Chapter 2: Lines and Angles (Geometry)

• Rihan (one point): he can draw countless (infinitely many) lines, in every direction, through a single point.
• Sheetal (two points): she can draw exactly one line through two given points.

Through one point: infinitely many lines. Through two points: only one line.

The path is L → M → P → Q → R.

• Line segments: $\overline{LM}$, $\overline{MP}$, $\overline{PQ}$ and $\overline{QR}$.
• On exactly one segment: the end points L (only on LM) and R (only on QR).
• On two segments: M (LM and MP), P (MP and PQ) and Q (PQ and QR).

Segments LM, MP, PQ, QR. L and R are on one segment each; M, P and Q are on two segments each.

Rays: $\overrightarrow{TA}$, $\overrightarrow{TN}$, $\overrightarrow{TB}$ (the same ray as TN, since N and B lie on it) and $\overrightarrow{NB}$.

No. T is the starting point of TA, TN and TB, but the ray $\overrightarrow{NB}$ starts at N.

Rays TA, TN (= TB) and NB. T is not the starting point of NB, which starts at N.

See the figures: two lines through O; ray XY crossing line PQ at M; line l through E and F with D off it; P between A and B on AB.

In Fig. 2.6, D, E, O and B lie on one line; one ray goes from O through C, and another ray goes upward from O.

a. Points: D, E, O, B and C.

b. A line: $\overleftrightarrow{DB}$ (it can also be named DE, DO, EO, EB or OB).

c. Rays: $\overrightarrow{OB}$, $\overrightarrow{OC}$, $\overrightarrow{OD}$ and $\overrightarrow{OE}$ (also $\overrightarrow{EB}$, $\overrightarrow{DB}$, …).

d. Line segments: $\overline{DE}$, $\overline{DO}$, $\overline{DB}$, $\overline{EO}$, $\overline{EB}$ (also $\overline{OB}$, $\overline{OC}$).

a. D, E, O, B, C b. line DB c. OB, OC, OD, OE d. DE, DO, DB, EO, EB (others also possible)

a. Yes. A ray is named by its starting point followed by any other point on it. The ray starts at O and B lies on it, so $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are the same ray.

b. No. $\overrightarrow{AO}$ would mean a ray that starts at A and goes through O, i.e. in the opposite direction. The order of letters matters: the first letter is always the starting point.

a. Yes: it starts at O and passes through B. b. No: AO would start at A and go the other way.

Yes, there are many angles in these pictures.

• Bicycle: where the frame bars meet, e.g. at D, the bars DB and DC form ∠BDC, with vertex D and arms $\overrightarrow{DB}$ and $\overrightarrow{DC}$. Other angles: ∠BAD at A, ∠ABC at B.
• Ladder: each step (rung) makes an angle with the side rail; the vertex is the point where the rung meets the rail.
• Bridge: the slanting bars of the truss make angles with the top and bottom beams, at the joints.
• Window grill: the slanting bars make angles with each other and with the frame.

Yes. For example, in the bicycle, ∠BDC has vertex D and arms DB and DC. Angles also appear where the rungs meet the rails of the ladder and where the bars meet in the bridge and the grill.

The two arms start at the same point S, so S is the vertex.

Draw two rays from S, through T and through R; this is ∠TSR (or ∠RST).

At P there are three rays, PA, PB and PC, so more than one angle has its vertex at P: ∠APB, ∠BPC and ∠APC. Writing just ∠P would not tell us which of these angles we mean. A single letter can be used only when there is exactly one angle at that vertex.

Three angles (∠APB, ∠BPC, ∠APC) share the vertex P, so ∠P would be unclear.

Both marked angles have vertex T and start from the arm TR: the smaller arc marks ∠QTR (between TR and TQ) and the larger arc marks ∠PTR (between TR and TP).

∠QTR (or ∠RTQ) and ∠PTR (or ∠RTP).

• Lines: 3: $\overleftrightarrow{AB}$, $\overleftrightarrow{BC}$ and $\overleftrightarrow{CA}$.
• Angles: 3: ∠BAC (= ∠CAB), ∠ABC (= ∠CBA) and ∠ACB (= ∠BCA), one at each point (left figure).

3 lines (AB, BC, CA) and 3 angles (∠BAC, ∠ABC, ∠ACB).

Lines: 6: AB, AC, AD, BC, BD and CD (each point is joined to the other 3, and each line is counted from both ends: $\frac{4\times 3}{2}=6$). See the right-hand figure above.

Angles: at each point, 3 lines meet, giving 3 angles between them. So there are $4\times 3=12$ angles:

6 lines (AB, AC, AD, BC, BD, CD) and 12 angles, 3 at each point.

Suppose the slanting fold EF goes from point E on the top side AB to point F on the bottom side DC of the rectangle ABCD. Four angles are formed: ∠AEF and ∠BEF at E, and ∠DFE and ∠CFE at F.

• At each end, one angle is bigger than a right angle and the other is smaller (together they make a straight angle).
• If the fold leans to the right, ∠AEF and ∠CFE are the larger (obtuse) angles, and ∠BEF and ∠DFE are the smaller (acute) ones; the two large angles are equal to each other, and so are the two small ones.
• If the fold is made straight across (from the middle of one side to the middle of the opposite side), all four angles are equal right angles.

The more slanting the fold, the smaller the smallest angle and the larger the largest angle. Compare your angles by placing one over the other (superimposition).

A slanting fold makes two equal smaller (acute) angles and two equal larger (obtuse) angles with the sides; a straight fold makes four right angles. The more the slant, the smaller the smallest angle.

In the figure, the rays from O are, in order: OA (top), OX, OY, then OB and OC along the bottom (B and C lie on the same ray).

a. ∠AOB is greater. ∠XOY lies completely inside ∠AOB: ∠AOB = ∠AOX + ∠XOY + ∠YOB.

b. ∠AOB is greater. ∠XOB is a part of ∠AOB (∠AOB = ∠AOX + ∠XOB).

c. They are equal. B and C lie on the same ray from O, so OB and OC are the same arm; ∠XOB and ∠XOC are the same angle.

a. ∠AOB b. ∠AOB c. Neither: ∠XOB = ∠XOC (B and C are on the same ray).

In this figure the two angles are not one inside the other; they overlap only partly (OA lies between OX and OY, and OY between OA and OB). So we cannot tell just by looking. We have to compare them by tracing one angle and placing it over the other (superimposition, with vertices and one arm matching), or by measuring both with a protractor.

It cannot be decided by just looking, since neither angle lies inside the other; we must compare them by superimposition or by measuring.

A rectangular window has 4 right angles at its corners, and more where the bars of the window frame cross: a window with one vertical and one horizontal bar has 4 + (4 × 2) + 4 = 16 right angles. (Count for the windows in your own classroom.)

Other right angles: the corners of the blackboard, door, books, desks, notice board and floor tiles, and where the walls meet the floor.

Each rectangular window has at least 4 right angles at its corners (more where its bars cross); doors, boards, books, desks and tiles also have right angles.

A straight angle at A needs the new line to continue BA in a straight line beyond A, in the direction opposite to B.

• Left grid (AB horizontal): join A to any of the 3 dots to the left of A in the same row.
• Right grid (AB slanting down to the right): join A to either of the 2 dots on the slanting line up and to the left of A.

Join A to the dots on line BA extended beyond A: 3 ways on the first grid, 2 ways on the second.

A right angle divides the straight angle at A into two equal parts, so the new line must be perpendicular to AB at A.

• Left grid: join A to any dot directly above A (2 dots) or directly below A (3 dots): 5 ways.
• Right grid: join A to any dot on the slanting line through A that goes up-right and down-left (perpendicular to AB): 2 dots up-right and 2 dots down-left, 4 ways.

On the line through A perpendicular to AB: 5 ways on the first grid, 4 ways on the second.

a. 4 right angles are formed where the two creases cross. When the paper is folded the second time, the two halves of the first crease lie exactly on top of each other. So the second crease divides the straight angle on the first crease into two equal parts: each is half of 180°, i.e. 90°. The other two angles on the opposite side are equal to these by the same fold, so all four are right angles (4 × 90° = 360°).

b. Steps:

1. Fold the paper in any slanting direction and press firmly. Unfold: this is the first crease.
2. Fold the paper again so that the fold passes through a point on the first crease and one part of the first crease lies exactly on top of the other part.
3. Press firmly and unfold. The new crease is perpendicular to the first one; the four angles at the crossing point are right angles.

a. 4 right angles; the second fold places one half of the first crease exactly on the other, so it halves the straight angle (180° ÷ 2 = 90°). b. Fold so that the first crease falls on itself, then unfold.

In the figures of this chapter:

• Right angles: the corners of the window, the classroom board, and the creases folded on themselves; the angle between the perpendicular lines on the dot grid.
• Straight angles: the angle formed at A when BA is extended (the dot grid), and the angle along a straight crease.
• Acute angles: the smaller angles made by a slanting fold, the open beak of each crane, the angles in the left group of the classifying-angles figure.
• Obtuse angles: the larger angles made by a slanting fold, the angles in the right group of the classifying-angles figure, the angle of a wide-open door.

Look for corners (right), straight edges and extended lines (straight), narrow openings (acute) and wide openings between 90° and 180° (obtuse).

Draw, for example:

• Acute: 30° opening to the right, 45° opening upward, 70° opening to the left, 20° tilted downward. (Each is smaller than the corner of a page.)
• Obtuse: 100° opening upward, 120° opening to the right, 150° opening downward, 135° tilted.

Check each with the corner of a page (a right angle): an acute angle fits inside the corner; an obtuse angle is wider than the corner but less than a straight line.

Acute angles (less than 90°) and obtuse angles (between 90° and 180°) can be drawn facing any direction; check with the corner of a page.

An acute angle has a narrow opening, so the corner it makes is pointed and sharp, like the tip of a needle, a knife or a pencil. An obtuse angle opens wide, so its corner is flattened and blunt, not pointed. The names describe how the corners look and feel.

Acute angles make narrow, pointed (sharp) corners; obtuse angles make wide, flattened (blunt) corners.

All the small triangles are equilateral, so every small angle is 60° (acute); two or three of them together make 120° (obtuse) or 180° (straight), which are not acute.

• Figure 1: the triangle has 3 acute angles (one at each corner).
• Figure 2: the triangle is divided into 4 by joining the midpoints of its sides. Each corner has 1 acute angle (3 in all), and each of the 3 midpoints has three 60° angles side by side (9 in all): $3+9=$ 12.
• Figure 3: the middle triangle is divided again in the same way, adding 3 new midpoints with 3 acute angles each: $12+9=$ 21.

Next figure: divide the newest middle triangle again by joining the midpoints of its sides; it will have $21+9=$ 30 acute angles.

Pattern: 3, 12, 21, 30, …: each figure has 9 more acute angles than the previous one (3 + 9 × number of divisions).

3, 12, 21; the next figure (divide the middle triangle again) has 30 acute angles. Each step adds 9.

Count the 1° units between the arms, using the long marks (every 10°) and medium marks (every 5°) to count in 10s and 5s.

a. ∠KAL = 30° b. ∠WAL = 50° c. ∠TAK = 120°

a. 30° b. 50° c. 120°

Reading each arm from the scale that starts at 0° on OP: OQ = 35°, OR = 95°, OS = 125°, OT = 160°, OU = 180°. The angle between any two arms is the difference of their readings:

15 angles, e.g. ∠POQ = 35°, ∠QOS = 90°, ∠ROS = 30°, ∠TOU = 20°, ∠POU = 180° (see the table).

Place the centre of the protractor on the vertex H and its base line along one arm, then read where the other arm crosses the scale that starts at 0° on the first arm:

(i) ∠IHJ ≈ 47° (ii) ∠GHK ≈ 23° (iii) ∠IHJ ≈ 108°

(Small differences of 1° or 2° are acceptable when measuring.)

About 47°, 23° and 108°.

Examples of what you may find:

• Corner of the blackboard, door, window or a book: 90°.
• A door opened partly: anything from 0° to about 100° (measure the angle between the door and the wall at the hinge).
• The angle between the two hands of the wall clock, the legs of a stool, the slope of a desk lid, or the arms of a pair of compasses.

Record each object and its measure in a table.

Corners of the board, doors, windows and books measure 90°; other angles (door openings, clock hands, compasses) vary; record your own measurements.

∠IHJ ≈ 42° (first figure) and ∠IHJ ≈ 116° (second figure).

The paper protractor made by folding has marks only at multiples of 22.5° (180° ÷ 8). It can tell us that 42° lies between 22.5° and 45°, and that 116° lies between 112.5° and 135°, but it cannot give the exact measures. A standard protractor with 1° marks is needed.

About 42° and 116°. The folded paper protractor (marks every 22.5°) cannot measure them exactly.

The marked angle is a reflex angle (more than 180°), so it cannot be read directly on a protractor, which goes only up to 180°. Instead:

1. Measure the unmarked angle between the two arms (the smaller side): it is 100°.
2. A full turn is 360°, so the marked angle = $360°-100°=$ 260°.

(Another way: extend one arm backwards to make a straight angle of 180°, measure the extra part beyond it, and add it to 180°.)

Measure the smaller angle (100°) and subtract it from 360°: the marked angle is 260°.

a. 80° b. 120° c. 60° d. 130° e. 130° f. 60°

a. 80° b. 120° c. 60° d. 130° e. 130° f. 60°

Reading from the scale that starts at 0° on XE: XC is at 85° and XB at 115°; XA is at 180°.

• ∠BXE = 115°
• ∠CXE = 85°
• ∠AXB = 180° − 115° = 65°
• ∠BXC = 115° − 85° = 30°

∠BXE = 115°, ∠CXE = 85°, ∠AXB = 65°, ∠BXC = 30°

Placing the protractor at Q with its base along QP:

• ∠PQR ≈ 45°
• ∠PQS ≈ 100°
• ∠PQT ≈ 150°

(So ∠RQS ≈ 55° and ∠SQT ≈ 50°.)

∠PQR ≈ 45°, ∠PQS ≈ 100°, ∠PQT ≈ 150°

Folding a square along a diagonal and then folding the corners up and down (steps 1–8) gives creases that are along the diagonals, along the middle lines and across the corners. When you unfold the paper and draw the creases, you will find mostly:

• 45° angles (between a diagonal crease and a side of the square),
• 90° angles (where the diagonal crease and the middle crease cross, and at the corners),
• 135° angles (a 90° and a 45° side by side) and 180° along each straight crease.

The creases mainly form angles of 45°, 90°, 135° and 180°, because each fold halves a right or straight angle.

Measure each angle carefully. For example, in triangle (b), which looks equilateral, each angle is about 60°: 60° + 60° + 60° = 180°. For (a) and (c) the three measures are different, but in each case the sum comes to 180° (or 1° or 2° away, because of small measuring errors).

Conjecture: the three angles of any triangle always add up to 180° (a straight angle).

In every triangle, the sum of the three angles is 180°.

The correct method: (1) place the centre of the protractor exactly on the vertex; (2) place the base line (0°–180° line) along one arm; (3) read the other arm on the scale that starts at 0° on the first arm.

• ∠U = 35°: the vertex is at the centre, but neither arm is on the base line: one arm is below the base line. The student read only where the upper arm crosses the scale. Correction: turn the protractor so that its base line lies along the lower arm, then read.
• ∠V = 80°: the protractor is tilted and its base line does not lie along an arm, so the reading does not start from 0°. Correction: rotate the protractor until the base line lies exactly along one arm.
• ∠W = 70°: the vertex of the angle is not at the centre of the protractor: the two arms meet near the edge of the scale. Correction: move the protractor so that its centre is on the vertex.
• ∠X = 150°: one arm is placed along the straight edge of the protractor instead of the base line, and the vertex is not at the centre. Correction: put the centre on the vertex and the base line on the arm; also read from the scale whose 0 is on that arm.
• ∠Y = 120°: the protractor is placed upright, with its base line along neither arm, and the vertex is not at the centre. Correction: place it as described above.
• ∠Z = 85°: the protractor is upside down; this is allowed if the base line is on one arm, but the student read the wrong scale (the one that does not start at 0° on the arm). Correction: read the scale that starts at 0° on the arm lying on the base line.

Errors: arms not on the base line (U, V, Y), vertex not at the centre (W, X, Y), arm along the edge instead of the base line (X), and reading the wrong scale (Z). Always put the centre on the vertex, the base line on one arm, and read from that arm's 0°.

a. A full turn of 360° is divided by the 12 numbers on the clock into 12 equal parts: $360°÷12=30°$. At 1 o'clock the hands point to 12 and 1, one part apart, so the angle is 30°.

b. 2 o'clock: $2\times 30°=60°$. 4 o'clock: $4\times 30°=120°$. 6 o'clock: $6\times 30°=180°$ (a straight angle).

c. 3 o'clock and 9 o'clock: 90° (right angle); 5 o'clock: 150°; 12 o'clock: 0° (the hands overlap); 10 o'clock: 60° (or 300° measured the other way).

a. 360° ÷ 12 = 30° between neighbouring numbers. b. 60°, 120°, 180°. c. 3 and 9 o'clock: 90°; 5 o'clock: 150°; 12 o'clock: 0°.

Yes. As the door turns on its hinges, the opening can be measured as an angle.

• Vertex: the line of the hinges (seen from above, the point where the door meets the door frame).
• Arms: the door in its open position and the door frame/wall (the closed position of the door).

Yes; the vertex is at the hinges, and the arms are the open door and the wall (the closed position of the door).

The angle is at the point where the swing's ropes are tied to the branch (the vertex). One arm is the rope hanging straight down (the rest position), and the other arm is the rope at the starting position, pulled back. The angle between these two positions is the angle Vidya starts with. This angle is not drawn anywhere, so it is hard to "see", but we can imagine the two positions of the rope.

The vertex is where the ropes are tied; the arms are the rope hanging straight down and the rope in the pulled-back starting position.

Yes. The slope of each slab can be described by the angle it makes with a horizontal line (or with the vertical side of the toy). For each angle, the vertex is the point where the slab is attached, one arm is the slab itself (visible) and the other arm is the horizontal line through that point (not visible; we have to imagine it). The greater the angle with the horizontal, the steeper the slab and the faster the ball rolls.

Yes: each slope is the angle between the slab (visible arm) and an imaginary horizontal line (invisible arm), with the vertex where the slab is fixed.

In Fig. 2.23 two horizontal lines (through A and B, and through C, D, L, S) are crossed by three slanting lines (CA, LP and SR). Angles are formed at A, P, R (top line) and at C, L, S (bottom line). At each crossing point there are four angles, for example at P: ∠APL, ∠LPR, ∠RP(upward) and the angle on the other side.

A list of the main angles: ∠CAP, ∠ACD, ∠APL, ∠DLP, ∠RPL, ∠SLP, ∠PRS, ∠LSR, ∠BRS, ∠CLP (and the angles opposite or next to them at each crossing).

Make a table with three columns: Angle, Guess, Measure. For example, at a crossing an angle that looks a little more than a right angle might be guessed as 100° and measured as 102°. Remember: the two angles next to each other on a straight line always add up to 180°, which helps you check your measures.

Angles are formed at A, P, R, C, L and S (e.g. ∠CAP, ∠ACD, ∠APL, ∠DLP, ∠RPL, ∠SLP, ∠PRS, ∠LSR, ∠BRS, ∠CLP). Guess, then measure, and compare in a table; neighbouring angles on a line add up to 180°.

Steps (example: 110°):

1. Draw a ray OA.
2. Place the centre of the protractor on O and the base line along OA.
3. On the scale that starts at 0° on OA, find 110° and mark a point B there.
4. Remove the protractor and join OB. ∠AOB = 110°.

Repeat with 40°, 75°, 112° and 134°. (40° and 75° are acute; 110°, 112° and 134° are obtuse.)

Draw a ray, place the protractor's centre at its start and the base along it, mark the required degree on the scale starting at 0°, and join.

Steps:

1. Measure the given angle ∠IHJ with a protractor (centre on H, base line along HJ). It measures about 116°.
2. Draw a ray PQ in your notebook.
3. Place the protractor with its centre on P and its base line along PQ.
4. Mark a point R at 116° on the scale that starts at 0° on PQ.
5. Join PR. Then ∠QPR = 116°, the same as the given angle.

Measure the given angle (about 116°), then draw a ray and use the protractor to mark and draw an angle of the same measure.

First grid (AB horizontal, pointing right):

• Acute: join A to a dot above or below the line, on the B side, e.g. the dot 1 up and 1 to the right of A (45°), or 2 up and 1 right (about 63°).
• Obtuse: join A to a dot on the other side, e.g. 1 up and 1 to the left of A (135°).
• Reflex: use any of the lines above, but mark the outer angle (going the long way round); e.g. with the dot 1 up and 1 right of A, the reflex angle is 360° − 45° = 315°.

Second grid (AB slanting down-right):

• Acute: join A to a dot that is close in direction to B, e.g. the dot directly to the right of A (45° with AB), or directly below A (45°).
• Obtuse: join A to a dot in the opposite direction, e.g. the dot directly to the left of A (135°) or directly above A (135°).
• Reflex: take any of these angles and mark the outer side, e.g. 360° − 45° = 315°.

Acute: join A to a dot near the direction of B (e.g. 1 up, 1 right: 45°). Obtuse: a dot on the far side (e.g. 1 up, 1 left: 135°). Reflex: mark the outside of any angle (e.g. 360° − 45° = 315°).

a. 30°, acute b. 60°, acute c. 102°, obtuse d. 258°, reflex

B, E and R lie on a straight line, so ∠BER = 180°.

$$\angle BET=180°-\angle TER=180°-80°=100°$$

∠SER is a right angle (90°), and ∠TER = 80° is a part of it:

$$\angle SET=90°-80°=10°$$

∠BET = 100° and ∠SET = 10°

Use the protractor steps from 2.10 for a, b, d and e.

For c. 195° (a reflex angle): draw a ray OA and extend it backwards to make a straight angle (180°). Now draw an angle of $195°-180°=15°$ beyond the straight line, on the other side (or draw $360°-195°=165°$ and mark the outer angle). The marked outer angle is 195°.

Draw a, b, d, e directly with the protractor; for 195°, draw 165° and mark the reflex (outer) angle, since 360° − 165° = 195°.

(Your estimates may differ; what matters is that you check them with a protractor. Measured values may vary by 1°–2°.)

a. ≈ 45°, acute b. ≈ 168°, obtuse c. ≈ 120°, obtuse d. ≈ 33°, acute e. ≈ 98°, obtuse f. ≈ 350°, reflex

One way is a zig-zag line whose corners have the required angles:

(Another way: a house-shaped figure or any drawing made of straight lines, with the angles marked.)

For example, a zig-zag line with corners of 40°, 120°, 60°, 90°, 150° and 50°.

The middle "V" of the M opens at 60°, so each slanting stroke makes 30° with the vertical. For the side angles to be 40°, the two outer strokes must lean slightly outward (10° from the vertical).

Draw the middle V with 60°, then draw each outer stroke at 40° to the slanting stroke at the top corners.

The three angles around the centre add up to 360° (150° + 60° + 150° = 360°). Draw the stem straight down and the two arms at 150° from the stem on each side; they are then 60° apart.

Stem straight down; each arm at 150° from the stem, so the two arms are 60° apart.

The 24 equally spaced spokes divide the full turn into 24 equal parts:

$$360°÷24=15°$$

The angles between any two spokes are multiples of 15°: 15°, 30°, 45°, 60°, 75°, 90°, … The largest one less than 90° is 75° (5 gaps).

15° between neighbouring spokes; the largest acute angle is 75°.

Let the angle be $x$ degrees.

• 4 times is still acute: $4x<90°$, so $x<22.5°$ (this also makes $2x$ and $3x$ acute).
• 5 times is obtuse: $5x>90°$, so $x>18°$ (and $5x<180°$ is then automatic).

So $18°<x<22.5°$. In whole degrees, the possibilities are 19°, 20°, 21° and 22°.

(Check with 22°: 44°, 66°, 88° are acute and 110° is obtuse. With 18°, 5 × 18° = 90° is a right angle, not obtuse.)

Any measure between 18° and 22.5°: in whole degrees, 19°, 20°, 21° or 22°.