Radians, Arcs and Sectors – A Level Maths Year 2

1 What a Radian Is

One radian is the angle at the centre of a circle subtended by an arc whose length is equal to the radius. Because the whole circumference is $2\pi r$, a full turn is $2\pi$ radians.

Converting

$2\pi \text{ rad} = 360^\circ \qquad \pi \text{ rad} = 180^\circ \qquad 1 \text{ rad} \approx 57.3^\circ$
degrees $\to$ radians: $\times\dfrac{\pi}{180}$ radians $\to$ degrees: $\times\dfrac{180}{\pi}$

Why bother? In radians, arc length and sector area have very simple formulae, and — crucially for Year 2 calculus — $\dfrac{\mathrm{d}}{\mathrm{d}x}(\sin x) = \cos x$ is only true when $x$ is in radians. From now on, assume radians unless you see a degree sign.

Worked Example 1 — Converting

(a) Convert $135^\circ$ to radians. (b) Convert $\dfrac{5\pi}{6}$ to degrees. (c) Convert $2.5$ rad to degrees.

①(a) $135 \times \dfrac{\pi}{180} = \dfrac{3\pi}{4}$.

②(b) $\dfrac{5\pi}{6} \times \dfrac{180}{\pi} = 150^\circ$. (Quick method: replace $\pi$ by $180^\circ$.)

③(c) $2.5 \times \dfrac{180}{\pi} = 143.2^\circ$ (1 d.p.).

2 Exact Values in Radians

$\theta$	$0$	$\dfrac{\pi}{6}$ ($30^\circ$)	$\dfrac{\pi}{4}$ ($45^\circ$)	$\dfrac{\pi}{3}$ ($60^\circ$)	$\dfrac{\pi}{2}$ ($90^\circ$)
$\sin\theta$	$0$	$\tfrac12$	$\tfrac{\sqrt2}{2}$	$\tfrac{\sqrt3}{2}$	$1$
$\cos\theta$	$1$	$\tfrac{\sqrt3}{2}$	$\tfrac{\sqrt2}{2}$	$\tfrac12$	$0$
$\tan\theta$	$0$	$\tfrac{\sqrt3}{3}$	$1$	$\sqrt3$	undefined

The symmetries of the graphs from Year 1 carry straight over, with $180^\circ$ replaced by $\pi$ and $360^\circ$ by $2\pi$:

Symmetry in radians

$\sin(\pi - \theta) = \sin\theta \qquad \cos(2\pi - \theta) = \cos\theta \qquad \tan(\pi + \theta) = \tan\theta$

Worked Example 2 — Exact values of larger angles

Find the exact values of $\sin\dfrac{5\pi}{6}$, $\cos\dfrac{4\pi}{3}$ and $\tan\dfrac{7\pi}{4}$.

①$\dfrac{5\pi}{6} = \pi - \dfrac{\pi}{6}$ (second quadrant, sine positive): $\sin\dfrac{5\pi}{6} = \sin\dfrac{\pi}{6} = \dfrac12$.

②$\dfrac{4\pi}{3} = \pi + \dfrac{\pi}{3}$ (third quadrant, cosine negative): $\cos\dfrac{4\pi}{3} = -\dfrac12$.

③$\dfrac{7\pi}{4} = 2\pi - \dfrac{\pi}{4}$ (fourth quadrant, tangent negative): $\tan\dfrac{7\pi}{4} = -1$.

Worked Example 3 — An equation in radians

Solve $\cos 2x = \tfrac12$ for $0 \leqslant x \leqslant 2\pi$.

①Let $u = 2x$. Then $0 \leqslant u \leqslant 4\pi$ — the interval doubles too.

②$\cos u = \tfrac12$: principal value $u = \dfrac{\pi}{3}$, and $u = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$.

③Add $2\pi$ to each: $u = \dfrac{7\pi}{3}, \dfrac{11\pi}{3}$ are also in range.

④Halve: $x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$.

If the interval is given in radians, your answers must be in radians. Set your calculator to RAD mode before you start.

3 Arc Length and Sector Area

A sector with angle $\theta$ is the fraction $\dfrac{\theta}{2\pi}$ of the whole circle. So:

Arc length and sector area — $\theta$ in radians

$s = \dfrac{\theta}{2\pi} \times 2\pi r = r\theta \qquad\qquad A = \dfrac{\theta}{2\pi} \times \pi r^2 = \tfrac12 r^2\theta$

Radians only. With $\theta$ in degrees you must go back to the GCSE formulae $\tfrac{\theta}{360} \times 2\pi r$ and $\tfrac{\theta}{360} \times \pi r^2$. Using $r\theta$ with $\theta = 60$ gives an arc sixty times the radius.

Worked Example 4 — Arc, perimeter and area

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

①Arc length: $s = 8 \times 1.2 = 9.6$ cm.

②Perimeter: two radii plus the arc $= 8 + 8 + 9.6 = 25.6$ cm.

③Area: $\tfrac12 \times 8^2 \times 1.2 = 38.4$ cm$^2$.

Worked Example 5 — Working backwards

A sector has perimeter $20$ cm and area $16$ cm$^2$. Find its radius and angle.

①$2r + r\theta = 20$, so $r\theta = 20 - 2r$.

②Area: $\tfrac12 r^2\theta = \tfrac12 r (r\theta) = \tfrac12 r(20 - 2r) = 16$.

③$10r - r^2 = 16 \Rightarrow r^2 - 10r + 16 = 0 \Rightarrow (r - 2)(r - 8) = 0$.

④$r = 2$ gives $\theta = \dfrac{16}{2} = 8$ rad — more than $2\pi$, impossible. Reject.

⑤$r = 8$ gives $\theta = \dfrac{4}{8} = 0.5$ rad. So $r = 8$ cm, $\theta = 0.5$.

4 Segments

A segment is the region between a chord and its arc. It is the sector minus the triangle $OAB$, and the triangle's area is $\tfrac12 ab\sin C = \tfrac12 r^2\sin\theta$.

Area of a segment

$A = \tfrac12 r^2\theta - \tfrac12 r^2\sin\theta = \tfrac12 r^2(\theta - \sin\theta)$

Worked Example 6 — An exact segment area

A chord of a circle of radius $6$ cm subtends an angle of $\dfrac{\pi}{3}$ at the centre. Find the exact area of the minor segment.

①$A = \tfrac12 \times 36\left(\dfrac{\pi}{3} - \sin\dfrac{\pi}{3}\right) = 18\left(\dfrac{\pi}{3} - \dfrac{\sqrt3}{2}\right)$

②$= 6\pi - 9\sqrt3$ cm$^2$ $\;(\approx 3.26$ cm$^2)$.

Worked Example 7 — Chord length and segment perimeter

For a circle of radius $7$ cm, a chord $AB$ subtends an angle of $2.4$ rad at the centre. Find the length of $AB$ and the perimeter of the minor segment.

①Split the isosceles triangle in half: $AB = 2r\sin\dfrac{\theta}{2} = 14\sin 1.2 = 13.05$ cm (2 d.p.).

②Arc length $= 7 \times 2.4 = 16.8$ cm.

③Perimeter $= 16.8 + 13.05 = 29.85$ cm.

The cosine rule gives the chord too: $AB^2 = 7^2 + 7^2 - 2(7)(7)\cos 2.4$. Both routes need the calculator in radians.

5 Quick Reference

Definition

$1$ rad: arc length equals radius.

Full turn

$2\pi$ rad $= 360^\circ$; $\pi$ rad $= 180^\circ$.

Convert

Deg $\to$ rad: $\times \tfrac{\pi}{180}$. Rad $\to$ deg: $\times\tfrac{180}{\pi}$.

Arc length

$s = r\theta$

Sector area

$A = \tfrac12 r^2\theta$

Segment area

$\tfrac12 r^2(\theta - \sin\theta)$

Chord

$2r\sin\tfrac{\theta}{2}$

Sector perimeter

$2r + r\theta$

Calculator

RAD mode for every radian question.

6 Practice Questions

Question 1

Convert $240^\circ$ to radians, and $\dfrac{7\pi}{12}$ to degrees.

▶ Show solution

$240 \times \dfrac{\pi}{180} = \dfrac{4\pi}{3}$.

$\dfrac{7 \times 180^\circ}{12} = 105^\circ$.

Question 2

Find the exact values of $\cos\dfrac{5\pi}{4}$, $\sin\left(-\dfrac{\pi}{6}\right)$ and $\tan\dfrac{2\pi}{3}$.

▶ Show solution

$\dfrac{5\pi}{4} = \pi + \dfrac{\pi}{4}$, third quadrant: $\cos\dfrac{5\pi}{4} = -\dfrac{\sqrt2}{2}$.

Sine is odd: $\sin\left(-\dfrac{\pi}{6}\right) = -\dfrac12$.

$\dfrac{2\pi}{3} = \pi - \dfrac{\pi}{3}$, second quadrant: $\tan\dfrac{2\pi}{3} = -\sqrt3$.

Question 3

A sector has radius $5$ cm and angle $0.8$ rad. Find its arc length and area.

▶ Show solution

$s = 5 \times 0.8 = 4$ cm.

$A = \tfrac12 \times 25 \times 0.8 = 10$ cm$^2$.

Question 4

A sector of radius $6$ cm has area $30$ cm$^2$. Find the angle of the sector and its perimeter.

▶ Show solution

$\tfrac12 \times 36 \times \theta = 30 \Rightarrow \theta = \tfrac{30}{18} = \tfrac53$ rad.

Arc $= 6 \times \tfrac53 = 10$ cm, so perimeter $= 6 + 6 + 10 = 22$ cm.

Question 5

Solve $\sin x = -\dfrac{\sqrt3}{2}$ for $0 \leqslant x \leqslant 2\pi$, giving exact answers.

▶ Show solution

The reference angle is $\dfrac{\pi}{3}$. Sine is negative in the third and fourth quadrants.

$x = \pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$ and $x = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$.

Question 6

Find the area of the minor segment cut off by a chord that subtends an angle of $2$ rad at the centre of a circle of radius $10$ cm.

▶ Show solution

$A = \tfrac12 \times 100 \times (2 - \sin 2) = 50(2 - 0.9093) = 54.5$ cm$^2$ (3 s.f.).

Question 7

A sector has perimeter $24$ cm and area $32$ cm$^2$. Find the two possible pairs of values of $r$ and $\theta$.

▶ Show solution

$r\theta = 24 - 2r$, and $\tfrac12 r(r\theta) = 32 \Rightarrow \tfrac12 r(24 - 2r) = 32$.

$12r - r^2 = 32 \Rightarrow r^2 - 12r + 32 = 0 \Rightarrow (r - 4)(r - 8) = 0$.

$r = 4$: $\theta = \tfrac{16}{4} = 4$ rad. $r = 8$: $\theta = \tfrac{8}{8} = 1$ rad.

Both angles are less than $2\pi$, so both are valid.

Question 8

Find the exact area of the segment of a circle of radius $4$ cm cut off by a chord subtending $\dfrac{2\pi}{3}$ at the centre.

▶ Show solution

$A = \tfrac12 \times 16\left(\dfrac{2\pi}{3} - \sin\dfrac{2\pi}{3}\right) = 8\left(\dfrac{2\pi}{3} - \dfrac{\sqrt3}{2}\right)$

$= \dfrac{16\pi}{3} - 4\sqrt3$ cm$^2$.

Question 9

A sector has a fixed perimeter $P$. Show that its area is $A = \tfrac12 r(P - 2r)$, and hence show that the area is greatest when $\theta = 2$ radians.

▶ Show solution

$P = 2r + r\theta \Rightarrow r\theta = P - 2r$, so $A = \tfrac12 r^2\theta = \tfrac12 r(r\theta) = \tfrac12 r(P - 2r)$ ✓

$A = \tfrac12 Pr - r^2$ is a quadratic in $r$ with a maximum where $\dfrac{\mathrm{d}A}{\mathrm{d}r} = \tfrac12P - 2r = 0$, i.e. $r = \tfrac{P}{4}$.

Then $r\theta = P - \tfrac{P}{2} = \tfrac{P}{2}$, so $\theta = \dfrac{P/2}{P/4} = 2$ rad. $\blacksquare$

Question 10

A car windscreen wiper pivots at a point $O$. The rubber blade lies along the arm from $15$ cm to $55$ cm from $O$, and the arm turns through an angle of $2.2$ radians.

(a) Find the distance travelled by the tip of the blade. (b) Find the area of windscreen wiped. (c) The manufacturer wants the wiped area to be at least $3500$ cm$^2$. Find the minimum angle, in radians and in degrees.

▶ Show solution

(a) The tip moves on an arc of radius $55$: $55 \times 2.2 = 121$ cm.

(b) Wiped region = large sector minus small sector:

$\tfrac12(55^2)(2.2) - \tfrac12(15^2)(2.2) = \tfrac12(3025 - 225)(2.2) = 1400 \times 2.2 = 3080$ cm$^2$.

(c) $1400\theta \geqslant 3500 \Rightarrow \theta \geqslant 2.5$ rad.

$2.5 \times \dfrac{180}{\pi} = 143.2^\circ$ (1 d.p.).

Radians, Arcs and Sectors · A Level Maths Year 2 · Created with MathJax