Chapter 2 — Lines and Angles
Complete, step-by-step solutions to every question with diagrams & explanations.
Points · Line Segments · Lines · Rays · Angles · Measuring & Drawing Angles
💡 In-text Questions
Answers to the exploratory prompts, “Think!” boxes and “Let’s Explore” questions inside the chapter.
(Section 2.6, Comparing Angles) Is it always easy to compare two angles?
No — it is not always easy to compare two angles just by looking. When two angles are very close in size, the eye cannot judge which is bigger.
For example, a \(89^\circ\) angle and a \(91^\circ\) angle look almost identical, so they cannot be compared reliably without measuring or overlapping (superimposition). For angles that differ a lot, a direct look is enough.
(Section 2.6) Where else do we use superimposition to compare?
Superimposition means placing one figure exactly over another to see which is larger. We use the same idea to compare the sizes of many things, not just angles.
(Section 2.8, Special Types of Angles) Is it possible to draw ray \(\overrightarrow{OC}\) such that the two angles are equal to each other in size?
Take the straight angle \(\angle AOB\) marked on the edge of a rectangular paper and fold it so that ray \(OB\) falls exactly on ray \(OA\).
(Section 2.8) If a straight angle is formed by half of a full turn, how much of a full turn will form a right angle?
A straight angle \(=\tfrac12\) of a full turn. A right angle is exactly half of a straight angle.
\[\text{Right angle}=\frac12\times\left(\frac12 \text{ turn}\right)=\frac14\ \text{turn}\]
(Section 2.9, Make your own protractor) A circle (full turn = \(360^\circ\)) is divided into \(1,2,3,4,5,6,8,9,10\) and \(12\) equal parts. What are the degree measures of the resulting angles?
Each equal part \(=\dfrac{360^\circ}{n}\), where \(n\) is the number of parts.
| Parts (n) | 1 | 2 | 3 | 4 | 5 | 6 | 8 | 9 | 10 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|
| Each angle | 360° | 180° | 120° | 90° | 72° | 60° | 45° | 40° | 36° | 30° |
For example \(\dfrac{360^\circ}{5}=72^\circ\) and \(\dfrac{360^\circ}{8}=45^\circ\).
(Section 2.9, labelled protractor — page 36) Name the different angles in the figure and write their measures.
Reading each ray from the protractor and finding the difference of the two readings gives each angle:
| Angle | Measure | Angle | Measure |
|---|---|---|---|
| ∠POQ | 35° | ∠QOT | 125° |
| ∠POR | 95° | ∠QOU | 145° |
| ∠POS | 125° | ∠ROS | 30° |
| ∠POT | 160° | ∠ROT | 65° |
| ∠QOR | 60° | ∠ROU | 85° |
| ∠QOS | 90° | ∠SOT | 35° |
| ∠SOU | 55° | ∠TOU | 20° |
(Section 2.9, Think! — page 40) In Fig. 2.19, \(\angle AOB=\angle BOC=\angle COD=\angle DOE=\angle EOF=\angle FOG=\angle GOH=\angle HOI = \underline{\ \ ?\ \ }\). Why?
The straight angle of \(180^\circ\) was folded into halves again and again until it was split into 8 equal parts. So each small angle is
\[\frac{180^\circ}{8}=22.5^\circ\]
(Section 2.11, Let’s Explore — page 53) In the figure, \(\angle TER = 80^\circ\). What is the measure of \(\angle BET\)? What is the measure of \(\angle SET\)?
Finding \(\angle BET\): Points \(B, E, R\) lie on a straight line, so \(\angle BER\) is a straight angle \(=180^\circ\). This is split into \(\angle BET\) and \(\angle TER\):
\[\angle BET = 180^\circ – \angle TER = 180^\circ – 80^\circ = 100^\circ\]
Finding \(\angle SET\): \(ES\) is perpendicular to \(ER\), so \(\angle SER = 90^\circ\). Now \(\angle SER\) is made of \(\angle SET\) and \(\angle TER\):
\[\angle SET = \angle SER – \angle TER = 90^\circ – 80^\circ = 10^\circ\]
📝 Exercise Questions
Detailed solutions to all the “Figure it Out” exercises, section by section.
Section 2.4 — Ray
Can you help Rihan and Sheetal find their answers?
Rihan (one point): Through a single point we can draw a line in any direction — turning it a little always gives a new line. So there is no limit.
Sheetal (two points): Two points fix a straight path completely, so only one straight line can pass through both.
Name the line segments in Fig. 2.4. Which of the five marked points are on exactly one of the line segments? Which are on two of the line segments?
Reading the joined points along the path L–M–P–Q–R gives four line segments. An end point of the whole path lies on only one segment; a turning point in the middle lies on two segments.
Points on exactly one line segment: L and R (the ends).
Points on two line segments: M, P and Q.
Name the rays shown in Fig. 2.5. Is T the starting point of each of these rays?
A ray is named by its starting point first, then any other point it passes through.
No. T is the starting point of \(\overrightarrow{TA},\overrightarrow{TB},\overrightarrow{TN}\), but not of \(\overrightarrow{NB}\) (that ray starts at N).
Draw a rough figure and write labels appropriately to illustrate each of the following:
a. \(\overleftrightarrow{OP}\) and \(\overleftrightarrow{OQ}\) meet at O. b. \(\overrightarrow{XY}\) and \(\overleftrightarrow{PQ}\) intersect at point M.
c. Line \(l\) contains points E and F but not point D. d. Point P lies on \(\overline{AB}\).
Each rough sketch simply shows the relationship described:
In Fig. 2.6, name: a. Five points b. A line c. Four rays d. Five line segments
All of D, E, O, B, C lie along one straight line (with O also the start of two more rays).
b. A line: \(\overleftrightarrow{DE}\) (also \(\overleftrightarrow{DO},\overleftrightarrow{DB},\overleftrightarrow{EO},\overleftrightarrow{EB},\overleftrightarrow{OB}\))
c. Four rays: \(\overrightarrow{OC},\overrightarrow{OB},\overrightarrow{OE},\overrightarrow{OD}\) (others are possible)
d. Five line segments: \(\overline{DE},\overline{DO},\overline{DB},\overline{EO},\overline{EB}\) (\(\overline{OB},\overline{OC}\) also possible)
Here is a ray \(\overrightarrow{OA}\) (Fig. 2.7). It starts at O and passes through the point A. It also passes through the point B.
a. Can you also name it as \(\overrightarrow{OB}\)? Why? b. Can we write \(\overrightarrow{OA}\) as \(\overrightarrow{AO}\)? Why or why not?
A ray is fixed by its starting point and the direction it travels.
b. No. \(\overrightarrow{OA}\) starts at O, while \(\overrightarrow{AO}\) starts at A. Different starting points mean different rays.
Section 2.5 — Angle
Can you find the angles in the given pictures? Draw the rays forming any one of the angles and name the vertex of the angle.
Wherever two rods (rays) meet, an angle is formed. In the bicycle frame, two rods meet at D.
Draw and label an angle with arms \(ST\) and \(SR\).
The two arms both start from the common point S, which becomes the vertex. One arm goes to T and the other to R.
Explain why \(\angle APB\) cannot be labelled as \(\angle P\).
At the point P, three different rays (\(\overrightarrow{PA}, \overrightarrow{PB}, \overrightarrow{PC}\)) start out. So several angles share the same vertex P — namely \(\angle APB\), \(\angle BPC\) and \(\angle APC\).
Name the angles marked in the given figure.
Both marked angles have vertex T, measured from arm \(\overrightarrow{TR}\).
Mark any three points (not on one line) as A, B, C. Draw all possible lines through pairs. How many lines do you get? How many angles can you name using A, B, C?
With 3 non-collinear points, each pair gives one line: (A,B), (B,C), (C,A).
We can name 3 angles: \(\angle ABC\ (=\angle CBA)\), \(\angle BCA\ (=\angle ACB)\) and \(\angle CAB\ (=\angle BAC)\).
Mark any four points so that no three are on one line, as A, B, C, D. Draw all possible lines through pairs. How many lines? How many angles can you name using A, B, C, D?
With 4 points, the number of lines is the number of pairs \(=\dbinom{4}{2}=6\).
We can name 12 angles: \(\angle BAC,\angle CAD,\angle BAD,\angle ADB,\angle BDC,\angle ADC,\angle DCA,\angle ACB,\angle DCB,\angle CBD,\angle DBA,\angle CBA\).
Section 2.6 — Comparing Angles
Fold a rectangular sheet of paper, then draw a line along the fold. Name and compare the angles formed between the fold and the sides of the paper. Which is the largest and smallest angle you made?
The slanting fold line meets the top side at F and the bottom side at E, making four angles.
Here \(\angle AEF\) and \(\angle CFE\) are the larger angles, while \(\angle BEF\) and \(\angle DFE\) are the smaller ones. (Fold in different ways to make different largest/smallest angles.)
In each case, determine which angle is greater and why: a. \(\angle AOB\) or \(\angle XOY\) b. \(\angle AOB\) or \(\angle XOB\) c. \(\angle XOB\) or \(\angle XOC\).
Compare by seeing whether one angle contains the other (a whole is bigger than its part).
b. \(\angle AOB\) is greater.
c. Neither — \(\angle XOB=\angle XOC\) (they are equal).
Which angle is greater: \(\angle XOY\) or \(\angle AOB\)? Give reasons.
Here neither angle is a part of the other and they look close in size.
Section 2.8 — Special Types & Classifying Angles
How many right angles do the windows of your classroom contain? Do you see other right angles in your classroom?
Each rectangular window pane has square corners.
Join A to other grid points by a straight line to get a straight angle. What are all the different ways of doing it?
A straight angle at A needs the two chosen points to be on the same straight line as A, one on each side (opposite directions), so the two rays open out to \(180^\circ\).
Now join A to other grid points to get a right angle. What are all the different ways of doing it? (Hint: extend the line so that a line through A divides the straight angle CAB into two equal parts.)
A right angle is exactly half of a straight angle. So at A we pick two directions that are perpendicular to each other.
• horizontal and vertical,
• the two diagonals (they are perpendicular to each other),
• or a slope like “1 up, 2 right” together with its perpendicular “2 up, 1 left”.
Each perpendicular pair gives a right angle at A.
Get a slanting crease on the paper. Now try to get another crease perpendicular to the slanting crease.
a. How many right angles do you have now? Justify why they are exact right angles.
b. Describe how you folded the paper so that anyone can follow.
Fold once to make the slanting crease. Fold again so the crease falls exactly on itself — the new crease is perpendicular to it.
b. How to fold: make a straight crease; then fold the paper so that this crease lies exactly on top of itself. Press and unfold — the second crease crosses the first at right angles.
Make a few acute angles and a few obtuse angles. Draw them in different orientations.
Any angle smaller than a right angle is acute; any angle bigger than a right angle but smaller than a straight angle is obtuse. Draw several pointing in different directions (up, down, sideways, slanting).
Acute means sharp and obtuse means blunt. Why do you think these words have been chosen?
Look at the opening between the two arms.
Find out the number of acute angles in each of the figures below. What will be the next figure and how many acute angles will it have? Do you notice any pattern?
Count the acute angles inside each figure:
Figure (i): \(3\) Figure (ii): \(12\) Figure (iii): \(21\)
The counts increase by \(9\) each time: \(3,\ 12,\ 21,\ \ldots\) So the next figure has \(21+9=30\).
A neat formula for the \(n\)-th figure is \(9n-6\): check \(9(1)-6=3,\ 9(2)-6=12,\ 9(3)-6=21,\ 9(4)-6=30\). (The extra triangles keep adding groups of angles as more inner triangles appear.)
Section 2.9 — Measuring Angles
Write the measures of the following angles from the protractor: a. \(\angle KAL\) b. \(\angle WAL\) c. \(\angle TAK\).
The vertex A is at the centre of the protractor, so we count the number of \(1^\circ\) units between the two arms (counting in 5s and 10s using the medium and long marks).
Yes — the long marks (every \(10^\circ\)) and medium marks (every \(5^\circ\)) let us count quickly in 5s or 10s.
Find the degree measures of the following angles using your protractor.
Place the centre of the protractor on the vertex H and read each angle.
Find the degree measures for the angles given below. Check if your paper protractor can be used here!
Measure with the standard protractor.
No, the handmade paper protractor cannot be used here — its creases only fall at multiples of \(22.5^\circ\) (\(0,22.5,45,67.5,90,\ldots\)), so it cannot measure \(42^\circ\) or \(116^\circ\) exactly.
How can you find the degree measure of the angle given below using a protractor?
A protractor only measures up to \(180^\circ\). For a reflex angle, first measure the smaller (unmarked) angle, then subtract from a full turn:
\[\text{Reflex angle}=360^\circ-\text{(smaller angle)}=360^\circ-100^\circ=260^\circ\]
Measure and write the degree measures for each of the following angles (a–f).
Placing the protractor centre on each vertex and reading the scale gives:
Find the degree measures of \(\angle BXE, \angle CXE, \angle AXB\) and \(\angle BXC\).
Read the ray positions from the protractor. Note \(A\)–\(X\)–\(E\) is a straight line (\(180^\circ\)).
\(\angle BXE = 115^\circ\); \(\angle CXE = 85^\circ\).
Then \(\angle AXB = 180^\circ – 115^\circ = 65^\circ\) and \(\angle BXC = \angle BXE – \angle CXE = 115^\circ – 85^\circ = 30^\circ\).
Find the degree measures of \(\angle PQR, \angle PQS\) and \(\angle PQT\).
Measure each angle from arm \(\overrightarrow{QP}\).
Measure all three angles of each triangle in Fig. 2.21, and add them up. What do you get? Make a conjecture.
On measuring and adding the three angles of each triangle:
\[\angle A + \angle B + \angle C = 180^\circ\]
The same total \(180^\circ\) comes up for every triangle you try.
Where are the Angles? (page 45)
Angles in a clock: a. At 1 o’clock the angle between the hands is \(30^\circ\). Why? b. What will be the angle at 2 o’clock? 4 o’clock? 6 o’clock? c. Explore other angles.
The full turn at the centre (\(360^\circ\)) is split by the 12 hour-marks into 12 equal parts:
\[\text{Angle between two numbers}=\frac{360^\circ}{12}=30^\circ\]
a. At 1 o’clock the hands are 1 part apart \(=30^\circ\).
b. \(2\ \text{o’clock}=2\times30^\circ=60^\circ\); \(4\ \text{o’clock}=4\times30^\circ=120^\circ\); \(6\ \text{o’clock}=6\times30^\circ=180^\circ\).
c. e.g. \(3\ \text{o’clock}=90^\circ\), \(9\ \text{o’clock}=270^\circ\) (or \(90^\circ\) the other way).
The angle of a door: Is it possible to express the amount by which a door is opened using an angle? What will be the vertex and the arms?
Yes. As a door swings open, it turns about its hinge.
Vidya on a swing: the greater the starting angle, the greater the speed. But where is the angle?
Think of the rope’s rest position and its highest position.
A toy with slanting slabs: the greater the slope, the faster the balls roll. Can angles describe the slopes? What are the arms? Which arm is visible and which is not?
A steeper slab makes a bigger angle with the horizontal, so the ball rolls faster.
An insect and its rotated version: Can angles describe the amount of rotation? What are the arms and the vertex? (Hint: observe the horizontal line touching the insects.)
The horizontal line touching both insects acts as the fixed starting arm.
Section 2.10 — Drawing Angles
In Fig. 2.23, list all the angles possible. Guess their measures, then measure with a protractor and compare.
Wherever two lines cross or meet, angles are formed. List every such angle, guess it, then check with a protractor.
Use a protractor to draw angles having the following degree measures: a. \(110^\circ\) b. \(40^\circ\) c. \(75^\circ\) d. \(112^\circ\) e. \(134^\circ\).
Steps for each angle: (1) draw a base ray; (2) place the protractor centre on its starting point and align the base with the \(0^\circ\) line; (3) mark the given degree; (4) join to form the angle.
Section 2.11 — Types of Angles & their Measures
In each grid, join A to another grid point by a straight line to get: a. an acute angle b. an obtuse angle c. a reflex angle. Mark the intended angles with curves.
Keep one ray fixed (say horizontal) and pick the second ray so that the opening is:
• Obtuse — opening between a right angle and a straight angle (\(90^\circ\!-\!180^\circ\)).
• Reflex — opening measured the long way round, more than a straight angle (\(>180^\circ\)).
Mark each intended angle with a small curve so it is clear which angle you mean.
Use a protractor to find the measure of each angle. Then classify each as acute, obtuse, right or reflex: a. \(\angle PTR\) b. \(\angle PTQ\) c. \(\angle PTW\) d. \(\angle WTP\).
Measure each from the correct arm. Note that \(\angle PTW\) and \(\angle WTP\) are the two ways round the same pair of arms, so they add to \(360^\circ\): \(102^\circ+258^\circ=360^\circ\).
b. \(\angle PTQ = 60^\circ\) — acute
c. \(\angle PTW = 102^\circ\) — obtuse
d. \(\angle WTP = 258^\circ\) — reflex
Draw angles with the following degree measures: a. \(140^\circ\) b. \(82^\circ\) c. \(195^\circ\) d. \(70^\circ\) e. \(35^\circ\). Then estimate & classify each.
Use the protractor to draw each angle from a base ray. For the reflex angle \(195^\circ\), draw the smaller \(360^\circ-195^\circ=165^\circ\) part first and mark the reflex side.
Estimate the size of each angle below, then measure it with a protractor and classify it as acute, right, obtuse or reflex.
How to classify at a glance: compare each opening with a right angle (\(90^\circ\)) and a straight angle (\(180^\circ\)):
• between \(90^\circ\) and \(180^\circ\) → obtuse • more than \(180^\circ\) → reflex
Angle (f) is drawn with a full round curve, so it is a reflex angle. First guess each value, then check with your protractor and see how close your guess was.
Make any figure with three acute angles, one right angle and two obtuse angles.
A ‘crown’ shape works nicely:
Draw the letter ‘M’ such that the angles on the sides are \(40^\circ\) each and the angle in the middle is \(60^\circ\).
Two slanting strokes make the sides (\(40^\circ\) each) and the middle V gives \(60^\circ\).
Draw the letter ‘Y’ such that the three angles formed are \(150^\circ, 60^\circ\) and \(150^\circ\).
The two upper arms make \(60^\circ\) between them, and each arm makes \(150^\circ\) with the stem. Check: \(150^\circ+60^\circ+150^\circ=360^\circ\) (a full turn about the joining point).
The Ashoka Chakra has 24 spokes. What is the degree measure of the angle between two spokes next to each other? What is the largest acute angle formed between two spokes?
The full turn \(360^\circ\) is divided equally by the 24 spokes:
\[\text{Angle between adjacent spokes}=\frac{360^\circ}{24}=15^\circ\]
Counting spokes apart: \(5\) gaps give \(5\times15^\circ=75^\circ\) (acute), but \(6\) gaps give \(6\times15^\circ=90^\circ\) (a right angle, not acute).
Puzzle: I am an acute angle. Doubling, tripling and quadrupling my measure still give acute angles, but multiplying by 5 gives an obtuse angle. What are the possibilities for my measure?
Let my measure be \(x\). We need:
\(4x<90^\circ\Rightarrow x<22.5^\circ\) (quadruple is still acute)
\(5x>90^\circ\Rightarrow x>18^\circ\) (five times is obtuse)
So \(18^\circ < x < 22.5^\circ\). (Also \(5x<180^\circ\Rightarrow x<36^\circ\) is automatically true.)
Educational solutions compiled from NCERT Ganita Prakash, Grade 6 · for practice & revision · @edugrown