Circular measure

📚The mathematics

What a radian is

One radian is the angle at the centre of a circle subtended by an arc equal in length to the radius. A full turn is an arc of \( 2\pi r \), so it is \( 2\pi \) radians:

Arcs, sectors, chords and segments

With \( \theta \) in radians:

The syllabus states that these formulas are not given. The last two are not new formulas at all: the chord comes from splitting the isosceles triangle \(OAB\) into two right-angled triangles, and the segment is the sector minus the triangle, whose area is \( \tfrac{1}{2}ab\sin C \) (that one is in the List of formulas). The chord can also come from the cosine rule.

Your calculator must be in radian mode for \( \sin\theta \) when \( \theta \) is in radians. \( \sin 1.2 \) in degree mode is 0.0209, not 0.932, and it is the most common way to lose every accuracy mark in a Paper 2 sector question.

Compound shapes

Break the shape into sectors, triangles and rectangles, and write the plan down before calculating. A perimeter only includes the edges on the outside: arcs and the straight edges that are actually boundaries, not the radii that are hidden inside the shape.

✏️Worked example (calculator)

In this question, lengths are in centimetres and angles are in radians. \(OAB\) is a sector of a circle with centre \(O\) and radius 8. Angle \( AOB = 1.2 \). (a) Find the perimeter of the sector. [2] (b) Find the area of the segment between the chord \(AB\) and the arc \(AB\). [3] (c) Find the perimeter of the segment. [2]

(a) Arc \( = 8 \times 1.2 = 9.6 \). Perimeter \( = 8 + 8 + 9.6 = 25.6 \) cm.

(b) Sector \( = \tfrac{1}{2}(8^{2})(1.2) = 38.4 \); triangle \( = \tfrac{1}{2}(8^{2})\sin 1.2 = 29.825\ldots \). Segment \( = 38.4 - 29.825 = 8.57 \) cm².

(c) Chord \( AB = 2(8)\sin 0.6 = 9.034\ldots \). Perimeter of the segment \( = 9.6 + 9.034 = 18.6 \) cm.

Check it. The chord must be shorter than the arc it cuts off: \( 9.03 < 9.6 \) ✓. And the segment should be a small fraction of the sector: \( 8.57 / 38.4 \approx 22\% \), plausible for an angle of about \( 69^{\circ} \).

Rounding too early. Using \( \sin 1.2 \approx 0.93 \) gives a triangle of 29.76 and a segment of 8.64 — wrong at 3 s.f. Keep the full calculator values until the final line.

📝Practise

Written in the style of the current Paper 1 (no calculator) and Paper 2 (calculator) questions.

1. (No calculator.) Convert \( 150^{\circ} \) to radians, and \( 0.4\pi \) radians to degrees. [2]

\( 150 \times \dfrac{\pi}{180} = \dfrac{5\pi}{6} \). \( 0.4\pi \times \dfrac{180}{\pi} = 72^{\circ} \).

2. (No calculator.) A sector has angle \( \tfrac{\pi}{3} \) and area \( 6\pi \) cm². Find its radius and its exact perimeter. [4]

\( \tfrac{1}{2}r^{2}\left(\tfrac{\pi}{3}\right) = 6\pi \Rightarrow r^{2} = 36 \Rightarrow r = 6 \). Arc \( = 6 \times \tfrac{\pi}{3} = 2\pi \). Perimeter \( = 12 + 2\pi \) cm.

3. (No calculator.) A chord of a circle of radius 6 cm subtends an angle of \( \tfrac{2\pi}{3} \) at the centre. Find the exact area of the minor segment. [4]

Sector \( = \tfrac{1}{2}(36)\left(\tfrac{2\pi}{3}\right) = 12\pi \). Triangle \( = \tfrac{1}{2}(36)\sin\tfrac{2\pi}{3} = 18 \times \tfrac{\sqrt{3}}{2} = 9\sqrt{3} \). Segment \( = 12\pi - 9\sqrt{3} \) cm².

4. (Calculator.) A sector of radius 8 cm has perimeter 30 cm. Find the angle of the sector and its area. [3]

Arc \( = 30 - 16 = 14 \), so \( \theta = \tfrac{14}{8} = 1.75 \) rad. Area \( = \tfrac{1}{2}(64)(1.75) = 56 \) cm².

5. (Calculator.) \(ABC\) is an equilateral triangle of side 10 cm. Three arcs are drawn: arc \(BC\) with centre \(A\), arc \(CA\) with centre \(B\) and arc \(AB\) with centre \(C\), each of radius 10 cm. Find the perimeter and the area of the curved shape they enclose. [6]

Each arc: \( 10 \times \tfrac{\pi}{3} \), so the perimeter is \( 10\pi = 31.4 \) cm. The area is the triangle plus three segments. Triangle \( = \tfrac{1}{2}(100)\sin\tfrac{\pi}{3} = 25\sqrt{3} \). Each segment \( = \tfrac{1}{2}(100)\left(\tfrac{\pi}{3} - \tfrac{\sqrt{3}}{2}\right) \). Total \( = 25\sqrt{3} + 150\left(\tfrac{\pi}{3} - \tfrac{\sqrt{3}}{2}\right) = 50\pi - 50\sqrt{3} = 70.5 \) cm².

6. (Calculator.) A sector of radius 5 cm has area 15 cm². Find the length of the chord joining the ends of its arc. [3]

\( \tfrac{1}{2}(25)\theta = 15 \Rightarrow \theta = 1.2 \). Chord \( = 2(5)\sin 0.6 = 5.65 \) cm.