# NCERT Solutions For Class 9 Maths Chapter 6 Lines And Angles (Ex 6.1) Exercise 6.1

Class 9 students do their best to ace mathematics examinations. There are a lot of tips that can help make the process a bit easier. One of the most important things is to be sure to understand how to use the theorems and formulas in the particular question that are covered in the textbook. This can be quickly done by reviewing the materials recommended by Extramarks as well as doing any practice problems that are available in the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1. In addition, following the procedures and methods that are used in the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, can be very beneficial for students who want to score high in their exams.

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Students go through several challenges and problems while solving Mathematical questions. To overcome such challenges students can take help from the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 which is proposed by Extramarks.

**NCERT Solutions For Class 9 Maths Chapter 6 Lines and Angles (Ex 6.1) Exercise 6.1 **

In the educational system of India, NCERT textbooks are recognised as a comprehensive guidebook for students’ study material. Among all the essential elements of the Indian educational system, the NCERT curriculum is the most important and is treated as a curriculum followed by most students. Its goal is to provide students with a complete understanding of the subjects they are studying in their academic courses. The NCERT curriculum may be profitable to students in various ways. Giving students a firm understanding of the topics it covers is the curriculum’s principal goal. Second, the syllabus is based on CBSE, which is used by educational institutions across the nation (India). The NCERT curriculum results in students being more equipped to do well on the CBSE exams. To incorporate the most recent changes to the syllabus, the NCERT curriculum is frequently modified. Last but not least, the curriculum is delivered in a manner that is approachable and interesting to students, making learning an enjoyable and exciting experience. However, Extramarks provides materials for NCERT Solutions, such as NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, in PDF format for students to download for later usage.

**NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.1) Exercise 6.1**

NCERT Solutions are a collection of solutions created by Extramarks’ experts to aid students in comprehending the ideas presented in the council-recommended textbooks. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 are created in line with the most recent CBSE syllabus and are accessible for students studying in classes 6 through 12. Both Hindi and English versions are available for them.

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**NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.1) Exercise 6.1**

Students will study Basic terms and definitions of Lines and Angles as well as other vital topics in Chapter 6, Exercise 6.1 of Class 9 in Mathematics subject. The following themes should be well prepared by students while solving Exercise 6.1:

- Terms and Definitions – Line Segment, Ray, Collinear and Non-collinear points
- Angles – Arms and Vertexes
- Types of Angles (7) – Acute Angle, Right Angle, Obtuse Angle, Straight Angle, Reflex Angle, Complementary Angle, and Supplementary Angle
- Linear Pair of Angles – Vertically opposite angles
- Intersecting Lines and Non-Intersecting Lines
- Pair of Angles – Linear (Theorems and Examples)

Exercise 6.1 has 6 questions in total that can be solved with the guidance of NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 emphasised by Extramarks.

**Chapter Wise NCERT Solutions For Class 9 Maths**

The chapter-wise NCERT Solutions for Class 9 can be found by students on the Extramarks mobile application or website. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, are a great resource for students to understand the concepts of Mathematics and to prepare for exams. These NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, are easy to follow and will help students score better grades. For students’ ease, Extramarks has provided the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 in PDF format, so that, students can download them and refer to them whenever required. Exercise 6.1 Class 9 Solutions, assists students in understanding the questions mentioned in Class 9 Maths Chapter 6 Exercise 6.1. The chapters that are covered in the Class 9 Mathematics subject course are as follows:

- Chapter 1 – Number Systems
- Chapter 2 – Polynomials
- Chapter 3 – Coordinate Geometry
- Chapter 4 – Linear Equations in Two Variables
- Chapter 5 – Introduction to Euclid’s Geometry
- Chapter 6 – Lines and Angles
- Chapter 7 – Triangles
- Chapter 8 – Quadrilaterals
- Chapter 9 – Areas of Parallelograms and Triangles
- Chapter 10 – Circles
- Chapter 11 – Constructions
- Chapter 12 – Heron’s Formula
- Chapter 13 – Surface Areas and Volumes
- Chapter 14 – Statistics
- Chapter 15 – Probability

Students can explore solutions for the above-mentioned Chapters by visiting the Extramarks website. On the other hand, if students are looking for a better understanding of Mathematical concepts in Lines and Angles and want to score well in the exams, they should check out Class 9 Maths Chapter 6 Exercise 6.1 Solution. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 are easy to follow and will help students to understand the concepts sufficiently.

**Class 9 Maths Chapter 6 Includes:**

In Class 9, Chapter 6 of Mathematics is all about Lines and Angles as the name of the chapter suggests. In this chapter, students will learn how to solve questions regarding Line segments, Types and Pairs of Angles, Intersecting and Non-Intersecting Lines, Parallel Lines and Transversal, Lines Parallel to the same lines, and other topics. These topics have several exercises incorporated into them that students can practice and get a deeper understanding of it. Similarly, if students are not able to solve any particular question, they can take help from the NCERT Solutions, like the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 advised by Extramarks.

**NCERT Solutions for Class 9**

Class 9 students will find the NCERT Solutions for Exercises questions from the Extramarks website which provides a comprehensive guide for completing their assignments and preparing for examinations. The solutions similar to NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 are well-organized and easy to understand, making them an ideal resource for students. The Exercises in the Class 9 Mathematics NCERT textbook have many questions that need a helping hand for students. The questions are carefully selected to cover all the important concepts. So, students must solve all the exercises carefully. They will have a more in-depth understanding of the ideas and will be able to study well for their examinations as a result. Students should use the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, as a great resource since they offer clear instructions and directions for finishing assignments and getting ready for exams. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, delivered by Extramarks are encouraged by academic scholars and subject-matter professionals.

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

**Ans**

$\begin{array}{l}\mathrm{Given}:\text{Lines AB and CE itersect at O and}\\ \angle \mathrm{AOC}+\angle \mathrm{BOE}=70\mathrm{\xb0}\text{and}\angle \mathrm{BOD}=40\mathrm{\xb0}\\ \mathrm{To}\text{find:}\angle \mathrm{BOE}\text{and reflex\hspace{0.17em}}\angle \mathrm{COE}.\\ \angle \mathrm{AOB}\text{is a straight angle, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=180\mathrm{\xb0}\\ \angle \mathrm{AOC}+\angle \mathrm{COE}+\angle \mathrm{BOE}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}70\mathrm{\xb0}+\angle \mathrm{COE}=180\mathrm{\xb0}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{COE}=180\mathrm{\xb0}-70\mathrm{\xb0}=110\mathrm{\xb0}\\ \angle \mathrm{COD}\text{is a straight angle, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{COD}=180\mathrm{\xb0}\\ \angle \mathrm{COE}+\angle \mathrm{EOB}+\angle \mathrm{BOD}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}110\mathrm{\xb0}+\angle \mathrm{BOE}+40\mathrm{\xb0}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOE}=180\mathrm{\xb0}-150\mathrm{\xb0}\\ =30\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Reflex}\text{}\angle \mathrm{COE}=360\mathrm{\xb0}-\angle \mathrm{COE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=360\mathrm{\xb0}-110\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=250\mathrm{\xb0}\end{array}$

**Q.2 **In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and *a *: *b *= 2 : 3, find *c*.

**Ans**

$\begin{array}{l}\mathrm{Given}:\text{ Lines XY and MN intersect at O.}\angle \text{POY}=90\mathrm{\xb0}\text{and a:b}=2:3\\ \mathrm{To}\text{find: c}\\ \text{Since, a:b = 2:3}\\ \text{so, let a}=\text{2x, b}=\text{3x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{POX}+\angle \mathrm{POY}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\mathrm{b}+90\mathrm{\xb0}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+3\mathrm{x}+90\mathrm{\xb0}=180\mathrm{\xb0}\\ 5\mathrm{x}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{180\mathrm{\xb0}}{5}=30\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=2\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\times 30\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60\mathrm{\xb0}\\ \mathrm{MN}\text{is a straight line, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{MOX}+\angle \mathrm{XON}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}}60\mathrm{\xb0}+\mathrm{c}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=180\mathrm{\xb0}-60\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=120\mathrm{\xb0}\end{array}$

**Q.3** In Fig. 6.15, ∠ PQR = ∠ PRQ, then prove that ∠PQS = ∠PRT.

**Ans**

$\begin{array}{l}\mathrm{Given}:\text{In}\mathrm{\Delta}\text{PQR,\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR=}\angle \text{PRQ}\\ \text{To Prove:}\angle \text{PQS=}\angle \text{PRT}\\ \text{Proof}:\text{}\angle \text{PQR}+\angle \text{PQS}=180\mathrm{\xb0}\left[\mathrm{Linear}\text{pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}=180\mathrm{\xb0}-\angle \text{PQS\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PRQ}+\angle \text{PRT}=180\mathrm{\xb0}\left[\mathrm{Linear}\text{pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PRQ}=180\mathrm{\xb0}-\angle \text{PRT\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{But}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}=\angle \text{PRQ}\left[\mathrm{Given}\right]\\ \Rightarrow 180\mathrm{\xb0}-\angle \text{PQS}=180\mathrm{\xb0}-\angle \text{PRT}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQS}=\angle \text{PRT}\\ \text{Hence Proved.}\end{array}$

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

**Ans**

$\begin{array}{l}\mathrm{Given}:\text{x}+\mathrm{y}=\mathrm{w}+\text{z}\\ \mathrm{To}\text{Prove: AOB is a line.}\\ \text{Proof:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\because \mathrm{x}+\mathrm{y}+\mathrm{w}+\mathrm{z}=360\mathrm{\xb0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Complete}\text{ angle}\right]\\ \mathrm{x}+\mathrm{y}+\mathrm{x}+\mathrm{y}=360\mathrm{\xb0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}[\because \text{x}+\mathrm{y}=\mathrm{w}+\text{z}]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2(\mathrm{x}+\mathrm{y})=360\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\mathrm{y}=\frac{360\mathrm{\xb0}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{AOB\hspace{0.17em}}=180\mathrm{\xb0}\\ \Rightarrow \mathrm{Since},\angle \text{AOB is a straight angle, so AOB is a straight line.}\end{array}$

**Q.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 = (1/2)(∠QOS – ∠POS).

**Ans**

$\begin{array}{l}\mathrm{Given}:\text{ POQ is a line.}\overrightarrow{\mathrm{OR}}\perp \mathrm{PQ}\text{and}\overrightarrow{\mathrm{OS}}\text{is another ray between}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\overrightarrow{\mathrm{OR}}\text{\hspace{0.17em}and}\overrightarrow{\mathrm{OP}}.\\ \mathrm{To}\text{Prove:}\angle \mathrm{ROS}\text{}=\text{}\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})\\ \text{Proof:R.H.S.}=\text{}\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\{(\angle \mathrm{QOR}+\angle \mathrm{ROS})-(\angle \mathrm{ROP}-\angle \mathrm{ROS})\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\{(90\mathrm{\xb0}+\angle \mathrm{ROS})-(90\mathrm{\xb0}-\angle \mathrm{ROS})\}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\angle \mathrm{QOR}=\angle \mathrm{ROP}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\mathrm{\xb0}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}(\angle \mathrm{ROS}+\angle \mathrm{ROS})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}(2\angle \mathrm{ROS})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\angle \mathrm{ROS}\end{array}$

**Q.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.

**Ans**

$\begin{array}{l}\mathrm{Given}:\text{}\angle \text{XYZ}=\text{64\xb0, XY is}\mathrm{}\text{produced to point P.Ray YQ bisects}\\ \angle \text{ZYP.}\\ \text{To find:}\angle \text{XYQ and reflex}\angle \text{QYP}.\\ \text{}\mathrm{Since},\text{}\overrightarrow{\text{YQ}}\text{bisects}\angle \text{ZYP, so}\angle \text{ZYQ}=\angle \text{PYQ}=\mathrm{a}\left(\mathrm{let}\right)\\ \text{\hspace{0.17em}}\angle \text{XYZ}+\angle \text{ZYP}=\text{180\xb0\hspace{0.17em}}\left[\mathrm{Linear}\text{pair of angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}64\mathrm{\xb0}+2\mathrm{a}=180\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}=180\mathrm{\xb0}-64\mathrm{\xb0}\\ =116\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{116\mathrm{\xb0}}{2}\\ =58\mathrm{\xb0}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{XYQ}=64\mathrm{\xb0}+\mathrm{a}\\ =64\mathrm{\xb0}+58\mathrm{\xb0}\\ =122\mathrm{\xb0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}reflex}\angle \text{QYP}=360\mathrm{\xb0}-\angle \text{QYP}\\ =360\mathrm{\xb0}-58\mathrm{\xb0}\\ =302\mathrm{\xb0}\end{array}$

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### 1. Are the NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 suggested by Extramarks available in the Hindi language as well?

Numerous students find it problematic to study Hindi solutions, while a similar number find studying English solutions difficult. However, Extramarks makes it simple for students having such dilemma, as it provides the NCERT Solutions in both the Hindi and the English versions. The NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 in Hindi are often more extensive and go into greater depth of knowledge about the topics stated in the chapter than the English versions. This is one of the primary variations between the two languages in NCERT Solutions. Contrary to that, the English NCERT Solutions tend to be shorter and simpler to read. If students are seeking a summary of the subject, English solutions can be an excellent choice. Therefore, both of the versions that are offered on the Extramarks website and mobile application must be tested out by students. An example of such a study guide can be NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, which are available on the Extramarks website.

### 2. What are the advantages of using NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1, and how can they assist students in preparing for their exams?

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