3.4 Radians, arcs and sectors

Degrees divide a full turn into 360 equal parts, a choice inherited from ancient astronomy rather than from any property of the circle. The radian measures an angle by comparing the length of an arc with the radius of its circle, which makes it the natural unit for any angle that comes from turning. With angles in radians, the length of an arc and the area of a sector have very short formulas, and in calculus the derivatives of $\sin x$ and $\cos x$ take their simple forms only when $x$ is in radians.

Arcs and sectors appear wherever something turns: the distance travelled by a point on a wheel, the region swept by a windscreen wiper, the slice of a pie chart, or the curved bend of a running track.

The parts of a circle

A circle is the set of points at a fixed distance, the radius $r$, from a fixed point, the centre $O$. Several parts of a circle have names that are used throughout this subtopic.

• A chord is a straight line segment joining two points on the circle. A chord through the centre is a diameter.

• An arc is a part of the circumference between two points. The shorter one is the minor arc and the longer one is the major arc.

• A sector is the region between two radii and an arc, shaped like a slice of pizza.

• A segment is the region between a chord and an arc.

The angle $\theta$ between the two radii, measured at the centre, is called the angle subtended by the arc at the centre. Arc lengths, sector areas and segment areas are all calculated from this angle and the radius.

Radian measure

Defining the radian

Take a circle of radius $r$ and mark off an arc whose length is exactly $r$. The angle at the centre between the radii to the ends of this arc is one radian.

Because the definition compares two lengths, the size of a radian does not depend on the size of the circle. A larger circle has a longer radius and a proportionally longer arc, so the angle is the same. This is why a radian is a pure number: an angle in radians is the ratio $\frac{\text{arc length}}{\text{radius}}$, and the units of length cancel.

How many radians in a full turn

The whole circumference has length $2\pi r$. The number of radius-lengths that fit around the circle is therefore

so a full turn is $2\pi$ radians, a little more than 6.28 rad. Since a full turn is also $360°$,

This single fact, $\pi$ radians equals $180°$, is all that is needed to convert in either direction. Dividing by $\pi$ gives the size of one radian in degrees:

Converting between degrees and radians

To convert, multiply by the ratio that cancels the unit being removed.

When the angle in degrees is a simple fraction of $180°$, the answer in radians is best left as an exact value, a fraction of $\pi$. The angles in the next table occur so often that it is worth knowing them by heart.

Radian mode on a calculator

A graphic display calculator (GDC) works in either degree mode or radian mode, and the same keystrokes give different answers in each. In degree mode $\sin 2$ means the sine of $2°$, about $0.0349$; in radian mode it means the sine of 2 radians, about $0.909$. Whenever an angle is given in radians, or whenever a formula such as the segment area formula mixes $\theta$ with $\sin\theta$, the calculator must be in radian mode.

Arc length

Deriving the formula

An arc that subtends an angle $\theta$ at the centre is a fraction of the full circumference. The fraction is the angle divided by a full turn, $\frac{\theta}{2\pi}$ in radians or $\frac{\theta}{360}$ in degrees. So the length of the arc is

The $2\pi$ cancels, leaving one of the shortest formulas in geometry. The derivation also agrees with the definition of a radian: when $\theta = 1$, the arc length is $r$.

The formula $l = r\theta$ can be rearranged to find either the radius or the angle: $r = \frac{l}{\theta}$ and $\theta = \frac{l}{r}$. The second of these is exactly the definition of radian measure, angle = arc length ÷ radius.

The perimeter of a sector

The boundary of a sector is made of two radii and one arc, so its perimeter is

Leaving out the two radii is the most common error with sector perimeters.

Arc length in context

A point on the rim of a wheel, a capsule on an observation wheel and the tip of a clock hand all move along arcs. The distance travelled is the arc length, whereas the straight-line distance from start to finish is the length of a chord, which is always shorter.

Arc length also converts angles into distances on the Earth. Along a line of longitude, the distance between two places is an arc of a circle whose radius is the Earth's radius, about $6371$ km. One degree of latitude is $\frac{\pi}{180}$ radians, so it corresponds to about $6371 \times \frac{\pi}{180} \approx 111$ km on the ground. This is the reason a nautical mile was originally defined as one-sixtieth of a degree of latitude.

Area of a sector

Deriving the formula

A sector with angle $\theta$ at the centre is the same fraction $\frac{\theta}{2\pi}$ of the whole disc as its arc is of the circumference. Its area is therefore

Both $l = r\theta$ and $A = \frac{1}{2}r^2\theta$ are in the formula booklet, with $\theta$ in radians. Combining them gives a useful link: since $r\theta = l$, the area can also be written as $A = \frac{1}{2}rl$. This mirrors the area of a triangle, $\frac{1}{2} \times \text{base} \times \text{height}$, with the arc as the base and the radius as the height; a thin sector is very nearly a triangle.

Finding the radius or the angle from an area

When the area is known, rearrange $A = \frac{1}{2}r^2\theta$. To find the angle, $\theta = \frac{2A}{r^2}$; to find the radius, $r = \sqrt{\frac{2A}{\theta}}$, taking the positive root because a radius is a length.

Sectors in context

The hands of a clock sweep out sectors. The minute hand turns through a full turn, $2\pi$ radians, every 60 minutes, so it turns through $\frac{\pi}{30}$ radians each minute. The hour hand turns through $2\pi$ every 12 hours, which is $\frac{\pi}{360}$ radians each minute.

Regions between two arcs

Some shapes are the difference between two sectors with the same centre and angle, such as the region swept by a windscreen wiper blade or a curved path around a flower bed. Their area is the area of the larger sector minus the area of the smaller one:

Area of a segment

Sector minus triangle

A minor segment is the part of a sector that lies outside the triangle formed by the two radii and the chord. Its area is therefore the area of the sector minus the area of triangle $OAB$. The triangle has two sides of length $r$ with the angle $\theta$ between them, so by the formula $\frac{1}{2}ab\sin C$ from 3.2 its area is $\frac{1}{2}r^2\sin\theta$.

This formula is not in the formula booklet, but it follows in one line from two that are. Both terms use the same angle $\theta$, so the calculator must be in radian mode when evaluating $\sin\theta$.

Major segments

The major segment is everything in the circle outside the minor segment, so its area is

The segment formula also works directly with the reflex angle $2\pi - \theta$. Since $\sin(2\pi - \theta) = -\sin\theta$, it gives $\frac{1}{2}r^2(2\pi - \theta + \sin\theta)$, which simplifies to the same expression. For a reflex angle the "triangle" term is negative and is added on, because the triangle now lies inside the major segment.

Segment area as a function of the angle

For a circle of radius 1, the segment area is $\frac{1}{2}(\theta - \sin\theta)$. It grows from $0$ when $\theta = 0$ to $\frac{\pi}{2}$ (half the disc) when $\theta = \pi$, and to $\pi$ (the whole disc) when $\theta = 2\pi$. The equation $\theta - \sin\theta = k$ cannot be rearranged to make $\theta$ the subject, so questions that give a segment area and ask for the angle are solved with a GDC, by graphing or with an equation solver.

Problems with unknown radius and angle

When two facts about a sector are given, such as its perimeter and its area, each gives an equation in $r$ and $\theta$. Solving them simultaneously finds both, and the solutions must then be checked against the geometry: $r > 0$ and $0 < \theta < 2\pi$.

The next type combines a length condition and an area condition, leading to a quadratic equation in $r$ (see 2.7). One of its roots often fails the condition $\theta < 2\pi$.

Why radians are used

Degrees remain convenient in triangle problems and for bearings, and most triangle questions in 3.2 and 3.3 use them. Radians are preferred elsewhere for two reasons. First, they make the circle formulas simple: $l = r\theta$ and $A = \frac{1}{2}r^2\theta$ need no factor of $\frac{\pi}{180}$. Second, in calculus the result $\frac{d}{dx}(\sin x) = \cos x$ holds only when $x$ is in radians; in degrees an awkward factor of $\frac{\pi}{180}$ appears in every derivative. For this reason the circular functions in 3.7 and all of the calculus in Topic 5 use radians.

For small angles, the radian also explains a useful approximation. The arc $r\theta$ and the half-chord $r\sin\theta$ become almost equal as $\theta$ gets close to zero, so $\sin\theta \approx \theta$ when $\theta$ is small and in radians. For example, $\sin 0.1 = 0.0998\ldots$, very close to $0.1$.

Exam-style questions on arcs, sectors and segments

The longer questions on this content combine arc length, sector area and triangle trigonometry (the cosine rule and $\frac{1}{2}ab\sin C$) in several linked parts. The two worked questions show both a calculator and a non-calculator question.

The first question is about two equal circles that overlap.

Self-check

• To do:

  I can state the definition of a radian and explain why $\pi$ radians equals $180°$.

• To do:

  I can convert angles between degrees and radians, giving exact answers in terms of $\pi$.

• To do:

  I can find an arc length with $l = r\theta$ and the perimeter of a sector with $2r + r\theta$.

• To do:

  I can find the area of a sector with $A = \frac{1}{2}r^2\theta$, and rearrange it to find $r$ or $\theta$

• To do:

  I can find the area of a segment as sector minus triangle, $\frac{1}{2}r^2(\theta - \sin\theta)$.

• To do:

  I can find the area of a region between two arcs, or of a region made of two segments.

• To do:

  I can solve simultaneous equations in $r$ and $\theta$ and reject values that do not give a valid sector.

• To do:

  I can set my GDC to the correct angle mode and use it to solve equations such as $\theta - \sin\theta = k$.

1. (a)

   Find the value of $\theta$.

   [2]2 marks

   Markscheme (a)

   arc length $\mathrm{AB}$ equals the circumference of the base: $10\theta = 2\pi \times 6$ (M1)

   $\theta = \frac{6\pi}{5} \; (= 3.77)$ A1

2. (b)

   Find the exact area of the sector $\mathrm{AOB}$.

   [2]2 marks

   Markscheme (b)

   EITHER

   $\frac{1}{2} \times 10^2 \times \frac{6\pi}{5}$

3. (c)

   Find the exact volume of the space enclosed by the hat and its circular base.

   [2]2 marks

   Markscheme (c)

   height $= \sqrt{10^2 - 6^2} = 8$

(a) $\frac{5\pi}{6} = \frac{5 \times 180°}{6} = 150°$

(b) $\frac{7\pi}{4} = \frac{7 \times 180°}{4} = 315°$

(c) $2.5 \times \frac{180}{\pi} = 143.239\ldots° \approx 143.2°$

(d) $0.8 \times \frac{180}{\pi} = 45.836\ldots° \approx 45.8°$

In (c) and (d) there is no $\pi$ in the angle, so these are not "nice" fractions of a turn and the answer is a decimal.

Triangle: $\frac{1}{2} \times 10^2 \times \sin\frac{2\pi}{3} = 50 \times \frac{\sqrt{3}}{2} = 25\sqrt{3}$, using the exact value $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ from 3.5.

Segment:

In a "show that", write the unrounded value before rounding to the given answer. The cosine rule in triangle $APB$ gives the same result: $\cos P\hat{A}B = \frac{5^2 + 6^2 - 5^2}{2 \times 5 \times 6} = \frac{36}{60} = 0.6$.

(b) By symmetry, $P\hat{A}Q = 2 \times 0.927295\ldots = 1.854590\ldots$ rad.

(c) The shaded region consists of two identical segments, one in each circle, each cut off by the chord $PQ$. One segment is sector $APQ$ minus triangle $APQ$.

(The triangle has base $PQ = 8$ cm and height $AM = 3$ cm, which confirms its area of $12$ cm².)

(d) The boundary is two equal arcs, each of radius $5$ cm and angle $1.854590\ldots$ rad.

Then $\theta = \frac{6}{4} = \frac{3}{2}$ radians. This lies between $0$ and $2\pi$, so it is valid.

(b) Segment area = sector area − triangle area.

This is the exact answer: $\sin\frac{3}{2}$ is not one of the special values, so it stays as it is. (With a calculator it is about $4.02$ cm², a useful check.)

OR

curved surface area $= \pi \times 6 \times 10$ (M1)

THEN

$= 60\pi$ (cm$^2$) A1

volume $= \frac{1}{3}\pi \times 6^2 \times 8 = 96\pi$ (cm$^3$) A1

Note: Award (A0)A0 for using $10$ as the height.

[3]3 marks