Radians, arc length and sector area

AS · P1 · 11 min

Degrees are an arbitrary choice: there is nothing special about $360$ . The radian measures an angle by the circle itself, and that makes the formulas for arc length and sector area as simple as possible. Circular measure is a short syllabus section, but it appears on almost every Paper 1, usually as a 5 to 7 mark question on a diagram of sectors, and radians are then used throughout trigonometry and calculus.

What a radian is

Take a circle of radius $r$ and walk a distance $r$ around its edge. The angle this arc makes at the centre is one radian, whatever the size of the circle.

Definition

One radian is the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle.

x^2 + y^2 = 1 (0, 0) -- (1, 0) (0, 0) -- (cos(1), sin(1))

The arc from $(1, 0)$ to $(\cos 1, \sin 1)$ on the unit circle has length $1$ , equal to the radius, so the angle at the centre is $1$ radian, about $57.3^\circ$ .

Converting between radians and degrees

The whole circumference is $2\pi r$ , which is $2\pi$ radius-lengths. So a full turn is $2\pi$ radians:

Key result

2\pi \text{ radians} = 360^\circ, \qquad \pi \text{ radians} = 180^\circ

Degrees to radians: multiply by $\dfrac{\pi}{180}$ .
Radians to degrees: multiply by $\dfrac{180}{\pi}$ .
$1$ radian $\approx 57.3^\circ$ .

Learn the common angles as fractions of $\pi$ , because exact answers use them:

Degrees	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$135^\circ$	$150^\circ$	$180^\circ$	$270^\circ$	$360^\circ$
Radians	$\tfrac{\pi}{6}$	$\tfrac{\pi}{4}$	$\tfrac{\pi}{3}$	$\tfrac{\pi}{2}$	$\tfrac{2\pi}{3}$	$\tfrac{3\pi}{4}$	$\tfrac{5\pi}{6}$	$\pi$	$\tfrac{3\pi}{2}$	$2\pi$

An angle written without a degree sign, such as $\theta = 1.2$ , is in radians. You do not need to write "rad", although " $1.2$ rad" is fine.

Arc length and sector area

A sector is the region between two radii and the arc joining them, like a slice of pizza. Its angle at the centre is $\theta$ .

x^2 + y^2 = 25 (0, 0) -- (5, 0) (0, 0) -- (5*cos(1.2), 5*sin(1.2))

The sector is the fraction $\dfrac{\theta}{2\pi}$ of the whole circle. So both its arc length and its area are that fraction of the circle's circumference and area:

s = \frac{\theta}{2\pi} \times 2\pi r = r\theta, \qquad A = \frac{\theta}{2\pi} \times \pi r^2 = \tfrac{1}{2}r^2\theta

The $\pi$ s cancel: that is the whole point of radians.

Key result

For a sector of radius $r$ and angle $\theta$ in radians:

\text{arc length } s = r\theta, \qquad \text{sector area } A = \tfrac{1}{2}r^2\theta

The perimeter of the sector is $2r + r\theta$ (two radii plus the arc).

A useful link between the two: $A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}r \times (r\theta) = \tfrac{1}{2}rs$ . If you know the arc length and the area, this gives $r$ immediately.

The formulas are only true in radians. In degrees they would be $s = \dfrac{\theta}{360} \times 2\pi r$ and $A = \dfrac{\theta}{360} \times \pi r^2$ ; if a question gives an angle in degrees, convert it first.

Solving a sector problem

Mark every length you know on the diagram. Remember that all radii of the same circle are equal.
Convert any angle in degrees to radians.
Write down the formulas you will use: $s = r\theta$ , $A = \tfrac{1}{2}r^2\theta$ , perimeter $= 2r + r\theta$ .
Substitute and solve. If there are two unknowns, use two equations (for example, area and perimeter).
Give exact answers in terms of $\pi$ when the angle is a fraction of $\pi$ and the question asks for exact values; otherwise give 3 significant figures.

Working with your calculator

Set your calculator to radian mode for any question in radians that uses $\sin$ , $\cos$ or $\tan$ . For $s = r\theta$ and $A = \tfrac{1}{2}r^2\theta$ alone, the mode does not matter, since there is no trigonometric function. The mistake to avoid is computing $\sin 1.2$ in degree mode, which gives $0.0209$ instead of $0.932$ .

Worked examples

Converting angles

(a) Express $135^\circ$ in radians, in terms of $\pi$ .

(b) Express $\dfrac{5\pi}{12}$ radians in degrees.

Solution

(a) $135 \times \dfrac{\pi}{180} = \dfrac{135\pi}{180} = \dfrac{3\pi}{4}$ .

(b) $\dfrac{5\pi}{12} \times \dfrac{180}{\pi} = \dfrac{5 \times 180}{12} = 75^\circ$ .

Perimeter and area of a sector

A sector has radius $8$ cm and angle $1.2$ radians. Find its perimeter and its area.

Solution

Arc length $= r\theta = 8 \times 1.2 = 9.6$ cm.

Perimeter $= 8 + 8 + 9.6 = 25.6$ cm.

Area $= \tfrac{1}{2}r^2\theta = \tfrac{1}{2} \times 64 \times 1.2 = 38.4\ \text{cm}^2$ .

An angle in degrees, exact answers

A sector of a circle has angle $150^\circ$ and area $15\pi\ \text{cm}^2$ . Find the radius, and the exact perimeter of the sector.

Solution

$150^\circ = \dfrac{5\pi}{6}$ radians. Then

\tfrac{1}{2}r^2 \times \frac{5\pi}{6} = 15\pi \quad\Rightarrow\quad \frac{5r^2}{12} = 15 \quad\Rightarrow\quad r^2 = 36 \quad\Rightarrow\quad r = 6 \text{ cm}

Arc length $= 6 \times \dfrac{5\pi}{6} = 5\pi$ cm, so the perimeter is $(12 + 5\pi)$ cm.

Radius and angle from area and arc length

A sector has area $48\ \text{cm}^2$ and arc length $12$ cm. Find its radius and its angle.

Solution

Two unknowns, two equations: $\tfrac{1}{2}r^2\theta = 48$ and $r\theta = 12$ .

Write the area as $\tfrac{1}{2}r(r\theta)$ and substitute $r\theta = 12$ :

\tfrac{1}{2}r \times 12 = 48 \quad\Rightarrow\quad r = 8 \text{ cm}

Then $\theta = \dfrac{12}{8} = 1.5$ radians.

The region between two arcs

Two sectors $OAB$ and $OCD$ have the same centre $O$ and the same angle $\theta$ radians, with $OA = 10$ cm and $OC = 6$ cm, so $C$ lies on $OA$ and $D$ on $OB$ . The region $ABDC$ between the two arcs has area $32\ \text{cm}^2$ . Find $\theta$ and the perimeter of $ABDC$ .

Solution

The region is the large sector minus the small one:

\tfrac{1}{2}(10^2)\theta - \tfrac{1}{2}(6^2)\theta = 32 \quad\Rightarrow\quad 32\theta = 32 \quad\Rightarrow\quad \theta = 1

The perimeter is two arcs and two straight pieces $CA$ and $DB$ , each $10 - 6 = 4$ cm:

10(1) + 6(1) + 4 + 4 = 24 \text{ cm}

Greatest area for a given perimeter

A sector of a circle has radius $r$ cm and angle $\theta$ radians, and its perimeter is $20$ cm.

(a) Show that the area $A\ \text{cm}^2$ of the sector is given by $A = 10r - r^2$ .

(b) Express $A$ in the form $a - (r - b)^2$ and hence find the greatest possible area and the corresponding value of $\theta$ .

Solution

(a) The perimeter gives $2r + r\theta = 20$ , so $r\theta = 20 - 2r$ . Then

A = \tfrac{1}{2}r^2\theta = \tfrac{1}{2}r(r\theta) = \tfrac{1}{2}r(20 - 2r) = 10r - r^2

(b) Completing the square:

10r - r^2 = -\left(r^2 - 10r\right) = -\left[(r - 5)^2 - 25\right] = 25 - (r - 5)^2

The greatest area is $25\ \text{cm}^2$ , when $r = 5$ . Then $5\theta = 20 - 10 = 10$ , so $\theta = 2$ radians.

Watch out

Using degrees in $s = r\theta$ . With $\theta = 60$ , $s = r\theta$ gives an arc sixty times the radius. Convert to $\tfrac{\pi}{3}$ first.

Forgetting the radii in a perimeter. The perimeter of a sector is $r\theta + 2r$ , not just the arc.

Squaring the wrong thing. The area is $\tfrac{1}{2}r^2\theta$ , with only $r$ squared.

Calculator in degree mode. Any $\sin\theta$ or $\cos\theta$ with $\theta$ in radians must be evaluated in radian mode.

Rounding $\pi$ too early. If the question asks for an exact answer, leave $\pi$ in: $12 + 5\pi$ , not $27.7$ .

Exam tip

"Exact" means in terms of $\pi$ (and surds, if a triangle is involved). A decimal answer loses the final mark.
Accuracy. Otherwise give lengths and areas to 3 significant figures, and angles in radians to 3 significant figures (or as asked).
Show the formula. Write $s = r\theta = 8 \times 1.2$ rather than just $9.6$ ; the method mark is for the substitution.
Read the diagram. Diagrams are not to scale. Radii of the same circle are equal; a line labelled as a tangent is perpendicular to the radius. Composite figures, with triangles and segments, are covered in Segments and composite regions.
"Show that" angles. If asked to show an angle equals, say, $\tfrac{\pi}{3}$ , give a full reason (an equilateral triangle, or a cosine value), not a measurement.

Summary

One radian is the angle at the centre subtended by an arc equal in length to the radius.
$\pi$ radians $= 180^\circ$ . Multiply by $\tfrac{\pi}{180}$ to convert to radians, by $\tfrac{180}{\pi}$ to convert to degrees.
Arc length $s = r\theta$ ; sector area $A = \tfrac{1}{2}r^2\theta$ ; sector perimeter $2r + r\theta$ . All need $\theta$ in radians.
$A = \tfrac{1}{2}rs$ links area and arc length.
Two unknowns need two equations; substitute $r\theta$ as a block.
Use radian mode for trigonometric functions of radian angles.

Practice questions

Question

(a) Express $210^\circ$ and $40^\circ$ in radians in terms of $\pi$ . (b) Express $\tfrac{3\pi}{8}$ radians and $1.2$ radians in degrees.
A sector has radius $12$ cm and area $54\ \text{cm}^2$ . Find its angle and its perimeter.
An arc of length $7$ cm subtends an angle of $0.35$ radians at the centre of a circle. Find the radius of the circle and the area of the sector.
A sector has radius $10$ cm and angle $\tfrac{2\pi}{5}$ . Find the exact arc length, area and perimeter.
A sector has perimeter $30$ cm and arc length $10$ cm. Find its radius, angle and area.
A sector has area $27\ \text{cm}^2$ and perimeter $24$ cm. Find the two possible pairs of values of the radius and the angle.
Two concentric sectors have radii $8$ cm and $5$ cm and the same angle $\theta$ . The region between their arcs has area $19.5\ \text{cm}^2$ . Find $\theta$ and the perimeter of the region.
The minute hand of a clock is $12$ cm long. Find the exact area swept by the hand in $20$ minutes, and the exact distance moved by its tip.
A sector has radius $r$ and angle $\theta$ , and its perimeter is $40$ cm. Show that its area is $20r - r^2$ , and find the greatest possible area and the value of $\theta$ at which it occurs.
A sector $OAB$ has radius $r$ and angle $\theta$ . A square has sides equal in length to the arc $AB$ . Given that the area of the square is equal to the area of the sector, find $\theta$ . Given also that the perimeter of the sector is $15$ cm, find $r$ .

Answers

(a) $210 \times \tfrac{\pi}{180} = \tfrac{7\pi}{6}$ ; $40 \times \tfrac{\pi}{180} = \tfrac{2\pi}{9}$ . (b) $\tfrac{3\pi}{8} \times \tfrac{180}{\pi} = 67.5^\circ$ ; $1.2 \times \tfrac{180}{\pi} = 68.8^\circ$ .
$\tfrac{1}{2}(144)\theta = 54$ , so $\theta = 0.75$ . Arc $= 12 \times 0.75 = 9$ , so the perimeter is $12 + 12 + 9 = 33$ cm.
$0.35r = 7$ , so $r = 20$ cm. Area $= \tfrac{1}{2}(400)(0.35) = 70\ \text{cm}^2$ .
Arc $= 10 \times \tfrac{2\pi}{5} = 4\pi$ cm. Area $= \tfrac{1}{2}(100)\left(\tfrac{2\pi}{5}\right) = 20\pi\ \text{cm}^2$ . Perimeter $= (20 + 4\pi)$ cm.
$2r + 10 = 30$ , so $r = 10$ cm. $\theta = \tfrac{10}{10} = 1$ radian. Area $= \tfrac{1}{2}rs = \tfrac{1}{2}(10)(10) = 50\ \text{cm}^2$ .
$r\theta = 24 - 2r$ , so the area is $\tfrac{1}{2}r(24 - 2r) = 12r - r^2 = 27$ . Then $r^2 - 12r + 27 = 0$ , $(r - 3)(r - 9) = 0$ .
- $r = 3$ : $r\theta = 18$ , $\theta = 6$ radians. This is less than $2\pi \approx 6.28$ , so it is possible (a sector that is almost a whole circle).
- $r = 9$ : $r\theta = 6$ , $\theta = \tfrac{2}{3}$ radians.
$\tfrac{1}{2}(64 - 25)\theta = 19.5$ , so $19.5\theta = 19.5$ and $\theta = 1$ . Perimeter $= 8(1) + 5(1) + 3 + 3 = 19$ cm.
In $20$ minutes the hand turns a third of a revolution: $\theta = \tfrac{2\pi}{3}$ . Area $= \tfrac{1}{2}(144)\left(\tfrac{2\pi}{3}\right) = 48\pi\ \text{cm}^2$ . Distance $= 12 \times \tfrac{2\pi}{3} = 8\pi$ cm.
$2r + r\theta = 40$ gives $r\theta = 40 - 2r$ , so $A = \tfrac{1}{2}r(40 - 2r) = 20r - r^2 = 100 - (r - 10)^2$ . The greatest area is $100\ \text{cm}^2$ at $r = 10$ ; then $10\theta = 20$ and $\theta = 2$ .
$(r\theta)^2 = \tfrac{1}{2}r^2\theta$ . Dividing by $r^2\theta$ (both non-zero): $\theta = \tfrac{1}{2}$ . Then $2r + \tfrac{1}{2}r = 15$ , so $r = 6$ cm.

What a radian is

Converting between radians and degrees

Arc length and sector area

Working with your calculator

Worked examples

Practice questions

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