Circular measure

• Arc equals radius: that is one radian. And the bracket in MF19 is part of the formula — \(s = r\theta\) and \(A = \tfrac12 r^2\theta\) only with \(\theta\) in radians.
• Arc equals radius is one radian; s = rθ and A = ½r²θ with θ in radians; segment = sector − triangle = ½r²(θ − sin θ), built, never looked up.

• 1.4.1 I can understand — understand the definition of a radian, and use the relationship between radians and degrees
• 1.4.2 I can use — use the formulae s = rθ and A = ½r²θ in solving problems concerning the arc length and sector area of a circle

Accuracy on this chapter. Exact where the question says exact: in terms of \(\pi\) and surds, such as \(\tfrac{50\pi}{3} - 25\sqrt3\). Otherwise lengths and areas to 3 significant figures, angles in radians to 3 significant figures and angles in degrees to 1 decimal place. Keep the angle and the trigonometric values unrounded through the working — 1.591206, 0.999792, 3.62358 — and round once, at the end.

• “\(r = 5\) and \(\theta = 60^\circ\), so \(s = r\theta = 300\).” Correct \(\theta\) in radians in \(s = r\theta\) and \(A = \tfrac12 r^2\theta\), every time. MF19 prints the condition in a bracket after each formula because it is part of the formula. Convert first: \(60^\circ = \tfrac{\pi}{3}\), so \(s = \tfrac{5\pi}{3}\).
• “The segment formula \(\tfrac12 r^2(\theta - \sin\theta)\) is on the formula sheet.” Correct The segment formula is not in MF19: it is sector minus triangle. MF19 gives only \(r\theta\) and \(\tfrac12 r^2\theta\). Write \(\tfrac12 r^2\theta - \tfrac12 r^2\sin\theta\) as the method line and the segment builds itself.
• “\(\sin 1.2 = 0.0209\).” Correct That is \(\sin 1.2^\circ\), from a calculator left in degree mode. \(\sin(1.2 \text{ rad}) = 0.9320\). Switch to radian mode before evaluating any trigonometric function of an angle in radians.
• “A radian is \(\dfrac{180}{\pi}\).” Correct A radian is an angle: the angle subtended at the centre of a circle by an arc equal in length to the radius. \(\dfrac{180}{\pi} \approx 57.3\) is how many degrees it contains.
• “Perimeter of the sector \(= r\theta\).” Correct \(r\theta\) is the arc only. A sector's perimeter adds the two radii: \(2r + r\theta\). A segment's perimeter adds the chord instead: \(r\theta + 2r\sin\tfrac{\theta}{2}\).
• “\(\theta = 1.591206\), call it 1.6.” Correct A segment is a small difference of two larger numbers, so early rounding is magnified: with 1.6 the segment in worked example 4 comes out as 14.7 instead of 14.5. Carry the unrounded angle; round once, at the end.
• “A radian is \(\dfrac{180}{\pi}\).” Repair A radian is an angle: the angle subtended at the centre of a circle by an arc equal in length to the radius. \(\dfrac{180}{\pi} \approx 57.3\) is how many degrees it contains.
• “\(150^\circ = 150\pi\) rad.” Repair Multiply by \(\dfrac{\pi}{180}\), not by \(\pi\): \(150^\circ = \dfrac{150\pi}{180} = \dfrac{5\pi}{6}\) rad.
• “\(s = r\theta\) with \(r = 5\), \(\theta = 60^\circ\), so \(s = 300\).” Repair \(\theta\) must be in radians: \(60^\circ = \tfrac{\pi}{3}\), so \(s = \tfrac{5\pi}{3}\). The answer 300 is longer than the whole circumference, \(10\pi \approx 31.4\).
• “\(A = \tfrac12 r^2\theta = \tfrac12 \times 6 \times 1.2 = 3.6\).” Repair \(r\) is squared: \(A = \tfrac12 \times 36 \times 1.2 = 21.6\).
• “\(\sin 1.2 = 0.0209\).” Repair That is \(\sin 1.2^\circ\), from a calculator in degree mode. \(\sin(1.2 \text{ rad}) = 0.9320\).
• “Perimeter of the sector \(= r\theta\).” Repair That is the arc; the perimeter adds the two radii: \(2r + r\theta\).
• “Area of the segment \(= \tfrac12 r^2\theta - \tfrac12 r^2\theta\).” Repair That is zero. The triangle's area is \(\tfrac12 r^2\sin\theta\), from \(\tfrac12 ab\sin C\), so the segment is \(\tfrac12 r^2(\theta - \sin\theta)\).
• “The segment formula \(\tfrac12 r^2(\theta - \sin\theta)\) is in MF19.” Repair MF19 gives only \(r\theta\) and \(\tfrac12 r^2\theta\). Build the segment as sector minus triangle, and write both pieces.
• “Chord \(AB = r\theta\).” Repair \(r\theta\) is the arc, which is always longer than the chord. The chord is the straight line: \(2r\sin\tfrac{\theta}{2}\), or from the cosine rule.
• “Perimeter of the segment \(= r\theta + 2r\).” Repair The segment's straight edge is the chord, not two radii: \(r\theta + 2r\sin\tfrac{\theta}{2}\).
• “\(\theta = 1.591206 \approx 1.6\), so the segment is \(24.5(1.6 - \sin 1.6) = 14.7\).” Repair Keep full precision: \(24.5(1.591206 - 0.999792) = 14.49\), so 14.5. A segment is a small difference of larger numbers, so early rounding is magnified.
• “Exact area \(= 12.3\) cm².” Repair Exact means in terms of \(\pi\) and surds: \(18\sqrt3 - 6\pi\). The decimal may follow it, but it cannot stand in for it.
• “Perimeter of the band \(= 7.2 + 4.5 + 5 + 5 + 3 + 3\).” Repair The inner radii are not edges of the band between two arcs. Its edges are the two arcs and the two straight pieces of length \(R - r\): \(7.2 + 4.5 + 3 + 3 = 17.7\).

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• The segment is on the arc side of the chord. The region between the chord and the centre is the triangle. A sketch that shades the triangle and calls it the segment leads straight to subtracting the wrong way round; check that your shaded region touches the arc.
• Interleave with the chapters that use this one. Chapter 5 (trigonometry) uses radians and exact values from the first page: when you reach it, re-answer retrieval questions 2–5. Chapter 14 (Paper 3) solves equations such as 2θ = 3 sin θ: re-do worked example 8 and then solve its equation there. Chapter 12 differentiates sin x, which needs radians: re-answer retrieval question 17. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: Paper 1 is assumed knowledge for every other paper, and an individual examination question may involve ideas and methods from more than one section of that paper’s content, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 1 content, examined in Paper 1. Paper 1 (Pure Mathematics 1) is compulsory for both the AS Level and the A Level: it is 60% of the AS Level and 30% of the A Level, and its content is assumed knowledge for every other paper. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• Paper 1 questions are structured, so a circular-measure question can be set on a printed diagram — a sector, a segment, a tangent, two arcs — and asked in parts that build on each other: an angle (for example “show that angle AOB \(= \ldots\)”, from a triangle), then a length or a perimeter, then an area. A show that part may instead ask you to derive an equation in \(\theta\) from an area or perimeter condition. The reasoning is carried by naming the region as sector, triangle and segment pieces before any number is written.
• MF19's Mensuration list prints “Arc length of circle \(= r\theta\) (\(\theta\) in radians)” and “Area of sector of circle \(= \tfrac12 r^2\theta\) (\(\theta\) in radians)”. It does not print the radian–degree relationship, \(\tfrac12 ab\sin C\), the cosine or sine rule, Pythagoras or the segment area: those you must know. See the MF19 card.
• Calculator in radian mode whenever \(\sin\theta\), \(\cos\theta\) or \(\tan\theta\) is evaluated with \(\theta\) in radians. “Exact” means in terms of \(\pi\) and surds (\(18\sqrt3 - 6\pi\), not 12.3). Otherwise carry the unrounded angle (1.591206, not 1.6) and round once: lengths and areas to 3 significant figures, angles in radians to 3 significant figures, angles in degrees to 1 decimal place. Write the method line — the cosine rule with numbers in, the sector minus the triangle — before the value.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.