Major and minor arcs, and why an arc's central angle is double what it subtends on the circle (Theorem 9)

What to assume they know

• Chords of equal length cut off equal central angles, and the converse (Theorems 2–3) — chords, the central angle, and the isosceles triangle a chord makes with the centre
• Turning an observation into a definition: the circle as a locus — circle, centre, radius, chord
• The exterior-angle theorem: an exterior angle of a triangle equals the sum of the two interior angles not adjacent to it
• That a triangle with two equal sides has two equal angles
• Reflex angles, and that a complete turn is 360°
• Measuring an angle with a protractor, including one past 180°

What they should be able to do

• Say what an arc is, and identify the two arcs a given pair of points determines
• Distinguish major and minor arc on a drawn circle, and use the printed three-letter naming that says which arc is meant
• Explain the chapter's swept-angle definition of the angle an arc makes at the centre, and use it to give a reading above 180° for a major arc
• Classify an arc as major or minor from its central angle, using 180° as the dividing line
• Say what it means for an arc to make an angle at a point of the circle beyond that arc, and mark such a point correctly on a figure
• Prove the doubling relation for the case where the auxiliary line through the centre meets the circle on the arc
• Rework the proof for the case where that line meets the circle off the arc, and say which sums become differences
• Explain why the swept-angle definition is what lets one statement serve both arcs

Where it usually goes wrong

• "An arc is named by its two ends, so arc AB is unambiguous." It is not — there are two arcs on A and B. The chapter's three-letter naming, with a middle letter taken from the arc itself, exists precisely to disambiguate, and a student who drops the middle letter cannot state Theorem 9 correctly.
• "An angle at a centre cannot be more than 180°." Under the chapter's swept-angle definition it certainly can, and for a major arc it must be. This is the single hardest idea in the section and the reason Fig. 5.18 is colour-coded.
• "The bigger arc has the bigger angle at the circle." It has the bigger angle at the centre. Its angle at a point of the circle off it is bigger too, but the two facts have to be kept apart, because the point in question moves to the other arc when you switch.
• "Half of a reflex angle is still reflex." Half of 290° is 145°, which is obtuse but not reflex. Students expect the halving to preserve the category.
• "Theorem 9 needs two cases because there are two arcs." No — the two cases are about where the line from D through the centre happens to re-cut the circle. The arc choice is handled once, by the definition.
• "The point where the angle is measured can be anywhere on the circle." It has to be off the arc in question. Put it on the arc and the configuration is different, and the theorem as stated does not apply.
• "Measuring three angles in the activity and finding them equal proves it." The chapter says outright at the top of p. 103 that many examples do not settle a claim, and then goes on to prove this one. The activity is there so the student has something to be surprised by.

Questions to check understanding

• Given a central angle, state the angle the arc makes at a point of the circle off it — the form End-of-Chapter Q2, p. 114 takes
• Given the angle at a point of the circle, find the central angle
• Classify a named arc as major or minor from its central angle
• Name an arc unambiguously using three letters, given a lettered figure
• Prove the doubling relation, for the case where the line through the centre meets the circle on the arc
• Rework the proof for the other case
• Measure the angles a drawn arc makes at the centre and at three points off it, and say what is observed
• Find the two readings a pair of points gives — one from each arc — and check that the two point-on-circle readings add to 180°

Examples worth working on the board

The chapter prints no answers, so anything marked verified is an added measurement or working, not the book's.

• Fig. 5.17 (p. 106; the caption names both arcs). Circle with centre O. A sits low on the left, B higher on the left, X on the left between them, Y at the top. So going from A to B the short way passes X, and the long way passes Y. The caption fixes the naming convention the rest of the section depends on: the middle letter says which route you mean.
• Fig. 5.18 (p. 106, read as the printed page). This is the figure the section turns on and it is colour-coded, which extraction cannot show. Circle with centre O; B on the left and A on the right, both a little above centre; X on the upper left arc, Y at the bottom. The arc running from A over the top through X to B is drawn green, with an arrowhead on it showing the direction of travel; the arc running from A down the right side through Y to B is drawn purple, again with an arrowhead. The radii OA and OB are black. At O sits a small shaded disc split the same two ways — green above, purple below — so the wedge colour tells you which arc's sweep produced it. The teacher must keep the two-colour coding; drawn in one colour the figure says nothing.
• Fig. 5.19 and its Exercise (p. 107). Circle with centre O, and a shaded wedge drawn at O. Six points are lettered on the circle: C near the top, B below C and to its left, K on the left, A low on the left, D at the bottom, L on the right. Only four segments run from O, to A, B, C and D. K and L are bare dots with nothing drawn to them — they are waypoints that name the two arcs, and drawing radii to them would bury what the exercise asks. The question asks for the angles that arc AKB and arc CLD make at O, states the rule that a reading under 180° means a minor arc and over 180° a major arc, and asks for each arc to be classified. Measured by me off the printed page, to a few degrees: arc AKB makes about 98° at O, so it is a minor arc; arc CLD makes about 204°, so it is a major arc. The figure is built so that one of each turns up — say that, because a student who gets two minor arcs has mis-traced a route.
• Fig. 5.20 and the Activity (p. 107). Circle with centre O; A at the top, K on the upper right, B on the right, and P, Q, R spaced round the rest — P upper left, Q left, R at the bottom. The chord AB and the radii to A and B are solid; the six segments from P, Q and R out to A and B are drawn dashed. The instruction is to measure the angle arc AKB makes at O, then measure the angles it makes at P, Q and R, and notice what happens. Verified, as a consistent set the explanation can use: if the central reading is 100°, the readings at P, Q and R are each 50°. Choose the numbers, but keep them consistent — the whole point is that the three agree.
• Fig. 5.21, Theorem 9's first case (p. 108). Circle with centre C. A at the top left; E then F on the upper right arc; B on the right; D at the lower left, the point off the arc. The arc in question is AFB. The construction is to join D to C and carry the line on until it cuts the circle again, and in this case it cuts at E, which lies on arc AFB. Given: the angle ACB is what arc AFB makes at the centre. To show: the angle at C is twice the angle at D, i.e. angle BCA = 2 × angle BDA.
• The first-case argument, as data. CB = CD, both radii, so triangle DCB is isosceles and its two base angles are equal. The angle BCE is exterior to triangle BCD, so by the exterior-angle theorem it equals the sum of those two equal base angles — that is, twice the angle BDC. Symmetrically, CA = CD makes triangle ADC isosceles and the angle ACE comes out as twice the angle CDA. Now the angle BCA splits as BCE plus ECA, and the angle BDA splits as BDE plus EDA; adding the two doubled pieces gives the result. The doubling happens in the exterior-angle step, and it happens twice.
• Fig. 5.22, Theorem 9's second case (p. 109). Same circle and centre C, but D has moved: A is at the top, F on the upper right, B on the right, D on the left, and the line from D through C now cuts the circle at E on the lower right, which is off arc AFB. Given and goal are as before. The argument changes shape: the two doubled angles ACE and BCE are still obtained the same way, but the angle ACB is now their difference rather than their sum, and likewise the angle ADB is the difference of the two angles at D. The doubling survives subtraction because both terms carry the same factor.
• A numerical spine for the whole topic (not in the book; the chapter prints no such example). Take a minor arc making 70° at the centre. Verified: it makes 35° at every point of the circle off it, and the major arc on the same two ends makes 290° at the centre and 145° at every point of the circle off it. Note that 35° and 145° add to 180°, which is the fact §5.8 will use for cyclic quadrilaterals — and it is invisible unless the major arc's reading is allowed to exceed 180°. This 70° figure is exactly the form End-of-Chapter Q2 on p. 114 takes.

Figures to have open

• Fig. 5.18 redrawn in two colours with sweep arrows and the split wedge at the centre. This is the chapter's own figure (p. 106) and it is the only place the swept-angle definition is made visible. A monochrome redraw destroys it.
• Fig. 5.17 redrawn: one circle, two points, the two routes traced separately. Chapter's own figure (p. 106).
• Fig. 5.19 redrawn with all six lettered points and both arcs traced — but with four segments from O only, to A, B, C and D; K and L stay bare. Chapter's own figure (p. 107).
• Fig. 5.20 redrawn keeping the solid/dashed distinction — chord and radii solid, the six viewing segments dashed. Chapter's own figure (p. 107); extraction gives no hint that any line is dashed.
• Figs. 5.21 and 5.22 redrawn side by side so the reader can see that only D has moved and only the sum has become a difference. These are two figures on two pages in the book (pp. 108, 109) and the comparison is the argument of section 11.
• A protractor graphic that reads past 180°. Standard schematic; the chapter states the rule in words at the Fig. 5.19 exercise and draws no scale.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 5, "I’m Up and Down, and Round and Round", §5.7, printed heading "Angles Subtended by an Arc", pp. 106–107, and §5.7.1, whose printed heading is the numbered line beginning "Angle subtended by an arc at a point on the circle" and continuing onto a second line, pp. 107–109.
• Figs. 5.17 and 5.18, p. 106. Fig. 5.19 with its Exercise, and Fig. 5.20 with its Activity, p. 107. Fig. 5.21, p. 108. Fig. 5.22, p. 109.
• Theorem 9 with its first-case argument, p. 108; the second-case argument, p. 109.
• Exercise Set 5.6, Q1, p. 110. End-of-Chapter Exercises Q2, p. 114.
• The chapter's statement that examples do not establish a general claim, p. 103.
• Chapter Summary, p. 117 — the doubling bullet, phrased there in terms of the remaining part of the circle rather than of an arc.