Cambridge A Level Mathematics 9709 · Pure 1

Circular measure

Circular measure uses radians instead of degrees, and with the angle in radians the arc length of a sector is $s = r\theta$ and its area is $A = \tfrac{1}{2} r^2 \theta$ . One radian is the angle subtended at the centre by an arc equal in length to the radius, so a full turn is $2\pi$ radians and $180^\circ = \pi$ . Cambridge 9709 Pure 1 section 1.4 has appeared on every one of the 40 papers Quanta has mapped, for 273 marks — about 9% of the total — and almost every question is the same task: a shape built from sectors and triangles, and a shaded region whose area or perimeter you must assemble from the parts.

Updated 15 September 2026

Radians

\pi \text{ rad} = 180^\circ \qquad 1 \text{ rad} = \frac{180^\circ}{\pi} \approx 57.3^\circ \qquad \theta^\circ = \theta \times \frac{\pi}{180} \text{ rad}

The exact values worth knowing without conversion: $\tfrac{\pi}{6} = 30^\circ$ , $\tfrac{\pi}{4} = 45^\circ$ , $\tfrac{\pi}{3} = 60^\circ$ , $\tfrac{\pi}{2} = 90^\circ$ . A P1 question gives angles in radians and expects answers in radians; if an angle appears as a decimal such as $1.2$ , it is radians. Set the calculator to radian mode at the start of the paper and leave it there.

Arc length and sector area

s = r\theta \qquad A = \tfrac{1}{2} r^2\theta \qquad (\theta \text{ in radians})

Both are fractions of the whole circle: the sector is $\theta / 2\pi$ of it, so its arc is that fraction of $2\pi r$ and its area that fraction of $\pi r^2$ . That is why the formulae only work in radians — in degrees the fraction would be $\theta/360$ and the $\pi$ would not cancel.

The perimeter of a sector is the arc plus two radii: $r\theta + 2r$ . Forgetting the radii is the single most frequent perimeter error.

Segments

A segment is the region between a chord and its arc: a sector minus the triangle formed by the two radii and the chord. With $\tfrac{1}{2}ab\sin C$ for the triangle,

A_{\text{segment}} = \tfrac{1}{2}r^2\theta - \tfrac{1}{2}r^2\sin\theta = \tfrac{1}{2}r^2(\theta - \sin\theta)

The chord itself, when a perimeter is asked, comes from the cosine rule or from splitting the isosceles triangle: $\text{chord} = 2r\sin\tfrac{\theta}{2}$ .

The shapes examiners build

Sector with a tangent. The tangent at the end of one radius meets the other radius extended. The tangent is perpendicular to the radius, so a right-angled triangle appears with $\tan\theta$ and $\cos\theta$ giving its sides.
Two sectors sharing a chord (two overlapping circles). The overlap is two segments; the angle in each is found from the isosceles triangle on the common chord.
Sector inside a triangle, or triangle inside a sector. The shaded region is one minus the other; the perimeter mixes an arc with straight edges.
Semicircle or quarter-circle on a rectangle. $\theta = \pi$ or $\tfrac{\pi}{2}$ .

In every case the method is the same: name each piece, write its area or its boundary length with the formula, and add or subtract. Write the plan before the numbers — “shaded = triangle OAC − sector OAB” earns the method mark even if a value slips.

Worked example

OAB is a sector of a circle, centre O, radius 8 cm, with angle AOB = 1.2 radians. The tangent to the circle at A meets OB extended at C. Find (i) the area and (ii) the perimeter of the region bounded by AC, CB and the arc AB.

Plan: the region is triangle OAC minus sector OAB. The tangent at $A$ is perpendicular to $OA$ , so triangle OAC has a right angle at $A$ .

(i) Area. In triangle OAC, $AC = 8\tan 1.2 = 20.58$ cm, so

\text{triangle } OAC = \tfrac{1}{2} \times 8 \times 8\tan 1.2 = 82.31 \text{ cm}^2

\text{sector } OAB = \tfrac{1}{2} \times 8^2 \times 1.2 = 38.4 \text{ cm}^2

\text{shaded area} = 82.31 - 38.4 = 43.9 \text{ cm}^2 \ (3 \text{ s.f.})

(ii) Perimeter. Three pieces: $AC$ , $CB$ and the arc $AB$ . $OC = \dfrac{8}{\cos 1.2} = 22.08$ cm, so $CB = OC - OB = 22.08 - 8 = 14.08$ cm. The arc is $8 \times 1.2 = 9.6$ cm.

\text{perimeter} = 20.58 + 14.08 + 9.6 = 44.3 \text{ cm} \ (3 \text{ s.f.})

Both answers keep four figures in the working and round only at the end — rounding $AC$ to 20.6 first shifts the perimeter to 44.28 and risks the accuracy mark.

Common mistakes

1.Calculator in degree mode
$\sin 1.2$ in degree mode is 0.0209; in radian mode it is 0.932. Every trig value in the question is wrong and the error is invisible until the answer is. Check the mode indicator before the first question.
2.Using degrees in s = rθ
The formulae need radians. If an angle is given in degrees, convert with $\times \pi/180$ first.
3.Perimeter without the radii or the chord
A perimeter is every edge of the region. List the pieces before adding: arcs, radii, chords, tangents.
4.Area of the wrong triangle
$\tfrac{1}{2}ab\sin C$ needs the angle between the two sides used. In the tangent shape the right angle is at the point of contact, not at the centre.
5.Rounding mid-working
Carry at least four significant figures through and round the final answer to three. Store intermediate values on the calculator.

Common questions

How do you convert degrees to radians?

Multiply by $\pi/180$ . So $60^\circ = 60 \times \pi/180 = \pi/3$ . To convert radians to degrees, multiply by $180/\pi$ .

What is the formula for the area of a sector?

$A = \tfrac{1}{2}r^2\theta$ with $\theta$ in radians. Arc length is $s = r\theta$ . Both come from the sector being $\theta/2\pi$ of the full circle.

How do you find the area of a segment?

Sector minus triangle: $\tfrac{1}{2}r^2(\theta - \sin\theta)$ , where $\theta$ is the angle at the centre in radians.

Why does Pure 1 use radians instead of degrees?

Because the arc and area formulae only take their simple forms in radians, and because the calculus of trigonometric functions — the derivative of $\sin x$ being $\cos x$ — is only true when $x$ is in radians. Pure 3 depends on it.

Practise circular measure against real mark schemes

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