Circular Measure

• Radians turn an angle into a ratio. Once the angle is a ratio, arc length is just \(r\) times it and sector area is just \(\tfrac12r^2\) times it. Every error in this topic is either a unit that was never converted or a boundary that was never traced.
• Every error in this chapter is silent. Nothing will tell you the angle was in degrees or the sine was in the wrong mode — only the habit of converting first, tracing the boundary, and comparing the answer against the whole circle.
• The radian is a ratio, not a unit. Count how many radii of arc the angle opens up: that count is the angle. Six radii and a bit go all the way round, which is why a full turn is \(2\pi\).
• One identity: \(180^\circ=\pi\). Into radians the number shrinks, so multiply by \(\pi/180\); into degrees it grows, so multiply by \(180/\pi\). Check every conversion against the right angle, \(90^\circ=\pi/2\).
• \(s=r\theta\) is the definition of a radian, rearranged. Convert the angle first, multiply second, and check the arc against the circumference last.
• \(A=\tfrac12r^2\theta\) is the fraction \(\theta/2\pi\) of \(\pi r^2\). If the arc is already known, \(A=\tfrac12rs\) is quicker. Test your memory of either at \(\theta=2\pi\): it must give \(\pi r^2\).
• A perimeter is traced, not quoted. \(2r+r\theta\) is what tracing an ordinary sector gives you; anything else needs tracing again. Internal edges belong to two regions and to no boundary.
• Segment \(=\) sector \(-\) triangle, and the triangle is the one on the two radii: \(\tfrac12r^2\theta-\tfrac12r^2\sin\theta=\tfrac12r^2(\theta-\sin\theta)\). Same \(\theta\), radian mode, and a perimeter of arc plus chord with no radius in it.

• Define the radian as the ratio of arc length to radius, explain why that makes it a pure number, and state that one radian is the angle for which the arc equals the radius. Section A
• State and use \(180^\circ=\pi\) radians to convert in both directions, keeping exact multiples of \(\pi\) when the question allows. Section B
• Use \(s=r\theta\) to find an arc length, and rearrange it to find \(r\) or \(\theta\) from the other two. Section C
• Use \(A=\tfrac12r^2\theta\) to find a sector area, and rearrange it to recover \(r\) or \(\theta\). Section D
• Find the perimeter of a sector as \(2r+r\theta\), and explain when that expression is not the perimeter of the region in front of you. Section E
• Derive and use \(A=\tfrac12r^2(\theta-\sin\theta)\) for a segment, and evaluate the sine in radian mode on the same angle. Section F
• Distinguish minor from major arcs, sectors and segments, and choose between the reflex angle and a subtraction from the whole circle. Section G
• Decompose a compound circular figure into sectors, triangles and segments, and trace only the exposed boundary when a perimeter is asked for. Section H
• Check every answer dimensionally, giving lengths in cm and areas in cm², and retain full precision until the final rounding. Comparisons
• Recognise and repair the standard errors of this topic, above all the degree-mode calculation that produces a plausible but wrong number. Mistake clinic

Two sentences that decide most of the answers in this topic. First: every formula on this page is false unless \(\theta\) is in radians — and none of them is printed in the examination formula list, so they must be recalled. Second: a perimeter is what the boundary of the region actually is, not what a formula says it usually is; in a compound figure you trace the edge, you do not quote \(2r+r\theta\) out of habit.

• Using \(s=r\theta\) or \(A=\tfrac12r^2\theta\) with \(\theta\) in degrees. Why it fails Both formulas were derived from the definition \(\theta=s/r\), which measures an angle in radii. A degree value is roughly 57 times too large, so the answer is roughly 57 times too big — and still looks like a length. Fix Convert first, on its own line, before either formula is written down.
• Taking a sine in degree mode on a radian angle. Why it fails \(\sin 1.2\) means the sine of \(1.2\) radians, which is \(0.932\). A calculator in degree mode returns \(\sin 1.2^\circ=0.0209\) — the sine is out by a factor of about \(45\). The segment answer that follows is not out by \(45\); it is out by rather less, because the sine sits inside a subtraction. For \(r=8\), \(\theta=1.2\) it returns \(37.7\ \mathrm{cm^2}\) instead of \(8.57\ \mathrm{cm^2}\), about four times too large; at other angles the factor is different again. What is constant is not the size of the error but its shape: the triangle almost vanishes, so the segment comes out as nearly the whole sector. Fix Set radian mode at the start of the topic. Two sanity checks: for \(0<\theta<\pi\), \(\sin\theta\) should be a healthy fraction, not something starting \(0.0\); and a segment should be a modest slice of its sector, never \(98\%\) of it.
• Quoting \(P=2r+r\theta\) as “the perimeter formula” for any shaded region. Why it fails That expression is the perimeter of one particular shape — a sector bounded by two full radii and an arc. A compound region may expose only part of a radius, or none of it, or extra straight edges the formula knows nothing about. Fix Put a finger on the boundary and walk all the way round it once. Add exactly what you traced.
• Subtracting the wrong triangle when finding a segment. Why it fails The triangle that must be removed from the sector is the one built on the two radii and the chord, with area \(\tfrac12r^2\sin\theta\). A right-angled triangle, or a triangle drawn to the chord’s midpoint, removes the wrong region. Fix Draw the triangle \(OAB\) explicitly before subtracting. Its two equal sides are radii; that is the check.
• Rounding \(\pi\), or an angle, partway through. Why it fails A segment is a small difference between two much larger quantities, so it is extremely sensitive to rounding. Using \(\pi=3.14\) and \(\theta\) to two decimals can move the third significant figure of the answer. Fix Keep everything on the calculator, in exact form or full precision, and round once, at the end.
• Using \(s=r\theta\) with \(\theta\) in degrees. Why it fails \(s=r\theta\) is the definition \(\theta=s/r\) rearranged, and that definition measures the angle in radii of arc. A degree is a different and much smaller unit, so a degree value is about \(57.3\) times the radian value and the arc comes out about \(57.3\) times too long. Fix Convert on a separate line first. Then check the arc against \(2\pi r\); an arc longer than the circumference is impossible.
• Using \(\tfrac12r^2\theta\) with \(\theta\) in degrees. Why it fails Same cause, same factor. The sector area comes out about \(57.3\) times too large, and will usually exceed \(\pi r^2\) — a region bigger than the circle containing it. Fix Convert first; then compare the answer with \(\pi r^2\).
• Using \(\dfrac{\pi r^2\theta}{360}\) while \(\theta\) has already been converted to radians. Why it fails This is the degree formula fed a radian value. The \(360\) in the denominator is only correct if the numerator angle is measured in degrees; with radians the correct denominator is \(2\pi\). The answer comes out about \(57.3\) times too small. Fix Choose one system before writing anything. In radians: \(\tfrac12r^2\theta\). In degrees: \(\dfrac{\theta}{360}\times\pi r^2\). Never a hybrid.
• Forgetting that \(180^\circ=\pi\) radians, and reaching instead for “\(360=\pi\)” or “\(90=\pi\)”. Why it fails \(\pi\) radians is a half turn, because the full turn is \(2\pi\) — which is itself a consequence of the circumference containing \(2\pi\) radii. Fix Anchor on the right angle instead, which is harder to misremember: \(90^\circ=\pi/2\). Doubling it gives \(180^\circ=\pi\).
• Converting degrees to radians by multiplying by \(\dfrac{180}{\pi}\). Why it fails That multiplier is about \(57.3\), so it makes the number bigger — but radians are the larger unit, so a given angle has fewer of them and the number must get smaller. Fix Test the multiplier on \(90^\circ\). It must produce \(\pi/2\approx1.57\), not \(5157\).
• Converting radians to degrees by multiplying by \(\dfrac{\pi}{180}\). Why it fails The mirror image of the previous entry: that multiplier is about \(0.0175\) and shrinks the number, whereas degrees are the smaller unit and there must be more of them. Fix Same test. \(\pi/2\) must come out as \(90\).
• Converting with the rounded constant \(57\) or \(57.3\). Why it fails \(180/\pi=57.29577\ldots\). Using \(57\) introduces a relative error of about \(0.5\%\), which lands squarely in the third significant figure — the one you are being asked to report. Fix Always convert with \(\pi\) itself, using the calculator’s \(\pi\) key rather than a typed decimal.
• Taking a sine in degree mode on a radian angle. Why it fails \(\sin1.2\) asks for the sine of \(1.2\) radians, roughly \(69^\circ\), which is \(0.932\). In degree mode the calculator returns \(\sin1.2^\circ=0.0209\), the sine of a nearly flat angle. Every segment computed from it is wrong by a wide margin. Fix Set RAD at the start of the topic. Recognise the signature: a sine near \(0.02\) for an angle that ought to be substantial.
• Rounding \(\pi\) to \(3.14\), or an intermediate angle to two decimals, and carrying that forward. Why it fails A segment is a small difference between two much larger quantities, so it magnifies rounding error. In Worked example F1, rounding \(\sin1.2\) to \(0.93\) changes the answer from \(8.57\) to \(8.64\) — wrong in the second significant figure. Fix Keep every intermediate value on the calculator or in exact form. Round once, at the final line.
• Writing \(\sin\theta\) as \(\sin\times\theta\), or trying to “cancel” the \(\theta\) in \(\theta-\sin\theta\). Why it fails \(\sin\) is not a quantity; it is an instruction that turns one number into another. \(\theta-\sin\theta\) is a difference of two numbers with no common factor to remove, and it does not simplify. Fix Evaluate \(\sin\theta\) to a number first, then subtract.
• Confusing an arc with a chord. Why it fails They join the same two points but by different routes: the arc curves along the circle (\(r\theta\)), the chord goes straight across (\(2r\sin\tfrac{\theta}{2}\)). They are never equal, and the arc is always the longer. Fix Use the comparison as a check: any answer in which the chord exceeds the arc is wrong.
• Confusing a sector with a segment. Why it fails A sector is bounded by two radii and an arc; a segment by a chord and an arc. For a minor region the segment is the smaller of the two, because it is the minor sector with the triangle \(OAB\) removed. Be careful not to over-extend that: on the major side the relationship reverses, because the major segment is the major sector plus that same triangle. A major segment is always larger than the major sector it stands on. Fix Ask what the straight edges are. Two radii means sector; one chord means segment. Then ask which side of the chord the region is on, because that decides whether the triangle is added or removed.
• Confusing minor with major, or using the minor angle for a requested major region. Why it fails The major region stands on the reflex angle \(2\pi-\theta\), not on \(\theta\). Using \(\theta\) computes the minor region and simply relabels it. Fix Compute the minor region and subtract it from \(\pi r^2\). A “major” answer below half the disc, or a “minor” answer above it, is wrong.
• Subtracting the wrong triangle when finding a segment. Why it fails The triangle that leaves a segment behind is \(OAB\), on the two radii and the chord, with area \(\tfrac12r^2\sin\theta\). A right-angled half-triangle, or one drawn to the midpoint of the chord, removes a different region entirely. Fix Draw \(OAB\) before subtracting and confirm both of its equal sides are radii.
• Finding a major segment as “major sector minus triangle”. Why it fails The triangle lies on the minor side of the chord, outside the major sector’s missing piece. To reach the major segment the triangle must be added to the major sector, not removed. Fix Use \(\pi r^2\) minus the minor segment, where no sign decision arises at all.
• Omitting the two radii from a sector perimeter. Why it fails The arc is one of three boundary pieces, not all of them. It feels like the answer because it is the piece that required a formula. Fix Count the pieces of the outline, then count the terms in your sum. They must match.
• Including an internal construction line in a perimeter. Why it fails An edge shared by two parts of one region lies inside it, so nothing outside the region ever touches it. Two adjoining sectors expose two radii between them, not three. Fix Trace the outline with a finger. If your finger does not travel along a line, that line is not in the sum.
• Using the full radius as a straight edge in a region between concentric arcs. Why it fails Only the part of each radius outside the inner arc is exposed; the inner \(r\) centimetres lie within the hole. The exposed piece is \(R-r\). Fix Mark both radii on the diagram and label the exposed piece with its own value before adding.
• Giving an area in cm, or a length in cm². Why it fails The unit follows the power of \(r\): one \(r\) is a length, two is an area. A mismatch means the formula itself was the wrong one, not merely that the label slipped. Fix Glance at the formula before evaluating and count the \(r\)s.
• Substituting the diameter where the radius belongs. Why it fails Every formula in this chapter is written in \(r\). Using \(d\) doubles every length and quadruples every area, and because later parts of a question reuse earlier answers, one slip corrupts all of them. Fix Write \(r=d/2\) as its own line whenever a diameter is given.
• Measuring an angle or a length off the printed diagram. Why it fails A diagram is there to tell you how the regions fit together, not how big they are, and where a figure is not to scale the paper says so. A value measured off the page is never data — only the stated radii and angles are. Fix Use the diagram to identify the regions and the labels to compute them. If the two disagree, the labels win.
• Calling a rounded decimal “exact”. Why it fails An exact answer keeps \(\pi\) and surds intact. \(5\pi\) cm is exact; \(15.7\) cm is a 3-significant-figure approximation to it, and a question asking for an exact answer is not answered by a decimal, however accurate. Fix Read the instruction. If it says exact or in terms of \(\pi\), no decimal appears in the final line.

• Write the decomposition down. A student who writes “shaded \(=\) sector \(-\) triangle” and then makes an arithmetic error has still shown a correct method; one who writes only a final number that happens to be wrong has shown nothing at all. In a topic where every step is one line long, the temptation to do the whole thing on the calculator and write only the answer is strong — and on Paper 2 it is a bad habit, because it hides every part of the reasoning that the arithmetic did not touch.
• What these three have in common. None of them can be started by choosing a formula. Each begins by asking which quantities are known and which formula contains exactly those — \(r\theta\) appearing in both equations in W1, the arc appearing in both givens in W3. Reading the question for the shared block, rather than for the shape, is what turns a forbidding problem into a two-line one.
• How much of a solution is method. Look at how the steps above divide. In questions 3, 4 and 5 the allocation puts three of its five points on method — naming the decomposition, choosing the formula, writing the substitution — and only two on the values those steps produce. That proportion is the shape of a circular-measure solution, and it is the reason for writing the substitution line down: a solution that shows its method and then slips in the arithmetic has still communicated the mathematics, while one that shows a single wrong number has communicated nothing. The labels are Academiq’s rather than Cambridge’s, but the proportion is the point.
• The one-line reminder to carry forward. Write it on the inside cover of your notes: radians, then radius, then trace the boundary. Those three steps in that order head off the angle-unit and boundary errors that make up most of the mistake clinic, and none of them requires remembering a formula.

• Cambridge IGCSE Additional Mathematics 0606 is assessed by two written papers of equal weight, taken by every candidate, and circular measure can appear in either. They differ in one respect that matters more in this topic than in almost any other: Paper 1 is a non-calculator paper, and Paper 2 requires a scientific calculator. That single difference splits the topic in two, and the split is the first thing to get straight.
• Two papers, one topic, two habits. For Paper 2 the habit is set radian mode and leave it there. For Paper 1 there is no mode to set, so the habit is the opposite one: never evaluate anything you could carry. Use the rule rather than a list: a worked example on this page whose answer keeps \(\pi\) or a surd is Paper 1 work, and one that evaluates a sine which is not a special value, or needs \(\pi\) as a decimal, is Paper 2 work. Several sit in neither camp — decimal angles, no sine, arithmetic you can do by hand either way. The mixed challenge is split the same way and labels each question with the paper it belongs to.
• A circular-measure question almost always arrives as a diagram with a shaded region and a short list of given lengths and angles. The work is then in three layers, and marks are distributed across all three:
• Layer 1 · Read Identify the centre, the radius, the central angle, and its unit. Convert if required. This layer is worth very little on its own and decides everything after it.
• Layer 2 · Decompose Say, in writing, what the shaded region is made of: “sector \(OAB\) minus triangle \(OAB\)”, or “big sector minus small sector”. This sentence is the method; without it the numbers that follow have nothing to explain them.
• Layer 3 · Compute Substitute, evaluate at full precision, round once at the end, and attach the correct unit.