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NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

Last updated date: 16th Sep 2024
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Complete Resource of NCERT Maths Chapter 6 Lines and Angles Class 9 - Free PDF Download

NCERT Solutions Class 9 Maths Chapter 6, "Lines and Angles," provides an in-depth study of important geometric ideas related to lines and angles. By concentrating on the behaviour of parallel lines, the characteristics of the angles created when a transversal crosses parallel lines, and the different kinds of angles, including adjacent, vertically opposite, complementary, and supplementary, this chapter defines the foundation for understanding deeper geometrical constructions. Check out Vedantu’s Class 9 Maths NCERT Solutions which is solved and explained by maths experts, to help you in your exams.

Table of Content
1. Complete Resource of NCERT Maths Chapter 6 Lines and Angles Class 9 - Free PDF Download
2. Glance of NCERT Solutions for Chapter 6 Maths Class 9: Lines and Angles | Vedantu
3. Access Exercise Wise NCERT Solutions for Chapter 6 Maths Class 9
4. Exercises Under NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles
5. Overview of Deleted Syllabus for Class 9 Maths Chapter 6 Lines and Angles
6. Class 10 Maths Chapter 6: Exercises Breakdown
7. Other Study Material for CBSE Class 9 Maths Chapter 6
8. Chapter-Specific NCERT Solutions for Class 9 Maths
FAQs

Glance of NCERT Solutions for Chapter 6 Maths Class 9: Lines and Angles | Vedantu

• Class 9th Maths Chapter 6 explains the difference between a line segment and a ray, collinear points, and the formation of angles.

• Focus on the connections between the angles that are created when a transversal passes between parallel lines. These connections include corresponding, alternating interior, and alternating exterior angles, all of which are essential for demonstrating that two lines are parallel.

• This chapter focuses more on the study of lines and angles, providing thorough notes on the characteristics of angles, proofs, and the use of geometric theorems like the Linear Pair Postulate and the Vertical Angle Theorem.

• The chapter also explains how vertically opposite angles are formed and how equal measurements are taken when two lines intersect.

• There are chapter notes, important questions, exemplar solutions, exercises links and video tutorials of chapter 6 class 9 explaining lines and angles for better understanding.

• There are two exercises (11 fully solved questions) in Class 9th Math, Chapter 6, Lines And Angles.

Access Exercise Wise NCERT Solutions for Chapter 6 Maths Class 9

 S.No. Current Syllabus Exercises of Class 9 Maths Chapter 6 1 NCERT Solutions of Class 9 Maths Triangles Exercise 6.1 2 NCERT Solutions of Class 9 Maths Triangles Exercise 6.2
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NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles
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Exercises Under NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

1. In the given figure, lines AB and CD intersect at O. if $\angle \text{AOC+}\angle \text{BOE=7}{{\text{0}}^{\text{o}}}$ and $\angle \text{BOD=4}{{\text{0}}^{\text{o}}}$find $\angle \text{BOE}$ and reflex $\angle \text{COE}$

Ans:

AB is a straight line, OC and OE are rays from O.

We know that a straight line covers ${{180}^{\circ }}$

$\text{ }\Rightarrow \angle \text{AOC+}\angle \text{COE+}\angle \text{BOE=18}{{\text{0}}^{\circ }}$

By clubbing $\angle \text{AOC and }\angle \text{BOE}$ together we can rewrite the above equation as

$\Rightarrow \left( \angle \text{AOC+}\angle \text{BOE} \right)+\angle \text{COE=18}{{\text{0}}^{\circ }}$

Putting $\text{ }\angle \text{AOC+}\angle \text{BOE=7}{{\text{0}}^{\circ }}$

$\text{ }\Rightarrow \text{7}{{\text{0}}^{\circ }}+\angle \text{COE=18}{{\text{0}}^{\circ }}$

$\text{ }\Rightarrow \angle \text{COE=18}{{\text{0}}^{\circ }}\text{-}{{70}^{\circ }}$

$\text{ }\Rightarrow \angle \text{COE=11}{{\text{0}}^{\circ }}$

Hence  reflex $\text{ }\angle \text{COE=36}{{\text{0}}^{\circ }}\text{-11}{{\text{0}}^{\circ }}$

$\text{ reflex}\angle \text{COE=25}{{\text{0}}^{\circ }}\text{ }$

Similarly CD is a straight line, OB and OE are rays from O.

We know that a straight line covers ${{180}^{\circ }}$

$\text{ }\Rightarrow \angle B\text{OD+}\angle \text{COE+}\angle \text{BOE=18}{{\text{0}}^{\circ }}$

$\text{ }\Rightarrow \text{ }{{40}^{\circ }}\text{+}{{110}^{\circ }}\text{+}\angle \text{BOE=18}{{\text{0}}^{\circ }}$

$\text{ }\Rightarrow \text{ }\angle \text{BOE=18}{{\text{0}}^{\circ }}\text{-}\left( \text{4}{{\text{0}}^{\circ }}\text{+11}{{\text{0}}^{\circ }} \right)$

$\text{ }\Rightarrow \text{ }\angle \text{BOE=18}{{\text{0}}^{\circ }}\text{-}{{150}^{\circ }}$

$\text{ }\Rightarrow \text{ }\angle \text{BOE=}{{30}^{\circ }}$

Hence $\angle \text{BOE=}{{30}^{\circ }}$ and reflex $\angle \text{COE=25}{{\text{0}}^{\circ }}$

2. In the given figure, lines XY and MN intersect at O. If $\angle \text{POY=9}{{\text{0}}^{\text{o}}}$ and a: b = 2:3, find c.

Ans:

Let the common ratio between a and b be x.

$\therefore$$\text{a = 2x}$ , and $\text{b = 3x}$

XY is a straight line, OM and OP are rays from O.

We know that a straight line covers ${{180}^{\circ }}$

$\angle \text{XOM +}\angle \text{MOP +}\angle \text{POY = 18}{{\text{0}}^{\text{o}}}$

Putting values for $\angle \text{XOM=b and}\angle \text{MOP=a}$

$\Rightarrow \text{ b + a +}\angle \text{POY = 18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ 3x + 2x +}\angle \text{POY = 18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ 5x = 9}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ x = 1}{{\text{8}}^{\text{o}}}$

$\therefore \text{ a = 2x}$

$\Rightarrow \text{ a = 2}\times \text{1}{{\text{8}}^{\circ }}$

$\text{ = 3}{{\text{6}}^{\circ }}$

$\therefore \text{ b = 3x}$

$\Rightarrow \text{ b = 3}\times \text{1}{{\text{8}}^{\circ }}$

$\text{ = 5}{{\text{4}}^{\circ }}$

Similarly MN is a straight line, OX is a ray from O

$\text{ }\therefore \text{b+c=18}{{\text{0}}^{\circ }}$

$\text{5}{{\text{4}}^{\text{o}}}\text{ + c = 18}{{\text{0}}^{\text{o}}}$

$\text{ }c=\text{ }{{180}^{\circ }}\text{ }-\text{ }{{54}^{\circ }}$

$\text{ }c=\text{ 12}{{\text{6}}^{\circ }}$

3. In the given figure,$\angle \text{PQR=}\angle \text{PRQ}$ , then prove that $\angle \text{PQS=}\angle \text{PRT}$.

Ans:

ST is a straight line, QP is a line segment from Q in ST to any point P

by Linear Pair property

$\angle \text{PQS+}\angle \text{PQR=18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{PQR=18}{{\text{0}}^{\text{o}}}\text{-}\angle \text{PQS}$ ......($1$ )

Similarly

$\angle PRT+\angle PRQ={{180}^{\circ }}$

$\Rightarrow \text{ }\angle \text{PRQ=18}{{\text{0}}^{\text{o}}}\text{-}\angle \text{PRT}$ ......($2$ )

Now in the question it is given that $\angle \text{PQR=}\angle \text{PRQ}$

Therefore on equating equation ($1$ ) and ($2$ ) we get

$\text{18}{{\text{0}}^{\text{o}}}\text{-}\angle \text{PQS=18}{{\text{0}}^{\text{o}}}\text{-}\angle \text{PRT}$

$\Rightarrow \text{ }\angle \text{PQS=}\angle \text{PRT}$

Hence proved

4. In the given figure, if $\text{x+y=w+z}$ then prove that AOB is a line.

Ans:

It can be observed that,

Since there are ${{360}^{\circ }}$ around a point therefore we can write

$x+y+z+w={{360}^{\circ }}$

It is given that,

$\text{x+y=w+z}$

Therefore writing$\text{x+y}$in place of $\text{w+z}$so that we can eliminate w and z, we get

$x+y+x+y={{360}^{\circ }}$

$2\left( x+y \right)={{360}^{\circ }}$

$x+y={{180}^{\circ }}$

Since x and y form a linear pair, hence we can say that AOB is a line.

Hence proved

5. In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

$\angle \text{ROS=}\frac{\text{1}}{\text{2}}\left( \angle \text{QOS-}\angle \text{POS} \right)$

Ans:

Since $\text{OR}\bot \text{PQ}$  therefore

$\angle \text{POR=9}{{\text{0}}^{\text{o}}}$

$\angle POS+ROS={{90}^{\circ }}$

$\Rightarrow \text{ }\angle \text{ROS=9}{{\text{0}}^{\text{o}}}\text{-}\angle \text{POS}$ ......   ($1$ )

Similarly $\angle \text{QOR=9}{{\text{0}}^{\text{o}}}$ (Since $\text{OR }\!\!\hat{\ }\!\!\text{ PQ}$)

$\therefore \angle \text{QOS-}\angle \text{ROS=9}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{ROS=}\angle \text{QOS-9}{{\text{0}}^{\text{o}}}$......($2$)

We can clearly see that on adding equation ($1$ ) and ($2$) ${{90}^{\circ }}$ get canceled out

$\Rightarrow \text{2}\angle \text{ROS=}\angle \text{QOS-}\angle \text{POS}$

Which can easily be written as

$\Rightarrow \angle \text{ROS=}\frac{\text{1}}{\text{2}}\left( \angle \text{ROS-}\angle \text{POS} \right)$

Hence proved

6. It is given that $\text{XYZ=6}{{\text{4}}^{\text{o}}}$ and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects $\text{ZYP}$, find $\text{XYQ}$ and reflex$\text{QYP}$.

Ans:

It is given that line YQ bisects $\angle \text{ZYP}$.

Hence, $\angle \text{QYP=}\angle \text{ZYQ}$

It can easily be understood that PX is a line, YQ and YZ being rays standing on it.

$\angle \text{XYZ+}\angle \text{ZYQ+}\angle \text{QYP=18}{{\text{0}}^{\text{o}}}$

From above relation $\angle \text{QYP=}\angle \text{ZYQ}$ we can write

$\text{6}{{\text{4}}^{\text{o}}}\text{+2}\angle \text{QYP=18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ 2}\angle \text{QYP=18}{{\text{0}}^{\text{o}}}\text{-6}{{\text{4}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{QYP=5}{{\text{8}}^{\circ }}$

Therefore $\angle \text{ZYQ=5}{{\text{8}}^{\circ }}$

Also Reflex $\angle \text{QYP=30}{{\text{2}}^{\circ }}$

Now we can write $\angle \text{XYQ}$ as below

$\angle \text{XYQ=}\angle \text{XYZ+}\angle \text{ZYQ}$

${{64}^{\circ }}+{{58}^{\circ }}={{122}^{\circ }}$

Therefore we found $\angle \text{XYQ=12}{{\text{2}}^{\circ }}$ and so the Reflex $\angle \text{QYP=30}{{\text{2}}^{\circ }}$

Exercise-6.2

1. In the given figure, if AB $\parallel$ CD, CD $\parallel$ EF and y: z = 3:7, find x.

Ans:

It is given that AB$\parallel$ CD and CD $\parallel$ EF

$\therefore$ AB$\parallel$CD$\parallel$ EF (Lines parallel to other fixed line are parallel to each other)

It can easily be understood that

$\text{x=z}$  (since alternate interior angles are equal) ...... ($1$)

It is given that y: z = 3: 7

Without any loss of generality we can let $\text{y=3a}$and $\text{z=7a}$

Also, $\text{x+y=18}{{\text{0}}^{\text{o}}}$ (Co-interior angles together sum up to ${{180}^{\circ }}$)

From equation ($1$) we can write z in place of x as shown

$\text{z+y=18}{{\text{0}}^{\text{o}}}$

It can further written as shown

$7a+3a={{180}^{\circ }}$

$\Rightarrow \text{ 10a=18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ a=1}{{\text{8}}^{\text{o}}}$

$\therefore \text{ x=7 }\!\!\times\!\!\text{ 1}{{\text{8}}^{\text{o}}}$

$\therefore \text{ x=12}{{\text{6}}^{\text{o}}}$

2. In the given figure, If AB$\parallel$CD, EF$\bot$ CD and$\angle \text{GED=12}{{\text{6}}^{\text{o}}}$ , find$\angle \text{AGE}$ , $\angle \text{GEF}$and $\angle \text{FGE}$.

Ans:

We are given that,

AB $\parallel$ CD and EF$\bot$ CD and

$\text{ }\angle \text{GED=12}{{\text{6}}^{\text{o}}}$

Which can be written as

$\Rightarrow \angle \text{GEF+}\angle \text{FED=12}{{\text{6}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{GEF+9}{{\text{0}}^{\text{o}}}\text{=12}{{\text{6}}^{\text{o}}}$

Hence we can obtain $\text{GEF}$as shown below

$\Rightarrow \text{ }\angle \text{GEF=12}{{\text{6}}^{\text{o}}}\text{-9}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{GEF=3}{{\text{6}}^{\circ }}$

$\angle \text{AGE=}\angle \text{GED=12}{{\text{6}}^{\text{o}}}$ (∠AGE and ∠GED are alternate interior angles)

But

$\angle \text{AGE+}\angle \text{FGE=18}{{\text{0}}^{\text{o}}}$ (because these form a linear pair)

$\Rightarrow \text{12}{{\text{6}}^{\text{o}}}\text{+}\angle \text{FGE=18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \angle \text{FGE=18}{{\text{0}}^{\text{o}}}\text{-12}{{\text{6}}^{\text{o}}}$

$\Rightarrow \angle \text{FGE=5}{{\text{4}}^{\text{o}}}$

Hence, we found $\angle \text{AGE=12}{{\text{6}}^{\text{o}}}$, $\angle \text{GEF=3}{{\text{6}}^{\text{o}}}$ , $\angle \text{FGE=5}{{\text{4}}^{\text{o}}}$

3. In the given figure, if PQ $\parallel$ ST, $\angle \text{PQR=11}{{\text{0}}^{\text{o}}}$ and$\angle \text{RST=13}{{\text{0}}^{\text{o}}}$ , find $\angle \text{QRS}$.

(Hint: Draw a line parallel to ST through point R.)

Ans:

In this question we will have some construction of our own, we draw a line XY parallel to ST and so parallel to PQ passing through point R.

$\angle \text{PQR+}\angle \text{QRX=18}{{\text{0}}^{\text{o}}}$ (Co-interior angles on the same side of transversal QR together sum up to ${{180}^{\circ }}$)

$\therefore \text{11}{{\text{0}}^{\text{o}}}\text{+}\angle \text{QRX=18}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{ }\angle \text{QRX=7}{{\text{0}}^{\text{o}}}$

Also,

$\angle \text{RST+}\angle \text{SRY=18}{{\text{0}}^{\circ }}$  (sum of Co-interior angles on the same side of transversal SR equals $\text{18}{{\text{0}}^{\circ }}$)

$\Rightarrow \text{ }\angle \text{SRY=18}{{\text{0}}^{\circ }}\text{-13}{{\text{0}}^{\circ }}$

$\Rightarrow \text{ }\angle \text{SRY=}{{50}^{\circ }}$

Now from the construction XY is a straight line. RQ and RS are rays from it.

$\angle \text{QRX+}\angle \text{QRS+}\angle \text{SRY=18}{{\text{0}}^{\circ }}$

$\Rightarrow \text{ 7}{{\text{0}}^{\circ }}+\angle QRS+{{50}^{\circ }}={{180}^{\circ }}$

Hence we found that

$\angle \text{QRS=6}{{\text{0}}^{\circ }}$

4. In the given figure, if AB $\parallel$ CD, $\angle APQ={{50}^{\circ }}$ and$\angle PRD={{127}^{\circ }}$ , find x and y.

Ans:

$\angle APR=PRD$ (since alternate interior angles are equal)

$\therefore \text{5}{{\text{0}}^{\circ }}+y={{127}^{\circ }}$ (since $\angle APR=\angle \text{APQ+}\angle \text{PQR}$)

$\Rightarrow \text{y=7}{{\text{7}}^{\circ }}$

similarly,

$\angle \text{APQ=PQR}$ ∠APQ = ∠PQR (since alternate interior angles are equal)

$\therefore \text{x=5}{{\text{0}}^{\circ }}$

Therefore we found that $\text{x=5}{{\text{0}}^{\circ }}$ and $\text{y=7}{{\text{7}}^{\circ }}$

5. In the given figure, 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 atC and again reflects back along CD. Prove that AB||CD.

Ans:

Let us construct BM$\bot$ PQ and CN$\bot$RS.

Since  PQ$\parallel$ RS, and so BM $\parallel$ CN

Therefore, CN and BM are two parallel lines and a transversal line BC cuts them at B and C respectively.

∠2 = ∠3......($1$) (since alternate interior angles are equal)

But, by laws of reflection in Physics

∠1 = ∠2 and ∠3 = ∠4

Now from equation ($1$)

∠1 = ∠2 = ∠3 = ∠4

Therefore

∠1 + ∠2 = ∠3 + ∠4

$\angle \text{ABC=}\sum \text{DCB}$

But, these are alternate interior angles.

$\therefore$ AB$\parallel$ CD

Hence  proved.

Overview of Deleted Syllabus for Class 9 Maths Chapter 6 Lines and Angles

 Chapter Dropped Topics Lines and Angles Exercise 6.5 6.5 - Parallel lines and a transversal 6.7 - Angle sum property of a triangle.

Class 10 Maths Chapter 6: Exercises Breakdown

 Exercise Number of Questions Exercise 6.1 6 Questions and Solutions (5 Short, 1 Long Question) Exercise 6.2 5 Questions and Solutions

Conclusion

NCERT Solutions for Chapter 6 Class 9 Math Lines and Angles, covers important concepts and theorems for mastering basic geometry. The chapter emphasizes understanding how lines and angles interact and proving lines parallel using corresponding, alternate, and co-interior angles. Mastering these concepts is crucial for future mathematics topics. The detailed solutions help students approach and solve geometric problems, enhancing their analytical and problem-solving skills. Mastering this chapter is essential not only for exams but also for a deeper understanding of geometry.

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles

1. How can NCERT Solutions for Class 9 Chapter 6 Maths Help in the Exam Preparation?

NCERT Solutions for Class 9 Chapter 6 make comprehensive study material and cover the concepts of the entire chapter to help students build a strong foundation. These solutions follow the latest CBSE curriculum and make students familiar with the types of questions being asked from the lines and angles chapter. Students can get in-depth knowledge about the topic, brief summary, solved questions, and more. Apart from reasoning and logical skills, these chapter solutions offer comprehensive learning to students. The difficult topics like types of angles, interior, and exterior angles are explained in a better way. Students can clear their tough concepts and can analyze their weak areas with the guidance of NCERT solutions class 6.

2. How are the Vertically Opposite Angles Equal when Two Lines Intersect each Other?

Let AB and CD be two lines intersecting at a single point, say, P. APC and BPD, and APD and BOC form two pairs of vertically opposite angles. Additionally, ray AP is perpendicular on line CD. It implies the sum of APC and APD is equal to 180o. Now, the sum of angles APD and BPD is equal to 180o. By using these two equations, it’s clear that the sum of angles APC and APD is equal to the sum of APD and BPD angles. It implies angle APC is equal to BPD. Similarly, it can be shown that angle APD is equal to angle BPC.

3. Are all the exercises important in Chapter 6 of Class 9 Maths?

All exercises of NCERT Class 9 Maths Chapter 6 are designed in a well-structured hierarchy of difficulty to test the student’s ability in problem-solving. They also help him or her bridge the knowledge gap. Practising all the exercises is of utmost importance to score well and to clearly understand all the necessary concepts. Hence it is advisable to not skip any exercise. All the solutions are also easily available on Vedantu website and the app.

4. What is the best Solution book for Chapter 6 of NCERT Class 9 Maths?

Vedantu is the place where you will get the solutions easily and for free. To download the solutions for Chapter 6 of NCERT Class 9 Maths, follow the steps mentioned below:

1. Visit the page of Chapter 6 of NCERT Solutions for Class 9 Maths.

2. The website for solutions for Chapter 6 of Class 9 Maths will open.

3. The solutions for the selected chapter will be displayed on the screen. One can also download these solutions by clicking on Download PDF to save it offline.

For more information regarding this, you can visit the Vedantu site or app and avail of their services to crack your entrances and perform well in all of your exams.

5. What do you mean by collinear and non-collinear points?

The collection of three or more points that occur on the same straight line is known as collinear points. On multiple planes, collinear points may occur, but not on different lines. Collinearity is the property of points being collinear. As a result, any three or more points are only collinear if they are in the same straight line. If any of the points are not on the same line, they are considered non-collinear as a group. The size of the triangle formed by the three non-collinear points will always be greater than zero.

6. What are the types of lines in Chapter 6 of Class 9 Maths?

In Geometry, there are two types of lines: straight and curved. Straight lines are divided into two categories: horizontal and vertical. Parallel, intersecting, and perpendicular lines are examples of other types of lines. Knowledge of all these types and the formulae related to them are very crucial and hence you must practice them judiciously.

7. What do you mean by a vertex?

A vertex is a point in geometry where two or more curves, lines, or edges intersect. As a result of this definition, vertices can be described as the junction where two separate lines tend to meet and make an angle, as well as the corners of polygons and polyhedra. In mathematics, it is an angular corner formed by the intersection of two or more lines or edges between two faces. A vertex may be found in 2D shapes such as pentagons and squares, as well as 3D shapes such as pyramids and cuboids.

8. What are lines and angles in brief?

Lines and angles are fundamental elements of geometry, forming infinitely long, straight paths. Understanding different line types and angles is crucial. Vedantu's solutions provide clear explanations, step-by-step breakdowns, and practice problems to help students understand these concepts, laying a strong foundation for future geometrical adventures.

9. What are the types of Line and Angles Class 9?

In the world of geometry of lines and angles class 9 solutions, angles come in various shapes and sizes, just like slices of pie! Acute angles are sharp turns less than 90 degrees, like a thin slice. Right angles are perfect 90-degree corners, like a closed book. Obtuse angles make wider turns between 90 and 180 degrees, like a large slice. Finally, straight angles span 180 degrees, forming a straight line when two lines extend opposite each other.

10. What's covered in the exercises of Maths Chapter 6 Class 9?

• Lines & Angles Identification: These exercises will teaches you to recognize lines (rays, segments) and various angles (acute, right, obtuse, straight).

• Measurement and Calculations: This exercise will make you become skilled with a protractor and be able to solve issues with missing angle measures, addition, and subtraction of angles.

• Real-World Application: Prepare yourself to solve word problems that evaluate your skills with lines and angles.