NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

Edited By Komal Miglani | Updated on May 02, 2025 08:13 PM IST

Have you ever looked at a long, straight road with no visible ending into the distance and wondered where it might end? Or noticed the hands of a clock at exactly 3 o’clock forming a perfect 90-degree angle? These are everyday examples where we unknowingly observe lines and angles around us. In class 9 Maths NCERT chapter 6, students will study lines, angles and their properties. The purpose of this article is to make learning easier for students and explain the NCERT Solutions thoroughly.

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This Story also Contains

Lines And Angles Class 9 Questions And Answers PDF Free Download
Lines And Angles Class 9 Solutions: Important Points
NCERT Solutions for Class 9 Maths Chapter 6: Exercise Questions
Class 9 Maths NCERT Chapter 6: Extra Question
Approach to Solve Questions of Lines and Angles Class 9
NCERT Solutions for Class 9 Mathematics: Chapter-wise
NCERT Solutions For Class 9: Subject Wise
NCERT Books and NCERT Syllabus

NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

Careers360 experts, experienced in this area, prepared the Class 9 NCERT solutions for lines and angles and adhered to the latest CBSE syllabus. There are various applications of lines and angles in our daily lives. So, having the concepts clear will be beneficial not only for the Class 9 board exams but also for day-to-day life. After going through all the NCERT Solutions for Class 9 Maths, students can practice the NCERT Exemplar Solutions Class 9 NCERT Maths Chapter Lines And Angles for a deeper understanding of the concepts. Along with this, using the NCERT Solutions for Class 9, learners can practise diagram-based questions and prove angle-based properties confidently.

Lines And Angles Class 9 Questions And Answers PDF Free Download

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Lines And Angles Class 9 Solutions: Important Points

Types of Angles:

Acute Angle: An acute angle measures between 0° and 90°.
Right Angle: A right angle is exactly equal to 90°.
Obtuse Angle: An angle that is greater than 90° but less than 180°.
Straight Angle: A straight angle is equal to 180°.
Reflex Angle: A reflex angle is greater than 180° but less than 360°.

Complementary Angles (x, y): x + y = 90°

Supplementary Angles (x, y): x + y = 180°

Adjacent Angles:

Adjacent angles are two angles that share a common side and a common vertex (corner point) without overlapping.

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Linear Pair:

A linear pair of angles is formed when two lines intersect. Two angles are considered linear if they are adjacent angles formed by the intersection of two lines. The measure of a straight angle is 180°, so a linear pair of angles must add up to 180°.

Vertically Opposite Angles:

Vertically opposite angles are formed when two lines intersect at a point. Vertically opposite angles are always equal.

Transversal:

A transversal is a line that intersects two or more given lines at distinct points. It is used to create various types of angles, including:
- Corresponding Angles
- Alternate Interior Angles
- Alternate Exterior Angles
- Interior Angles on the Same Side of the Transversal

NCERT Solutions for Class 9 Maths Chapter 6: Exercise Questions

Class 9 Maths Chapter 6 Question Answer: Exercise: 6.1

Total Questions: 6

Page number: 76-77

Question 1: In Fig. 6.13, lines AB and CD intersect at O. If $∠$ AOC + $∠$ BOE = 70° and $∠$ BOD = 40°, find $∠$ BOE and reflex $∠$ COE.

1640149435860

Answer:

1595609937996
Given that,

AB is a straight line. Lines AB and CD intersect at O.

$∠ A O C + ∠ B O E = 70 °$ and $∠$ BOD = $40 °$

Since AB is a straight line,

$∴$ $∠$ AOC + $∠$ COE + $∠$ EOB = $180 °$

$\Rightarrow ∠ C O E = 180 ° - 70 ° = 110^{0}$

[since $∠ A O C + ∠ B O E = 70 °$ ]

So, reflex $∠$ COE = $360 ° - 110^{0} = 250 °$

It is given that AB and CD intersect at O.

Therefore, $∠$ AOC = $∠$ BOD [vertically opposite angle]

$\Rightarrow ∠ C O A = 40 °$ [ GIven $∠$ BOD = $40^{0}$ ]

Also, $∠ A O C + ∠ B O E = 70 °$

So, $∠$ BOE = $30 °$

Question 2: In Fig. 6.14, lines XY and MN intersect at O. If $∠ P O Y = 90^{o}$ and a: b = 2 : 3, find c.

1640149490726

Answer:

1595610210463

Given that,

Line XY and MN intersect at O and $∠$ POY = $90^{0}$

Also, $a : b = 2 : 3 \Rightarrow b = \frac{3 a}{2}$ ..............(I)

Since XY is a straight line

Therefore, $a + b + ∠ P O Y = 180^{0}$

$\Rightarrow a + b = 180^{0} - 90^{0} = 90^{0}$ ...........(ii

Thus, from eq (i) and eq (ii), we get,

$\Rightarrow \frac{3 a}{2} + a = 90^{0}$

$\Rightarrow a = 36^{0}$

So, $b = 54^{0}$

Since $∠$ MOY = $∠$ c [vertically opposite angles]

$∠$ a + $∠$ POY = c

$126^{0} = c$

Question 3: In Fig. 6.15, $∠$ PQR = $∠$ PRQ, then prove that $∠$ PQS = $∠$ PRT.

1640149622015

Answer:

1595610294843

Given that,

ABC is a triangle such that $∠$ PQR = $∠$ PRQ, and ST is a straight line.

Now, $∠$ PQR + $∠$ PQS = $180^{0}$ {Linear pair}............(I)

Similarly, $∠$ PRQ + $∠$ PRT = $180^{0}$ ..................(ii)

Equating the equation (i) and eq (ii), we get,

$∠ P Q R + ∠ P Q S = ∠ P R T + ∠ P R Q$ {but $∠$ PQR = $∠$ PRQ }

Therefore, $∠$ PQS = $∠$ PRT

Hence, it is proved.

Question 4: In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

1640149658149

Answer:

1595610388882

Given that,

$x + y = z + w$ ..............(I)

It is known that the sum of all the angles at a point = $360^{0}$

$∴$ $x + y + z + w = 360^{0}$ ..............(ii)

From eq (i) and eq (ii), we get,

$2 (x + y) = 360^{0}$

$\Rightarrow x + y = 180^{0}$

Hence, proven that AOB is a line.

Question 5: In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $∠ ROS = \frac{1}{2} (∠ QOS - ∠ POS)$

1640149689300

Answer:

1595610457579

Given that,

POQ is a line, OR $⊥$ PQ and $∠$ ROQ is a right angle.

Now, $∠$ POS + $∠$ ROS + $∠$ ROQ = $180^{0}$ [since POQ is a straight line]

$\Rightarrow ∠ P O S + ∠ R O S = 90^{0}$

$\Rightarrow ∠ R O S = 90^{0} - ∠ P O S$ .............(I)

and, $∠$ ROS + $∠$ ROQ = $∠$ QOS

$∠ R O S = ∠ Q O S - 90^{0}$ ..............(ii)

Adding the eq (i ) and eq (ii), we get,

$∠ ROS = \frac{1}{2} (∠ QOS - ∠ POS)$

Hence, proved.

Question 6: It is given that $∠$ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects $∠$ ZYP, find $∠$ XYQ and reflex $∠$ QYP.

Answer:

1595611505816

Given that,

$∠$ XYZ = $64^{0}$ and XY produced to point P and Ray YQ bisects $∠$ ZYP $\Rightarrow ∠ Q Y P = ∠ Z Y Q$

Now, XYP is a straight line.

So, $∠$ XYZ + $∠$ ZYQ + $∠$ QYP = $180^{0}$

$\Rightarrow 2 (∠ Q Y P) = 180^{0} - 64^{0} = 116^{0}$

$\Rightarrow (∠ Q Y P) = 58^{0}$

Thus reflex of $∠$ QYP = $360^{0} - 58^{0} = 302^{0}$

Since $∠$ XYQ = $∠$ XYZ + $∠$ ZYQ

[ $∵$ $∠ Q Y P = ∠ Z Y Q$

$∠ X Y Q$ = $64^{0} + 58^{0} = 122^{0}$

Class 9 Maths Chapter 6 Question Answer: Exercise: 6.2

Total Questions: 5

Page number: 80-81

Question 1: In Fig. 6.29, if AB $| |$ CD, CD $| |$ EF and y : z = 3 : 7 , find x .

1640149786744

Answer:

1595610656146

Given AB || CD and CD || EF and $y : z = 3 : 7$

$\Rightarrow z = \frac{7}{3} y$

Therefore, AB || EF and $x = z$ (alternate interior angles)..............(i)

Again, CD || AB

$\Rightarrow x + y = 180^{0}$

$\Rightarrow z + y = 180^{0}$ .............(ii)

Putting the value of $z$ in equation (ii), we get,

$\frac{10}{3} y = 180^{0}$

$\Rightarrow y = 54^{0}$

Then $z = 180^{0} - 54^{0} = 126^{0}$

By equation (i), we get the value of $x = 126^{0}$

Question 2: In Fig. 6.30, if AB $| |$ CD, EF $⊥$ CD and $∠$ GED = 126°, find $∠$ AGE, $∠$ GEF and $∠$ FGE.

Answer:

1595610747454

Given AB || CD, EF $⊥$ CD and $∠$ GED = $126^{0}$

In the above figure,

GE is transversal.

So, that $∠$ AGE = $∠$ GED = $126^{0}$

[Alternate interior angles]

Also, $∠$ GEF = $∠$ GED - $∠$ FED = $126^{0} - 90^{0} = 36^{0}$

$\Rightarrow ∠ G E F = 36^{0}$

Since AB is a straight line.

Therefore, $∠$ AGE + $∠$ FGE = $180^{0}$

So, $∠$ FGE = $180^{0} - ∠ A G E = 180^{0} - 126^{0}$

$\Rightarrow ∠ F G E = 54^{0}$

Question 3: In Fig. 6.31, if PQ $| |$ ST, $∠$ PQR = 110° and $∠$ RST = 130°, find $∠$ QRS. [Hint: Draw a line parallel to ST through point R.]

1640149893960

Answer:

Draw a line EF parallel to the ST through R.

Since PQ || ST and ST || EF

$\Rightarrow$ EF || PQ
1595610821237

$∠$ PQR = $∠$ QRF = $110^{0}$ (Alternate interior angles)

$∠$ QRF = $∠$ QRS + $∠$ SRF .............(i)

Again, $∠$ RST + $∠$ SRF = $180^{0}$ (Interior angles of two parallels ST and RF)

$\Rightarrow ∠ S R F = 180^{0} - 130^{0} = 50^{0}$ ( $∠$ RST = $130^{0}$ , given)

Thus, $∠$ QRS = $110^{0} - 50^{0} = 60^{0}$

Question 4: In Fig. 6.32, if AB $| |$ CD, $∠$ APQ = 50° and $∠$ PRD = 127°, find x and y.

1640150269148

Answer:

1595610871500

Given, AB || CD, $∠$ APQ = $50^{0}$ and $∠$ PRD = $127^{0}$

PQ is a transversal.

So,

$∠$ APQ = $∠$ PQR= $50^{0}$ (alternate interior angles)

$∴ x = 50^{0}$

Again, PR is a transversal.

So,

$∠$ y + $50^{0}$ = $127^{0}$ (Alternate interior angles)

$\Rightarrow y = 77^{0}$

Question 5: In Fig. 6.27, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B; the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects along CD. Prove that AB $| |$ CD.

Answer:

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Draw a ray BL $⊥$ PQ and CM $⊥$ RS

Since PQ || RS (Given)

So, BL || CM and BC is a transversal.

$∴$ $∠$ LBC = $∠$ MCB (Alternate interior angles).............(I)

It is known that angle of incidence = angle of reflection

So, $∠$ ABL = $∠$ LBC and $∠$ MCB = $∠$ MCD

$\Rightarrow ∠ A B L = ∠ M C D$ ..................(ii)

Adding equations (i) and (ii), we get,

$∠$ ABC = $∠$ DCB

Both the interior angles are equal

Hence AB || CD

Class 9 Maths NCERT Chapter 6: Extra Question

Question: In the figure, $∠ P O R$ and $∠ Q O R$ form a linear pair. If a° - b° = 90° find the values of a and b.

Answer:

Given: $∠ P O R$ + $∠ Q O R$ = 180°

It means a° + b° = 180°..................(i)

And, a° - b° = 90° ..................(ii)

Now, add (i) and (ii), We get;

2 a° = 270°

Therefore, a° = 135°

And, thus b° = 180° - 135° = 45°

Approach to Solve Questions of Lines and Angles Class 9

1. Understand different types of angles: Acute, obtuse, straight, reflex, and right angles need revision as a starting point to understand angle relationships in lines.

2. Learn angle pair relationships: Students need to understand complementary angles along with supplementary angles, whereas adjacent angles and linear pairs and vertically opposite angles are necessary for solving angle problems.

3. Know the angle properties of intersecting lines: When two lines intersect, the resulting vertically opposite angles will always be equal in measure to each other.

4. Apply properties of parallel lines with a transversal: You will discover the required behaviours of alternate interior angles, corresponding angles, and consecutive interior angles in cases where parallel lines intersect with a transversal.

5. Use angle theorems for reasoning and proof: When proving lines parallel, it is essential to use angle theorems, especially the “If two lines are parallel, alternate interior angles are equal” theorem, together with its converse proof.

6. Solve problems step-by-step using angle relationships: The solution of angle-based questions progresses easily through known angle pairs combined with postulates to create exact answers.

NCERT Solutions for Class 9 Mathematics: Chapter-wise

NCERT Solutions for Class 9 Mathematics Chapter 1: Number Systems

NCERT Solutions for Class 9 Mathematics Chapter 2: Polynomials

NCERT Solutions for Class 9 Mathematics Chapter 3: Coordinate Geometry

NCERT Solutions for Class 9 Mathematics Chapter 4: Linear Equations in Two Variables

NCERT Solutions for Class 9 Mathematics Chapter 5: Introduction to Euclid’s Geometry

NCERT Solutions for Class 9 Mathematics Chapter 6: Lines and Angles

NCERT Solutions for Class 9 Mathematics Chapter 7: Triangles

NCERT Solutions for Class 9 Mathematics Chapter 8: Quadrilaterals

NCERT Solutions for Class 9 Mathematics Chapter 9: Circles

NCERT Solutions for Class 9 Mathematics Chapter 10: Heron's Formula

NCERT Solutions for Class 9 Mathematics Chapter 11: Surface Areas and Volumes

NCERT Solutions for Class 9 Mathematics Chapter 12: Statistics

NCERT Solutions For Class 9: Subject Wise

These are links to the solutions of other subjects, which students can check to revise and strengthen those concepts.

NCERT Books and NCERT Syllabus

Before planning a study schedule, always analyse the latest syllabus. Here are the links to the latest NCERT syllabus and some of the important books that will help students in this cause.

Frequently Asked Questions (FAQs)

1. What are the important topics in NCERT solution for maths chapter 6 class 9?

Pair of angles, Parallel and Transversal Lines, Angle sum property of a triangle are the important topics covered in this chapter. Students can practice NCERT solutions for ch 6 maths class 9 to command these concepts.

2. How many chapters are there in the CBSE class 9 maths ?

There are 15 chapters starting from the number system to probability in the CBSE class 9 maths. which are listed below

Number Systems
Polynomials
Coordinate Geometry
Linear Equations in Two Variables
Introduction to Euclid's Geometry
Lines and Angles
Triangles
Quadrilaterals
Areas of Parallelograms and Triangles
Circles
Constructions
Heron's Formula
Surface Areas and Volumes
Statistics
Probability

3. Where can I find the complete NCERT solutions for class 9 maths chapter 6 ?

Here you will get the detailed NCERT solutions for class 9 maths. For ease students can study lines and angles class 9 pdf both online and offline mode. practicing these solutions help them indepth understanding the cocepts which ultimately lead to score well in the exam.

Also Read

NCERT Solutions for Class 9 Maths Chapter 11 Exercise 11.1 - Surface Area and Volumes

NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.3 - Circles

NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.1 - Circles

NCERT Solutions for Class 9 Maths Chapter 10 Exercise 10.1 - Heron's Formula

NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.2 - Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 8 Exercise 8.1 - Quadrilaterals

NCERT Class 9 Science Notes - Download Chapter-wise PDFs

NCERT Class 9 Notes - Download Subject Wise PDF Notes

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

Triangles Class 9th Notes - Free NCERT Class 9th Maths Chapter 7 Notes - Download PDF

NCERT Exemplar Class 9 Solutions - Subject Wise Problems with Solutions

NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials

NCERT Exemplar Class 9 Maths Solutions Chapter 6 Lines and Angles

NCERT Exemplar Class 9 Maths Solutions Chapter 5 Introduction to Euclids Geometry

NCERT Exemplar Class 9 Science Solutions Chapter 15 Improvement in Food Resources

NCERT Exemplar Class 9 Science Solutions Chapter 14 Natural Resources

NCERT Books for Class 9 Maths 2025-26; Download Free PDF

NCERT Books for Class 9 All Subjects - Download Free PDF

NCERT Books for Class 9 Social Science 2025-26: Download PDF

NCERT Books for Class 9 Hindi 2025-26: Download PDF

NCERT Books for Class 9 English 2025-26: Download All Chapters PDF

NCERT Books for Class 9 Science 2025-26; Download PDF (All Chapters)

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