Circular Measure

Two formulas, one unit rule, and some of the most reliable marks in the paper. Circular measure questions look intimidating because of their diagrams, but they decompose into the same few pieces every session.

Radians: the unit the formulas demand

A radian is the angle for which arc length equals radius; $\pi$ radians $= 180°$. Conversions: degrees $\times \frac{\pi}{180} \to$ radians; radians $\times \frac{180}{\pi} \to$ degrees. The exact-value angles come as the standard set, $\frac{\pi}{6} = 30°$, $\frac{\pi}{4} = 45°$, $\frac{\pi}{3} = 60°$, $\frac{\pi}{2} = 90°$, and on Paper 1 answers stay as exact multiples of $\pi$.

The non-negotiable: check the angle’s units before any computation. Mixing degrees into radian formulas is one of the most-reported errors in 0606, and it’s catchable in two seconds.

The two formulas

For radius $r$ and angle $\theta$ in radians:

• Arc length: $s = r\theta$
• Sector area: $A = \frac{1}{2}r^2\theta$

Both on the memorise list. Perimeter of a sector $=$ arc $+$ two radii: $s + 2r$, forgetting the two straight edges is a classic single-mark leak.

The real exam skill: composite regions

0606 doesn’t ask “find the arc length” for long. It draws a sector with a triangle removed, two touching circles, or a shaded segment, and asks for perimeter and area. The universal method:

1. Decompose the region into named pieces: sector(s), triangle(s), arc(s), straight edges. Annotate the diagram, labelled decomposition is visible method.
2. Compute each piece: sector pieces from the two formulas; triangle pieces from $\frac{1}{2}ab\sin C$ (the bridging formula this topic shares with trigonometry); straight edges from the geometry or coordinate tools.
3. Add or subtract according to the shading.

The keystone case, the segment (region between chord and arc):

Area of segment $=$ area of sector $-$ area of triangle $= \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta$ Perimeter of segment $=$ arc $+$ chord, with chord $= 2r\sin(\theta/2)$

Derive these from the decomposition rather than memorising blind, the derivation is the working the mark scheme wants to see.

Worked exam-style question

A sector $OAB$ has radius $8$ cm and angle $1.2$ radians. Find the perimeter and area of the segment cut off by the chord $AB$.

Arc $AB = r\theta = 8 \times 1.2 = 9.6$ cm (M, A) Chord $AB = 2 \times 8 \times \sin 0.6 = 16\sin 0.6 \approx 9.03$ cm (M, calculator in radian mode!) Perimeter $\approx 18.6$ cm (A) Segment area $= \frac{1}{2}r^2(\theta - \sin\theta) = \frac{1}{2} \times 64 \times (1.2 - \sin 1.2) \approx 32 \times 0.268 \approx$ $8.58$ cm$^2$ (M, A)

Note the mode discipline: $\sin 1.2$ with the calculator in degrees gives nonsense that looks plausible. Setting radian mode is part of the method.

Common mistakes in this topic

• Degrees fed into $s = r\theta$ or $A = \frac{1}{2}r^2\theta$
• Calculator left in degree mode for $\sin\theta$ with $\theta$ in radians
• Sector perimeter without the two radii
• Segment area attempted as sector minus wrong triangle (the triangle is $OAB$ with the included angle $\theta$, use $\frac{1}{2}r^2\sin\theta$)
• Rounding intermediate values, then the final answer drifting outside tolerance, carry precision, round once

Radian fluency built here is load-bearing for trig equations with radian ranges and for calculus, where trig derivatives assume radians. A small topic that underwrites two big ones.

Composite-region questions click fast with guided practice, usually one session. Free 1-hour trial with Teacher Rig: message us on WhatsApp.