Circular measure

1.4 Circular measure

What a radian is

One radian is the angle at the centre of a circle made by an arc equal in length to the radius. A whole circle has circumference $2\pi r$, so it contains $2\pi$ radians.

$\pi \text{ radians} = 180^\circ$

To convert degrees to radians, multiply by $\dfrac{\pi}{180}$. To convert radians to degrees, multiply by $\dfrac{180}{\pi}$.

Arc length and sector area

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

Both come from taking a fraction $\dfrac{\theta}{2\pi}$ of the whole circle.

$s = 6 \times 1.2 = 7.2 \text{ cm} \qquad A = \tfrac{1}{2} \times 36 \times 1.2 = 21.6 \text{ cm}^2$

The perimeter includes the two radii: $6 + 6 + 7.2 = 19.2$ cm.

Segments and chords

A chord joins two points on the circle. A segment is the region between a chord and its arc: the sector with the triangle removed.

The triangle has two sides of length $r$ with angle $\theta$ between them, so its area is $\tfrac{1}{2}r^2\sin\theta$. Therefore:

Cutting the triangle in half gives two right-angled triangles, which give the chord length:

$\text{chord} = 2r\sin\frac{\theta}{2}$

$\text{chord} = 12\sin 0.6 = 6.78 \text{ cm}$

$\text{segment} = \tfrac{1}{2} \times 36 \times (1.2 - \sin 1.2) = 18 \times 0.26796 = 4.82 \text{ cm}^2$

Tackling harder questions

Most questions combine sectors with triangles, tangents or several circles. These habits help:

• Mark every right angle. A tangent makes a right angle with the radius, so a tangent question usually has a right-angled triangle in it, and $\tan\theta$ gives the tangent length.
• Express the area or perimeter as a sum or difference of sectors, triangles and straight lengths before calculating.
• Find the missing angle using trigonometry with a right-angled triangle, or the cosine rule with a triangle of known sides. The answer must be in radians before you use it in $r\theta$.
• Keep exact values when asked. With $\theta = \dfrac{\pi}{3}$, the area of the triangle uses $\sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$.

Angle $AOT = \dfrac{\pi}{3}$ and angle $OAT$ is a right angle, so $TA = 5\tan\dfrac{\pi}{3} = 5\sqrt{3}$.

The kite $OATB$ is two right-angled triangles, each of area $\tfrac{1}{2} \times 5 \times 5\sqrt{3}$, so its area is $25\sqrt{3}$.

The sector has area $\tfrac{1}{2} \times 25 \times \dfrac{2\pi}{3} = \dfrac{25\pi}{3}$.

$\text{Area} = 25\sqrt{3} - \frac{25\pi}{3}$