Circular Measure

— Pure Mathematics 1 · AS & A Level (9709) · Chapter 4 —

★ Exam Tip: Radians, arc length and sector area come up in Paper 1 almost every session. The end-of-chapter questions are real 9709 Paper 11 questions (2011–2016) — typically a shaded composite region asking for perimeter + area, worth 5–8 marks. Set the calculator to RADIAN mode!

You will learn to: ① understand what a radian is & convert radians ⇄ degrees ② use s = rθ and A = ½r²θ for the arc length & sector area of a circle.

0. Warm-up — Already Know This

From / (degrees version of sectors):

arc = θ360 × 2πr area = θ360 × πr² perimeter of sector = arc + 2r
Pythagoras, SOH-CAH-TOA | sine rule, cosine rule | area of triangle = ½ab sin C

Check your skillsanswers

1. Sector r = 6 cm, angle 30° :

arc = 30360 × 2π × 6 = π ≈ 3.14 cmperimeter = 12 + π ≈ 15.1 cmarea = 30360 × π × 6² = 3π ≈ 9.42 cm²

2. Right-angled triangle with sides 5 & 12 (angle x° opposite the 12):

y² = 5² + 12² ⇒ y = 13tan x° = 125 ⇒ x = 67.4

3. Triangle with sides 6 & 8 and the 40° angle between them:

x² = 6² + 8² − 2(6)(8) cos 40°x = 5.14 cmarea = ½(6)(8) sin 40° = 15.4 cm²

1. Another Measure for Angles

Babylonians: 6 equilateral △s, each angle at O split into 60 parts

Until now angles were in degrees. Why 360° in one revolution? The real reason is unknown — theories:

Astronomers: the Sun moves ≈ 1° per day along its path; a solar year ≈ 360 days.
Babylonians: split circle into 6 equilateral triangles, then each angle at O into 60 parts ⇒ 6 × 60 = 360.
360 has many factors ⇒ circle divides up easily.

Degrees are not the only way. Radian measure is the “natural unit” of angle — used all over maths because it simplifies many formulae & calculations (e.g. s = rθ, A = ½r²θ, and later calculus/trig).

2. Radians (4.1)

arc length = r ⇒ angle = 1 rad

Notation: 1 radian = 1 rad — but usually no symbol is written for angles in radians.

KEY POINT 4.1 — DefinitionAn arc equal in length to the radius subtends an angle of 1 radian at the centre.

Circumference = 2πr is an arc of length 2πr ⇒ it subtends 2π radians at the centre.

KEY POINT 4.22π radians = 360°
π radians = 180°

When an angle is written in terms of π, we usually omit the word radian (rad). So just write π = 180°.

① Degrees → Radians

Since 180° = π : 90° = π2, 45° = π4 … For other angles use the rule:

KEY POINT 4.3degrees → radians : multiply by π180

② Radians → Degrees

Since π = 180° : π6 = 30°, π10 = 18° … Otherwise:

KEY POINT 4.4radians → degrees : multiply by 180π

Remember: 1 radian = 1 × 180π ≈ 57° — a bit less than 60°, so ≈ 1 rad is an angle slightly smaller than an equilateral triangle’s.

π/180 to go “into” radians, 180/π to come back

Worked Example 4.1convert

a) Change 30° to radians (in terms of π). b) Change 5π9 radians to degrees.

a) Method 1180° = π rad(180 ÷ 6)° = π ÷ 630° = π6 rad

a) Method 230° = (30 × π180) rad30° = π6 rad

b) Method 1π rad = 180°π9 rad = 20°5π9 rad = 100°

b) Method 25π9 × 180π= 100°

③ Angles to learn by heart

Degrees	0°	30°	45°	60°	90°	180°	270°	360°
Radians	0	π6	π4	π3	π2	π	3π2	2π

Get other angles from these, e.g. 120° = 2 × 60° = 2 × π3 = 2π3; 135° = 3 × 45° = 3π4.

upper half of the circle: each ray labelled degrees = radians (add π to each for the lower half)

Common mistake: calculator in the wrong mode! sin 30 = 0.5 in degree mode but sin 30 = −0.988 in radian mode. For sin, cos, tan of an angle in radians do not convert to degrees — just switch the calculator to RAD.

Calculator examples (Exercise 4A style):

28° → 28 × π180 = 0.489 rad (3 s.f.) | 1.2 rad → 1.2 × 180π = 68.8° (1 d.p.)
sin(0.7), tan(1.5), cos(π/2), sin(π/3) … all computed in RAD mode (e.g. cos π2 = 0, sin π3 = √32).

Ex 4A Q7trig with radians

Right-angled △PQR: PQ = 5 cm, angle at P = 1 rad. Find QR.

tan 1 = QR5QR = 5 tan 1 (RAD mode)QR = 7.79 cm

④ Full conversion tables (Ex 4A Q5, completed)

Degrees	0	45	90	135	180	225	270	315	360
Radians	0	π4	π2	3π4	π	5π4	3π2	7π4	2π

Deg	0	30	60	90	120	150	180	210	240	270	300	330	360
Rad	0	π6	π3	π2	2π3	5π6	π	7π6	4π3	3π2	5π3	11π6	2π

3. Chord · Arc · Sector · Segment (Explore 4.1)

Vocabulary you need before the formulae. In a circle with centre O and chord AB subtending angle θ at the centre:

chord & arcs

sector = O + 2 radii + arc

segment = chord + arc

Chord — straight line joining 2 points on the circle | Arc — part of the circumference.
Sector — region between 2 radii and an arc (a “pizza slice”) | Segment — region between a chord and an arc.
Minor = the smaller one (angle < π) | Major = the larger one (angle > π) — for arc, sector and segment.

Chord trick: isosceles △OAB — drop ⟂ from O, it bisects θ and AB. Half-chord = r sin(θ/2).

AB = 2 r sin(θ/2) — same formula if θ in degrees or radians (the calculator must match!)

Find (minor, angle θ, radius r)	θ in degrees	θ in radians
length of arc AB	θ360×2πr	rθ
perimeter of sector AOB	θ360×2πr + 2r	rθ + 2r
length of chord AB	2r sin(θ°/2)	2r sin(θ/2)
perimeter of segment AOB	arc + chord = θ360×2πr + 2r sin(θ°/2)	rθ + 2r sin(θ/2)
area of sector AOB	θ360×πr²	½r²θ
area of segment AOB	sector − triangle = θ360πr² − ½r² sin θ°	½r²θ − ½r² sin θ = ½r²(θ − sin θ)

Note: radian versions are much cleaner — that is why we use radians. (Entries from the Explore 4.1 discussion; the sector formulae are proved in 4.2 and 4.3 below.)

4. Length of an Arc (4.2)

θ rad ⇒ arc = r × θ

Why: an arc subtending 1 rad has length r (definition). So an arc subtending θ rad is θ times as long: rθ.

KEY POINT 4.5Arc length s = rθ (θ in radians)

Remember: perimeter of a sector = arc + 2 radii = rθ + 2r. Don’t forget the straight edges!

If the angle is in degrees, convert to radians first (× π/180).

Worked Example 4.2arc length, exact

Arc subtends π3 rad at the centre of a circle, radius 15 cm. Find the arc length in terms of π.

arc = rθ= 15 × π3= 5π cm

Worked Example 4.3find the radius

A sector has angle 1.5 rad and arc length 12 cm. Find the radius.

arc = rθ12 = r × 1.5r = 8 cm

Same idea for the angle: θ = arc ÷ r (Ex 4B Q3).

Worked Example 4.4arc + triangle

△ABC isosceles, AC = CB = 8 cm. CD is an arc of a circle centre B, ∠ABC = 0.9 rad. Find a) arc CD b) AD c) perimeter of shaded region.

AB = 2 × (8 cos 0.9): drop ⟂ from C — it bisects AB in the isosceles △

Worked Example 4.4 (working)

a) arc CD (centre B, r = BC = 8)arc = rθ= 8 × 0.9 = 7.2 cm

b) ADAB = 2 × 8 cos 0.9 = 9.9457…AD = AB − DB= 9.9457… − 8 = 1.95 cm

c) perimeter of shaded regionP = DC + CA + AD= 7.2 + 8 + 1.945…= 17.1 cm (3 s.f.)

Ex 4B Q4real-life (chapter-opener)

High Roller Ferris wheel, diameter 158.5 m. Distance moved by a capsule as the wheel turns through π16 rad?

r = 158.5 ÷ 2 = 79.25 ms = rθ = 79.25 × π16s = 15.6 m (3 s.f.)

Careful: diameter given — halve it to get r.

5. Area of a Sector (4.3)

fraction of the circle = θ ÷ 2π

Derivation: sector is a fraction of the whole circle:

area of sectorarea of circle = angle in the sectorcomplete angle at the centre

θ in radians, whole angle = 2π, whole area = πr²:

area of sectorπr² = θ2πarea = θ2π × πr²area = ½ r² θ

KEY POINT 4.6Area of sector A = ½ r² θ (θ in radians)

Link: arc = rθ and area = ½r²θ = ½ × r × (rθ) = ½ × radius × arc length (like ½ × base × height).

Worked Example 4.5area, exact

Sector, radius 9 cm, angle π6 rad. Area in terms of π.

A = ½r²θ= ½ × 9² × π6= 27π4 cm²

Worked Example 4.6segment area

Circle radius 6 cm, centre O. AB is a chord, ∠AOB = 1.2 rad. Find a) area of sector AOB b) area of △AOB c) area of the shaded segment.

sector AOB

− triangle AOB

= segment

Worked Example 4.6 (working)

a) sector AOBA = ½r²θ= ½ × 6² × 1.2 = 21.6 cm²

b) triangle AOBA = ½ab sin C= ½ × 6 × 6 × sin 1.2= 16.7767… = 16.8 cm²

c) shaded segmentsegment = sector − triangle= 21.6 − 16.7767…= 4.82 cm² (3 s.f.)

Common mistake: for the triangle use sin 1.2 with the calculator in RAD — and don’t use ½r²θ for the triangle (that is the sector). Segment = sector − triangle.

Worked Example 4.7harder · circle in square + quarter circle

A circle is inscribed in a square of side 10 cm. A quarter circle, radius 10 cm, is drawn with a vertex O of the square as centre. Find the shaded area (inside the small circle, outside the quarter circle).

I = centre of the inscribed circle. P, Q = where the two circles meet. R, S = points on the two arcs between P and Q.

θ at O (so ∠POQ = 2θ) · α, β at I · shaded = segment PQR − segment PQS

6. Strategy for Shaded-Region Questions

Perimeter of a shaded region: walk round its edge — curved bits (rθ) + straight bits (radii, chords, tangent lengths). Include every edge.
Area: split into sectors, triangles, rectangles — then add or subtract (segment = sector − triangle).
Tangent ⟂ radius ⇒ right-angled triangle ⇒ Pythagoras / tan (Ex 4B Q6, 4C Q5–7).
Two radii make an isosceles triangle ⇒ cosine rule for chord, or drop a ⟂ (chord = 2r sin θ/2).
Angle given in degrees? Convert to radians before using rθ or ½r²θ.
“Exact” or “in terms of π / √3” ⇒ no decimals. Otherwise 3 s.f. (don’t round mid-way).

7. End-of-Chapter Past-Paper Questions — Solved

9709/11 Q5 June 2015optimisation with sector

Wire of length 24 cm is bent into the perimeter of a sector of radius r cm. i) Show A = 12r − r². ii) Write A as a − (r − b)². iii) r varies: greatest A and the corresponding angle.

i)perimeter: 2r + rθ = 24rθ = 24 − 2rA = ½r²θ = ½ × r × (rθ)A = ½r(24 − 2r) = 12r − r²

ii) complete the squareA = −(r² − 12r)= −[(r − 6)² − 36]A = 36 − (r − 6)² (a = 36, b = 6)

iii)(r − 6)² ≥ 0 ⇒ A ≤ 36max A = 36 cm² when r = 6rθ = 24 − 12 = 12 ⇒ 6θ = 12θ = 2 rad

9709/11 Q3 June 2013sector + semicircle

OAB sector, centre O, radius 8 cm, ∠BOA = α rad. OAC is a semicircle on diameter OA. Area of semicircle = 2 × area of sector OAB. i) α in terms of π. ii) perimeter of whole figure.

i)semicircle: ½π × 4² = 8πsector: ½ × 8² × α = 32α8π = 2 × 32αα = π8

ii)OB = 8arc AB = 8α = πsemicircle arc = π × 4 = 4πP = 8 + 5π

Ch-4 Review Q1equilateral △ + arc

Equilateral △PQR, side 5 cm; M = midpoint of QR. Arc, centre P, touches QR at M and meets PQ at X, PR at Y. Find, in terms of π and √3, a) total perimeter of the shaded region b) total area.

Key: PM ⟂ QR ⇒ radius PM = 5 sin 60° = 5√3⁄2; ∠XPY = π/3.

Ch-4 Review Q1 (working)

a) perimeterarc XY = rθ = 5√32 × π3 = 5√3π6QX = RY = 5 − 5√32QM + MR = 5P = 5√3π6 + 2(5 − 5√32) + 5P = 15 − 5√3 + 5√3π6

b) area△PQR = ½ × 5 × 5 × sin 60° = 25√34sector PXY = ½r²θ = ½ × 754 × π3 = 25π8shaded = △ − sectorA = 25√34 − 25π8

8. Quick Recap — Full Chapter Cheat-Sheet

✓ Radians & degrees

1 radian = angle at the centre subtended by an arc of length r.
π rad = 180° (2π = 360°) | 1 rad ≈ 57°.
deg → rad : × π180 rad → deg : × 180π
Learn: 30° = π/6, 45° = π/4, 60° = π/3, 90° = π/2, 180° = π, 270° = 3π/2, 360° = 2π.
Write “in terms of π” without the word rad. Calculator → RAD mode for sin/cos/tan of radian angles.

Arc length & sector area (θ in radians!)arc length s = rθ
area of sector A = ½ r² θ
perimeter of sector = rθ + 2r

Chord & segment (from Explore 4.1)chord AB = 2r sin(θ/2)
area of segment = ½r²θ − ½r² sin θ = ½r²(θ − sin θ)
perimeter of segment = rθ + 2r sin(θ/2)

Old degree versions (for reference)arc = θ°360 × 2πr area = θ°360 × πr² triangle area = ½ab sin C

Don’t lose marks: ① calculator mode ② perimeter of a sector includes 2 radii ③ diameter given → halve ④ convert degrees → radians first ⑤ exact answers keep π and surds ⑥ don’t round until the end.

Did you know? Lines of latitude (horizontal circles) and longitude (vertical circles through the poles) form the ‘grid’ of coordinates used on Earth’s surface. Web link: try “Where are you?” and “Belt” on the Underground Mathematics website (Belt = Ex 4B Q10).