correct piece and to combining the pieces with the right signs; the accuracy mark needs the final value to 3 significant figures from unrounded working.
Forming equations from a sector's perimeter and area, then solving them: a Show that ending in a printed quadratic in $r$ , followed by a Hence find for $r$ and $θ$ Perimeter written as $2 r + r θ$ with both radii, and area as $\frac{1}{2} r^{2} θ$ ; $θ$ eliminated; every line of algebra written down until the printed equation appears exactly. Then each value of $r$ paired with its own $θ$ , and any $θ$ that is not between $0$ and $2 π$ rejected.
Exact answers in terms of $π$ and surds on the non-calculator paper: a Find the exact perimeter or area, where the angle is a fraction of $π$ Arcs and sectors written with $π$ (for example $6 π$ and $36 π$ ), trigonometric values replaced by exact values such as $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$ , and the answer left in a form such as $36 π - 72$ . A rounded decimal is not an exact answer.
Showing that an angle in radians has a given value: a two-mark Show that, for example that an angle equals a given value to 3 decimal places A correct trigonometric statement for the angle, such as $\sin \frac{θ}{2} = \frac{4}{5}$ , and the angle worked to more decimal places than printed, such as $1.85459$ , before it is rounded to the given $1.855$ .
Converting an angle between degrees and radians: a one-mark Write down, or a step inside a longer arc or sector question Degrees multiplied by $\frac{π}{180}$ , with the result simplified to an exact multiple of $π$ such as $\frac{5 π}{6}$ ; radians to degrees by $\frac{180}{π}$ , to 1 decimal place.

Words to know

Radian: The angle at the centre of a circle made by an arc whose length equals the radius. $π$ radians $= 180^{\circ}$ .
Subtend: An arc or chord subtends an angle at the centre when the lines from its two ends to the centre make that angle.
Arc: Part of the circumference of a circle. With $θ$ in radians, its length is $r θ$ .
Sector: The region between two radii and the arc that joins their ends. With $θ$ in radians, its area is $\frac{1}{2} r^{2} θ$ .
Minor and major sector: The two sectors made by two radii: the minor sector has the smaller angle $θ$ , the major sector has the angle $2 π - θ$ .
Chord: A straight line that joins two points on a circle.
Segment: The region between a chord and an arc: sector minus triangle, $\frac{1}{2} r^{2} (θ - \sin θ)$ .
Exact answer: An answer left in terms of $π$ , surds or fractions with no rounding, for example $12 π - 9 \sqrt{3}$ .

All Add Maths lessons Next: 10 Trigonometry

9 Circular measure

What a radian is

Converting between degrees and radians

Arc length: s=rθ

Sector area: A=12r2θ

When both r and θ are unknown

Chords, triangles and segments

Compound shapes: name every piece first

Worked examples

Where people go wrong

Check yourself

Exam questions

How this is examined

Words to know

Arc length: $s = r θ$

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

When both $r$ and $θ$ are unknown