Radian measure, arc length and sector areaAQA A-Level Maths: Revision notes

A radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. The full circle has arc length $2\pi r$, so $2\pi\text{ radians}=360^\circ,\qquad \pi\text{ radians}=180^\circ.$ To convert: degrees to radians multiply by $\frac{\pi}{180}$; radians to degrees multiply by $\frac{180}{\pi}$. Learn the common values: $30^\circ=\frac{\pi}{6}$, $45^\circ=\frac{\pi}{4}$, $60^\circ=\frac{\pi}{3}$, $90^\circ=\frac{\pi}{2}$, $120^\circ=\frac{2\pi}{3}$, $180^\circ=\pi$. For example $1.2\text{ rad}=1.2\times\frac{180}{\pi}=68.8^\circ$. A radian is about $57.3^\circ$.

A sector of radius $r$ and angle $\theta$ (in radians) takes the fraction $\frac{\theta}{2\pi}$ of the whole circle. Its arc length is therefore $s=\frac{\theta}{2\pi}\times2\pi r=r\theta.$ Example: radius $5$ cm, $\theta=1.2$: $s=5\times1.2=6$ cm. Rearranged, $\theta=\frac{s}{r}$ and $r=\frac{s}{\theta}$. The perimeter of the sector includes the two radii: $P=2r+r\theta$.

The sector takes the fraction $\frac{\theta}{2\pi}$ of the circle's area $\pi r^2$, so the area of a sector is $A=\frac{\theta}{2\pi}\times\pi r^2=\frac12r^2\theta.$ Since $s=r\theta$, you can also write $A=\frac12rs$. Example: $r=5$, $\theta=1.2$: $A=\frac12(25)(1.2)=15$ cm$^2$. If $A=24$ and $s=8$ then $r=\frac{2A}{s}=6$ and $\theta=\frac{s}{r}=\frac43$.

A segment is the region between a chord and an arc. For a minor segment, subtract the triangle from the sector: $A_{\text{segment}}=\frac12r^2\theta-\frac12r^2\sin\theta=\frac12r^2(\theta-\sin\theta).$ The chord length is $2r\sin\frac{\theta}{2}$ (or use the cosine rule $c^2=2r^2-2r^2\cos\theta$). The perimeter of the segment is the arc plus the chord. Example: $r=10$, $\theta=\frac{\pi}{3}$: triangle $OAB$ is equilateral, chord $=10$, arc $=\frac{10\pi}{3}$, area $=\frac{50\pi}{3}-25\sqrt3$.

Most exam questions combine these ideas with other geometry:

• Use the sector (or $\frac12r^2\sin\theta$ for the triangle) to find composite areas by adding or subtracting.
• A ring-shaped sector between radii $R$ and $r$ has area $\frac12\theta(R^2-r^2)$.
• Given a perimeter, eliminate $\theta$: if $2r+r\theta=20$ then $r\theta=20-2r$, and $A=\frac12r(r\theta)=10r-r^2$, which has maximum $25$ at $r=5$ (so $\theta=2$).
• The outer edge at radius $r$ travels an arc $r\theta$ in the time taken, which gives its speed. Check your calculator is in radian mode whenever $\theta$ is in radians.