Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 3 Geometry and trigonometry · 3.4 The circle: radians, arcs and sectors
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Explain what one radian is | SL, HL | the reason behind every formula on this page |
| Convert between degrees and radians, exactly in multiples of π or as decimals | SL, HL | "Write 150° in radians" (1 mark), Paper 1 |
| Find the length of an arc, l = rθ | SL, HL | "Find the length of arc AB" (2 marks) |
| Find the area of a sector, A = ½r²θ | SL, HL | "Find the area of the shaded sector" (2 marks) |
| Find the perimeter of a sector, and work with minor and major sectors | SL, HL | "Find the perimeter of the major sector" (3 marks) |
| Work backwards from an arc, area or perimeter to r or θ | SL, HL | a pair of equations in r and θ, often a quadratic (5 to 6 marks) |
| Find the area of a segment, combining sector area with ½ab sin C from 3.2 | SL, HL | "Find the area of the shaded segment" (3 to 4 marks), Paper 2 |
Before you start
You need the area and circumference of a circle, πr² and 2πr, and the words arc, sector, chord and segment, all prior learning. You need the area rule ½ab sin C from 3.2 for segments, and solving a quadratic for the questions that work backwards. The formula booklet gives l = rθ and A = ½r²θ, each with "θ is the angle measured in radians". It does not give the conversion between degrees and radians; learn that 180° = π radians and everything follows.
1The idea in one paragraph
Degrees split a full turn into 360 parts, a number chosen by ancient astronomers, not by the circle. A radian is measured by the circle itself: it is the angle at the centre that cuts off an arc exactly as long as the radius. A full turn contains 2π of them, because the circumference is 2π radii long, so 2π radians = 360° and π radians = 180°. Measured this way, the arc cut off by an angle θ is simply r times θ, and the sector it cuts off has area ½r²θ. Every formula in this subtopic comes from asking "what fraction of the whole circle is this?", and radians make the answer tidy. They make calculus tidy too, which is why the rest of the course assumes them.
2What a radian is
One radian is the angle at the centre of a circle subtended by an arc equal in length to the radius.
Figure 1 draws it. Take a circle of radius r, walk a distance r around its edge, and join both ends to the centre. The angle between those radii is 1 radian, about 57.3°. It does not depend on the size of the circle: a bigger circle has a bigger radius and a proportionally bigger arc, so the angle stays the same.
How many radians in a full turn? The circumference is 2πr, which is 2π lots of r, so 2π arcs of length r fit around the circle. A full turn is therefore 2π radians, about 6.28.
Radians are pure numbers. An angle in radians is a length divided by a length (arc ÷ radius), so the units cancel. That is why you will often see an angle written 1.2 with no unit at all: no unit means radians. Degrees always carry their ° sign.
3Converting between degrees and radians
Since π rad = 180°:
Degrees to radians: multiply by π/180. Radians to degrees: multiply by 180/π.
The guide says radian measure may be written as exact multiples of π or as decimals, and both are correct. On Paper 1 you will usually work with exact multiples, and the angles in Figure 2 are the ones worth knowing on sight. Each is a simple fraction of the half turn π: 30° is one sixth of 180°, so it is π/6.
| Degrees | 30° | 45° | 60° | 90° | 120° | 135° | 150° | 180° | 270° | 360° |
|---|---|---|---|---|---|---|---|---|---|---|
| Radians | π/6 | π/4 | π/3 | π/2 | 2π/3 | 3π/4 | 5π/6 | π | 3π/2 | 2π |
Calculator mode. On examination papers, radian measure should be assumed unless a question says otherwise. Check the mode before every trigonometric calculation: sin 1.2 in radian mode is 0.932, and in degree mode it is 0.0209. If a question gives angles in degrees, either switch to degree mode or convert.
4Arc length
An arc is part of the circumference. An angle θ at the centre is the fraction θ/2π of a full turn, so its arc is that fraction of the whole circumference 2πr, as Figure 3(a) shows.
Arc length l = rθ and sector area A = ½r²θ, with θ in radians · both in the formula booklet.
This is the pay-off for radians. In degrees the same formula would be l = (θ/360) × 2πr, with an awkward constant in it.
Exact (Paper 1). A circle has radius 9 cm. Find the length of the arc cut off by an angle of 2π/3 at the centre: l = 9 × 2π/3 = 6π cm.
Decimal. Radius 12 cm, angle 1.4 radians: l = 12 × 1.4 = 16.8 cm. No π anywhere, because 1.4 is already a number of radians.
Degrees given. Radius 8 cm, angle 75°. Convert first: 75° = 75π/180 = 5π/12. Then l = 8 × 5π/12 = 10π/3 ≈ 10.5 cm. Putting 75 straight into rθ gives 600 cm, an arc longer than the whole circumference, which a sketch would expose at once.
Backwards. An arc of length 15 cm on a circle of radius 6 cm: θ = l ÷ r = 15 ÷ 6 = 2.5 rad. This is how the definition works in practice: the angle in radians is how many radii fit along the arc.
5Sector area
A sector is the region between two radii and the arc joining them, a slice of pizza. By the same fraction argument, Figure 3(b):
Exact (Paper 1). Radius 9 cm, angle 2π/3: A = ½ × 81 × 2π/3 = 27π cm².
Decimal. Radius 12 cm, angle 1.4 rad: A = ½ × 144 × 1.4 = 100.8 cm².
Backwards. A sector has angle 0.8 rad and area 40 cm². Then ½r²(0.8) = 40, so r² = 100 and r = 10 cm. Reject r = −10: a radius is a length.
Perimeter of a sector. The boundary of a sector is two radii and an arc, so perimeter = 2r + rθ. For radius 9 and angle 2π/3 this is 18 + 6π cm. Forgetting the two radii is the commonest slip in perimeter questions.
Minor and major sectors. Two radii cut a circle into two sectors. The smaller is the minor sector, with angle θ; the larger is the major sector, with angle 2π − θ. Figure 4 shows a circle of radius 6 with θ = π/3.
6Segments: sectors meet triangles
A segment is the region between a chord and its arc. Figure 5 shows why its area is a subtraction: the sector contains the segment plus an isosceles triangle with two sides r and the angle θ between them. The triangle's area comes from 3.2, ½ab sin C with a = b = r.
Worked example (Paper 2). A chord AB of a circle of radius 10 cm subtends an angle of 1.2 radians at the centre O. Find the area of the minor segment and the length of the chord.
The chord can also be found as 2r sin(θ/2) = 20 sin 0.6 = 11.3 cm, by splitting the isosceles triangle into two right-angled triangles. Either route is fine; the cosine rule is the one you already have.
This is where a calculator in degree mode does real damage: sin 1.2 in degrees is 0.0209, which makes the triangle almost nothing and the segment almost the whole sector.
7Working backwards with two unknowns
Examiners like to give two facts about a sector and ask for both r and θ. The method: write each fact as an equation, eliminate θ, solve for r, then check that θ makes sense (it must be positive and less than 2π for a sector).
Worked example (Paper 1). A sector has perimeter 20 cm and area 16 cm². Find its radius and its angle.
So r = 8 cm and θ = 0.5 radians. Check: perimeter 16 + 4 = 20 and area ½ × 64 × 0.5 = 16. The rejected root is the mark examiners are testing: a quadratic gives two candidates, and the context decides which survive. Sometimes both do, and then you give both.
8A context: the windscreen wiper
Real shapes are often a sector with a smaller sector removed. A windscreen wiper blade runs from 12 cm to 55 cm from its pivot and sweeps through 120°. Find the area of glass it cleans. Figure 6 shows the region.
The question gave degrees, so the first line converts. Leaving the answer as 2881π/3 cm² would be the exact form on Paper 1.
9Where marks are lost
Using degrees in l = rθ or A = ½r²θ. Both formulas need θ in radians. Convert 75° to 5π/12 first; putting in 75 gives nonsense.
Calculator in the wrong mode. Radians are the default. A segment question with sin θ evaluated in degree mode loses every mark after the sector.
Forgetting the radii in a perimeter. The perimeter of a sector is 2r + rθ, not rθ.
Using the minor angle for a major sector. The major sector's angle is 2π − θ. Read which region is shaded.
Treating a segment as a sector. A segment is bounded by a chord, not by two radii. Subtract the triangle.
Keeping an impossible root. When solving for r, check θ: a sector angle above 2π, or a negative radius, is rejected, with the reason stated.
Writing π as 3.14 on Paper 1. Exact means in terms of π: 27π, not 84.8 and not 84.78.
Mixing units. An angle of 1.4 is 1.4 radians, not 1.4°. Only write the degree sign when you mean degrees.
10Work it right
- Before any formula, make sure θ is in radians. If the question gives degrees, write the conversion as your first line.
- Sketch the circle, shade the region asked for, and label r and θ. Decide: arc, sector, perimeter, segment, minor or major.
- Write the formula with the numbers substituted: l = rθ, A = ½r²θ, perimeter = 2r + rθ, segment = ½r²(θ − sin θ).
- On Paper 1 keep π; on Paper 2 give 3 significant figures with units (cm, cm²).
- For two unknowns, eliminate θ by using rθ as a block, solve the quadratic, then test every root against 0 < θ < 2π.
- Check with the whole circle: a sector's area must be less than πr², and minor plus major must give the whole.
11Try it
Marks in brackets. Q1 and Q2 are Paper 1 style, no calculator. Q3 to Q5 are Paper 2 style, with a GDC.
Q1. (a) Write 150° in radians, as an exact multiple of π. 1 mark
(b) Write 7π/4 radians in degrees. 1 mark
(c) A sector of a circle has radius 6 cm and angle 5π/6. Find, in terms of π, the arc length and the area of the sector. 4 marks
Q2. A sector of a circle has radius r cm and angle θ radians. Its perimeter is 28 cm and its area is 48 cm².
(a) Show that r² − 14r + 48 = 0. 3 marks
(b) Find the two possible pairs of values of r and θ. 3 marks
Q3. A chord AB of a circle with centre O and radius 7 cm subtends an angle of 1.9 radians at O.
(a) Find the length of the chord AB. 2 marks
(b) Find the area of the minor segment cut off by AB. 3 marks
(c) Find the area of the major segment. 2 marks
Q4. A pendulum 60 cm long swings so that its tip moves along an arc of length 24 cm. Find the angle it swings through, in radians and in degrees. 3 marks
Q5. A flower bed is in the shape of a sector of a circle of radius 4.5 m and angle 110°.
(a) Find the area of the flower bed. 3 marks
(b) Edging is laid along the whole boundary of the bed. Find the length of edging needed. 3 marks
12In one breath
A radian is the angle at the centre that cuts off an arc as long as the radius, so a full turn is 2π radians and π radians is 180°: multiply by π/180 to get radians and by 180/π to get degrees, and an angle written without a degree sign is in radians, which every paper assumes. Because the angle in radians is the number of radii along the arc, the arc length is l = rθ and the sector area is ½r²θ, both straight from "what fraction of the circle is this?". The perimeter of a sector adds the two radii, a major sector uses 2π − θ, and a segment is a sector minus the triangle, ½r²(θ − sin θ). When two facts fix a sector, eliminate θ, solve for r, and reject any root that gives an impossible angle. Check the calculator is in radians before every sin.
Answers
Q1. (a) 150 × π/180 = 5π/6. A1.
(b) (7π/4) × 180/π = 315°. A1.
(c) Arc = 6 × 5π/6 = 5π cm. Area = ½ × 36 × 5π/6 = 15π cm². M1 for rθ, A1 for 5π, M1 for ½r²θ, A1 for 15π. Decimal answers (15.7, 47.1) score A0 each, since π was asked for.
Q2. (a) Perimeter: 2r + rθ = 28, so rθ = 28 − 2r. Area: ½r²θ = 48, so ½r(rθ) = 48. Substituting, ½r(28 − 2r) = 48, so 14r − r² = 48, which gives r² − 14r + 48 = 0. M1 for both equations, M1 for eliminating θ, A1 for rearranging to the given quadratic. "Show that": each step must be seen.
(b) (r − 6)(r − 8) = 0, so r = 6 or r = 8. With r = 6, θ = (28 − 12) ÷ 6 = 8/3 ≈ 2.67 rad. With r = 8, θ = (28 − 16) ÷ 8 = 1.5 rad. Both angles lie between 0 and 2π, so r = 6, θ = 8/3 and r = 8, θ = 3/2 are both valid. A1 for r = 6 and 8, A1 for θ = 8/3, A1 for θ = 3/2. Rejecting a valid pair without reason loses its A1.
Q3. (a) AB² = 7² + 7² − 2(7)(7) cos 1.9 = 129.68…, so AB = 11.4 cm (3 s.f.); or AB = 14 sin 0.95. M1 for the cosine rule or 2r sin(θ/2), A1 for 11.4.
(b) Sector = ½(49)(1.9) = 46.55. Triangle = ½(49) sin 1.9 = 23.18…. Minor segment = 23.4 cm² (3 s.f.). M1 for the sector area, M1 for the triangle area with sin 1.9 in radians, A1 for 23.4.
(c) Major segment = circle − minor segment = 49π − 23.365… = 131 cm² (3 s.f.). M1 for subtracting their minor segment from πr², A1 for 131. The alternative, major sector ½(49)(2π − 1.9) plus the triangle, is equally valid.
Q4. θ = l ÷ r = 24 ÷ 60 = 0.4 rad, and 0.4 × 180/π = 22.9° (3 s.f.). M1 for θ = l ÷ r, A1 for 0.4, A1 for 22.9°.
Q5. (a) θ = 110 × π/180 = 1.9198… rad. Area = ½(4.5²)(1.9198…) = 19.4 m² (3 s.f.). M1 for converting to radians, M1 for ½r²θ, A1 for 19.4. Using 110 directly in ½r²θ scores M0 for both method marks.
(b) Perimeter = 2(4.5) + 4.5 × 1.9198… = 9 + 8.639… = 17.6 m (3 s.f.). M1 for the arc length rθ, M1 for adding the two radii, A1 for 17.6.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 3.4 The circle: radians, arcs and sectors. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.