# NCERT Solutions for Class 9 Mathematics Chapter 6- Lines and angles

Mathematics is becoming more and more significant  due to its wide range of applications in our daily lives. The scope of the subject is also increasing due to rapid demand for engineering-related fields where professionals like coders, site engineers etc., use Mathematics in various ways to interpret their results. It helps us to improvise our reasoning, analytical skills  and  problem solving skills.Hence, it is turning out to be a crucial subject in the primary and secondary syllabus.

Lines and angles are more of a Geometry oriented chapter. This  chapter and NCERT Chapter 5  are  directly interrelated.. Hence, one should cover NCERT Chapter 5 before starting Class 9 Mathematics Chapter 6. The core concepts of this chapter include an introduction to various geometrical instruments like point, line, ray, line segment etc., which will be used greatly in higher classes. .

Extramarks is one the  leading online educational  platforms , and has a repository of  study materials for students which are well-renowned and sought after in the industry. Our academic research team has prepared the study materials based on the latest CBSE syllabus and as per the NCERT guidelines. Lakhs of students rely on authentic and reliable study material provided by Extramarks.   These notes are meticulously prepared, concise, and to the point for  the students to be well prepared and confident ahead of the exams. No wonder students have complete faith and trust in Extramarks.

We have included definitions of all the basic concepts along with examples in our NCERT Solutions for Class 9 Mathematics Chapter 6. The way the solutions are written will help you to grasp the topics completely and will lay a strong foundation for Mathematics in  higher classes. .. Students are advised to make full use of resources to get the most of it.

Extramarks’ website is the perfect platform for all the study material for primary and secondary grades. You can get a detailed analysis of the chapter as well as structurally formed solutions on our website easily.

## Key Topics Covered In NCERT Solutions for Class 9 Mathematics Chapter 6

In the last chapter, you have studied the introduction to Euclid’s Geometry which covers all the basic concepts, definitions, axioms and postulates required in the study of Geometry. In this chapter, you will  learn how to apply all the fundamental knowledge from NCERT Chapter 5 to the  geometrical tools and assets of NCERT Chapter 6. In short, you will learn  how all the terminologies are connected with one another.

It is hard to distinguish and recognise different Geometrical  tools if students don’t know what definitions and keywords to remember properly. As a result, most of the students find it difficult to cope with this chapter and for some it’s a nightmare.  We have designed NCERT Solutions for Class 9 Mathematics Chapter 6 to deal with the challenging and advanced concepts with problem solving and reasoning skills available  on the Extramarks’ website.

After completing this chapter, students will be able to deduce proof on their own and thereby find significant improvement in their  reasonable and analytical thinking skills.

Introduction

You have already learnt in Chapter 5 that two points make up a line. In this chapter, you will get to know about the properties of the angles formed when two lines intersect each other and the properties of the angles formed when a line intersects two or more parallel lines at distinct points. Further, you will use these properties to logically interpret some statements.

Basic terms and definitions

The basic terms and definitions related to straight lines, line segments, and angles are included in this section. Also, you will learn about the different types of angles which have been covered in detail in our NCERT Solutions for Class 9 Mathematics Chapter 6.

Intersecting lines and Non-intersecting lines

When two lines are drawn in such a way that they cut each other at a single point, they are called  intersecting lines, whereas when two lines are drawn in such a way that they don’t cut each other at a single point, they are called Non-intersecting lines. You can refer to examples given in the NCERT textbook based on this concept.

Pair of Angles

In this section, you will read about the different types of angles like complementary angles, supplementary angles, interior angles, linear pairs of angles, adjacent angles etc.

You can get to know about each pair of angles in detail, along with their axioms and theorems, in the NCERT Solutions for Class 9 Mathematics Chapter 6.

Parallel Lines and a Transversal

When two lines are drawn opposite to each other, they are known as parallel lines. When a line intersects two or more lines at distinct points, it is called a Transversal.

The pair of parallel lines and a  transversal aids in formulating different types of angles. The angles formed are corresponding angles, alternate angles etc. You can read about each angle in detail in the NCERT Solutions for Class 9 Mathematics Chapter 6.

Line Parallel to the Same Line

Is it necessary that if two lines are parallel to each other, the third line will also be parallel to these lines. This is what you will read about in this section.

Refer to NCERT Solutions for Class 9 Mathematics Chapter 6 for an in-depth explanation.

Angle sum property of a triangle

In this section, you will get knowledge of two important theorems of Geometry with a few examples. The theorems covered in this section are-

Theorem 1: The sum of the angles of a triangle is 180°.

Theorem 2: If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

### NCERT Solutions for Class 9 Mathematics Chapter 6 Exercise &  Solutions

You can find solutions to all the questions in the NCERT textbook in our NCERT Solutions for Class 9 Mathematics Chapter 6. We have included answers to all the exercises in Class 9 Mathematics Chapter 6. Also, you will find references for solving the same questions in many different ways. You can opt for the method that you understand the best to find quick and accurate answers. Get the NCERT Solutions on the Extramarks’ website now and  prepare ahead of the examinations to get excellent results.

The following links have exercise-specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 6:

•  Chapter 6: Exercise 6.1 Question and answers
•  Chapter 6: Exercise 6.2 Question and answers
• Chapter 6: Exercise 6.3 Question and answers

Along with NCERT Solutions for Class 9 Mathematics Chapter 6, students can explore NCERT Solutions on our Extramarks website for all primary and secondary classes.

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT solutions Class 10
• NCERT solutions Class 11
• NCERT solutions Class 12

#### NCERT Exemplar for Class 9 Mathematics

NCERT Exemplar is a set of books specially designed for various competitive examinations. After completing the chapter from the NCERT textbooks, students can refer to NCERT Exemplar for Class 9 Mathematics for further  practice.

The book contains extra questions from the NCERT Class 9 Mathematics textbook. The questions cover the entire syllabus whether it’s the in-text or end- text questions, students will get enough practice to improvise their mathematical skills. . The difficulty of the questions is from basic to advanced, thereby making students capable of solving  l different types of questions.

When students include NCERT Solutions and NCERT Exemplar in their study material, they can be rest assured  that nothing remains untouched and every example solution, exercise has been covered and hence they are  confident of their preparation to ace the exam with excellent results.  NCERT Exemplar for Class 9 Mathematics can easily be accessed from the Extramarks’ website.

#### Key Features of NCERT Solutions for Class 9 Mathematics Chapter 6

You can’t excel and see yourself among the top rankers  if you don’t have the right resources. Hence, NCERT Solutions for Class 9 Mathematics Chapter 6 focuses on the fact that you have the best resources. The following are the  key features: :

• Question banks compromise   the variety  of questions with varying degrees  of difficulty.
•  Important points  highlighted   to brush up on all the previous concepts you have learnt so far.
• Points to ponder at the end of the chapter to quickly revise before  examinations.

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{°}\text{and}\angle \mathrm{BOD}=40\mathrm{°}\\ \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{°}\\ \angle \mathrm{AOC}+\angle \mathrm{COE}+\angle \mathrm{BOE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}70\mathrm{°}+\angle \mathrm{COE}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{COE}=180\mathrm{°}-70\mathrm{°}=110\mathrm{°}\\ \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{°}\\ \angle \mathrm{COE}+\angle \mathrm{EOB}+\angle \mathrm{BOD}=180\mathrm{°}\\ \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{°}+\angle \mathrm{BOE}+40\mathrm{°}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOE}=180\mathrm{°}-150\mathrm{°}\\ =30\mathrm{°}\\ \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{°}-\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{°}-110\mathrm{°}\\ \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{°}\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{°}\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{°}\\ \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{°}=180\mathrm{°}\\ \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{°}=180\mathrm{°}\\ 5\mathrm{x}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{180\mathrm{°}}{5}=30\mathrm{°}\\ \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}}=2×30\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60\mathrm{°}\\ \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{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}60\mathrm{°}+\mathrm{c}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=180\mathrm{°}-60\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=120\mathrm{°}\end{array}$

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

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

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

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

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-

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

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-

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

Q.7 In Fig. 6.28, find the values of x and y and then show that AB || CD. Ans-

$\begin{array}{l}\mathrm{x}+50\mathrm{°}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{pair of angles}\right]\\ ⇒\mathrm{x}=180\mathrm{°}-50\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=130\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}130\mathrm{°}=\mathrm{y}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Vertical}\text{opposite angles.}\right]\\ \mathrm{Since},\text{​ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}=\text{y}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}\parallel \text{CD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Alternate}\text{interior angles}\right]\end{array}$

Q.8 In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x. Ans-

$\begin{array}{l}\mathrm{Since},\text{y:z}=\text{3:7}\\ \mathrm{Let}\text{y}=3\mathrm{a}\text{and z}=\text{7a}\\ \text{Since,}\mathrm{AB}\parallel \mathrm{CD}\text{and CD}\parallel \mathrm{EF}\\ \mathrm{so},\text{}\mathrm{AB}\parallel \mathrm{EF}\\ \mathrm{x}=\mathrm{z}\left[\mathrm{Alternate}\text{interior angles.}\right]\\ \because \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}\parallel \mathrm{CD}\\ \mathrm{x}+\mathrm{y}=180\mathrm{°}\left[\text{Cointerior angles}\right]\\ \mathrm{z}+\mathrm{y}=180\mathrm{°}\left[\because \mathrm{x}=\mathrm{z}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}7a}+3\mathrm{a}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10\mathrm{a}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{180\mathrm{°}}{10}=18\mathrm{°}\\ \mathrm{So},\text{x}=\text{7a}\\ =\text{7}\left(18\mathrm{°}\right)\\ =126\mathrm{°}\end{array}$

Q.9 In Fig. 6.30, if AB||CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE. Ans-

$\begin{array}{l}\mathrm{Given}:\mathrm{AB}\parallel \mathrm{CD},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{EF}\perp \mathrm{CD}\text{and}\angle \text{GED=126°.}\\ \text{To find:}\angle \mathrm{A}\text{GE,}\angle \text{GEF and}\angle \mathrm{F}\text{GE}\\ \text{AB}\parallel \text{CD}\\ \angle \text{AGE}=\angle \text{GE}\mathrm{D}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=126\mathrm{°}\\ \angle \text{GED}=\angle \text{GEF}+\angle \text{FED}\\ \text{126°}=\angle \text{GEF}+90\mathrm{°}\\ \angle \text{GEF}=126\mathrm{°}-90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=36\mathrm{°}\\ \angle \mathrm{A}\text{GE}+\angle \mathrm{F}\text{GE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}126\mathrm{°}+\angle \mathrm{F}\text{GE}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{F}\text{GE}=180\mathrm{°}-126\mathrm{°}\\ =54\mathrm{°}\end{array}$

Q.10 In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS. Ans-

$\begin{array}{l}\mathrm{Given}:\mathrm{PQ}\parallel \mathrm{ST},\text{}\angle \text{PQR}=\text{11}0\mathrm{°},\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{RST}=\text{13}0\mathrm{°}\\ \mathrm{To}\text{prove:}\angle \text{QRS}\\ \text{Construction:\hspace{0.17em}Draw line}\mathrm{l}\text{paralllel to ST through R.}\\ \text{Since, ST}\parallel \stackrel{↔}{\mathrm{l}}\text{, then}\\ \angle \text{\hspace{0.17em}}\mathrm{TSR}+\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Co}-\mathrm{interior}\text{​ angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}130\mathrm{°}+\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{°}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{\hspace{0.17em}}\mathrm{SRW}=180\mathrm{°}-130\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=50\mathrm{°}\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\parallel \mathrm{ST}\text{and ST}\parallel \stackrel{↔}{\mathrm{l}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\parallel \stackrel{↔}{\mathrm{l}}\\ \angle \mathrm{PQR}=\angle \mathrm{SRW}\left[\mathrm{Alternate}\text{interior angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}110\mathrm{°}=\angle \mathrm{QRS}+\angle \mathrm{SRW}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\angle \mathrm{QRS}+50\mathrm{°}\\ \angle \mathrm{QRS}=110\mathrm{°}-50\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60\mathrm{°}\end{array}$

Q.11 In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y. Ans-

$\begin{array}{l}\mathrm{Given}:\mathrm{AB}\parallel \mathrm{CD},\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{APQ}=\text{5}0\mathrm{°}\text{\hspace{0.17em}and}\mathrm{}\text{\hspace{0.17em}}\angle \text{PRD}=\text{127°}\\ \text{To find: x and y}\\ \text{Since, AB}\parallel \text{CD}\\ \text{So,}\angle \text{APR}=\angle \text{PRD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}50°}+\text{y}=127\mathrm{°}\\ ⇒\mathrm{y}=127\mathrm{°}-50\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=77\mathrm{°}\\ \text{And, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{APQ}=\angle \text{PQR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}50\mathrm{°}=\mathrm{x}\\ \mathrm{Thus},\text{x}=\text{50° and y}=\text{77°.}\end{array}$

Q.12 In Fig. 6.33, 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 back along CD. Prove that AB || CD. Ans-

$\begin{array}{l}\mathrm{Given}:\mathrm{PQ}\parallel \mathrm{RS},\text{\hspace{0.17em}AB is incident ray to PQ and BC is incident ray}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}to RS.}\\ \text{To Prove: AB}\parallel \text{CD}\\ \text{Construction: Draw MB}\perp \text{PQ and NC}\perp \text{RS.}\\ \text{Proof}:\text{}\angle \text{ABM}=\angle \mathrm{MBC}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{incident}\text{angle= reflected angle}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BCN}=\angle \mathrm{NCD}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{incident}\text{angle= reflected angle}\right]\end{array}$ $\begin{array}{l}\text{Since, perpendiculars to parallel lines are parallel.So,}\\ \text{\hspace{0.17em}\hspace{0.17em}MB}\parallel \text{NC}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{M}\text{BN}=\angle \mathrm{BCN}\\ \text{\hspace{0.17em}}\frac{1}{2}\angle \mathrm{A}\text{BC}=\frac{1}{2}\angle \mathrm{BCD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}\text{BC}=\angle \mathrm{BCD}\\ \mathrm{Since},\text{alternate interior angles are equal, so}\\ \text{\hspace{0.17em}\hspace{0.17em}AB}\parallel \text{CD.}\end{array}$

Q.13 In Fig. 6.39, sides QP and RQ of Δ PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ. Ans-

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\text{}\mathrm{\Delta }\text{PQR,\hspace{0.17em}}\mathrm{Ext}.\angle \text{SPR=135° and}\mathrm{Ext}.\angle \text{PQT=110°}\\ \text{To Find:}\angle \text{PRQ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}+\angle \text{PQT}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}+110\mathrm{°}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}=180\mathrm{°}-110\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PQR}=70\mathrm{°}\\ \mathrm{Ext}.\text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{S}\text{PR}=\angle \text{PQR}+\angle \text{PRQ}\\ \text{\hspace{0.17em}135°}=70\mathrm{°}+\angle \text{PRQ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PRQ}=135\mathrm{°}-70\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=65\mathrm{°}\end{array}$

Q.14 In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠ YOZ. Ans-

$\begin{array}{l}\mathrm{Given}:\text{\hspace{0.17em}}\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta }\text{XYZ,\hspace{0.17em}\hspace{0.17em}}\angle \text{X}=\text{62}\mathrm{°},\angle \text{\hspace{0.17em}XYZ}=\text{54}\mathrm{°}\text{, YO}\mathrm{}\text{\hspace{0.17em}and ZO are the}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}bisectors of\hspace{0.17em}}\angle \text{XYZ and}\angle \text{XZY}\mathrm{}\text{\hspace{0.17em}respectively.}\\ \mathrm{To}\text{find:}\angle \text{OZY and}\angle \text{YOZ}\\ \text{In}\mathrm{\Delta }\text{XYZ,}\\ \text{}\angle \text{\hspace{0.17em}XYZ}+\angle \text{\hspace{0.17em}}\mathrm{Y}\text{XZ}+\angle \text{\hspace{0.17em}YZX}=180\mathrm{°}\\ \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}}54\mathrm{°}+62\mathrm{°}+\angle \text{\hspace{0.17em}YZX}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \text{\hspace{0.17em}YZX}=180\mathrm{°}-116\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=64\mathrm{°}\\ \text{\hspace{0.17em}}\angle \text{\hspace{0.17em}OZY}=\frac{1}{2}\angle \text{\hspace{0.17em}YZX}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×64\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=32\mathrm{°}\\ \text{\hspace{0.17em}}\angle \text{\hspace{0.17em}OYZ}=\frac{1}{2}\angle \text{\hspace{0.17em}XYZ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}×54\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=27\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta }\text{XYZ,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{\hspace{0.17em}OYZ}+\angle \text{\hspace{0.17em}OZY}+\angle \text{\hspace{0.17em}}\mathrm{Y}\text{OZ}=180\mathrm{°}\\ \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}}27\mathrm{°}+32\mathrm{°}+\angle \text{\hspace{0.17em}}\mathrm{Y}\text{OZ}=180\mathrm{°}\\ \angle \text{\hspace{0.17em}}\mathrm{Y}\text{OZ}=180\mathrm{°}-59\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=121\mathrm{°}\end{array}$

Q.15 In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE. Ans-

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\text{\hspace{0.17em}}\mathrm{AB}\parallel \mathrm{DE},\text{\hspace{0.17em}}\angle \text{BAC}=\text{35}\mathrm{°}\text{and}\mathrm{}\text{\hspace{0.17em}}\angle \text{CDE}=\text{53}\mathrm{°}\\ \mathrm{To}\text{find:\hspace{0.17em}}\angle \text{DCE}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sol:\hspace{0.17em}Since, AB}\parallel \text{CD, so}\\ \angle \text{DE}\mathrm{A}=\angle \text{BAE}\left[\mathrm{Alternate}\text{​​ interior angles}\right]\\ \angle \text{DE}\mathrm{A}=35\mathrm{°}\\ ⇒\angle \text{CED}=35\mathrm{°}\\ \mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta }\text{CDE,}\\ \angle \text{CDE}+\angle \mathrm{D}\text{CE}+\angle \text{CED}=\text{180°\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{angle sum property.}\right]\\ \text{\hspace{0.17em}}53\mathrm{°}+\angle \mathrm{D}\text{CE}+35\mathrm{°}=\text{180°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{D}\text{CE}=180\mathrm{°}-88\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{D}\text{CE}=92\mathrm{°}\end{array}$

Q.16 In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT. Ans-

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\angle \text{PRT}=\text{4}0\mathrm{°},\angle \text{RPT}=\text{95}°\mathrm{}\text{and\hspace{0.17em}\hspace{0.17em}}\angle \text{TSQ}=\text{75}\mathrm{°}\\ \mathrm{To}\text{find:}\angle \text{SQT}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{In}\text{}\mathrm{\Delta }\text{PRT,}\\ \angle \text{PRT}+\angle \text{PTR}+\angle \text{RPT}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}40\mathrm{°}+\angle \text{PTR}+95\mathrm{°}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PTR}=180\mathrm{°}-135\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=45\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{QTS}=\angle \text{PTR}\left[\mathrm{Vertical}\text{opposite angels}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{QTS}=45\mathrm{°}\\ \mathrm{In}\text{}\mathrm{\Delta }\text{QST,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{QST}+\angle \text{SQT}+\angle \text{QTS}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}75\mathrm{°}+\angle \text{PTR}+45\mathrm{°}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PTR}=180\mathrm{°}-120\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=60\mathrm{°}\end{array}$

Q.17 In Fig. 6.43, if PQ ⊥ PS, PQ||SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y. Ans-

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\text{PQ}\perp \text{PS},\text{PQ}||\text{SR},\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{SQR}=\text{28}°\text{\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}\angle \text{QRT}=\text{65}\mathrm{°}\\ \mathrm{To}\text{find: x and y}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Sol}:\text{\hspace{0.17em}}\mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQ}\parallel \mathrm{ST}\\ \angle \mathrm{PQR}=\angle \mathrm{PRT}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+28\mathrm{°}=65\mathrm{°}\\ \text{\hspace{0.17em}}\mathrm{x}=65\mathrm{°}-28\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=37\mathrm{°}\\ \mathrm{In}\text{}\mathrm{\Delta }\text{PQS,}\\ \text{\hspace{0.17em}}\angle \mathrm{QPS}+\angle \mathrm{PSQ}\text{\hspace{0.17em}}+\angle \mathrm{PQR}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{angle sum property.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}90\mathrm{°}+\mathrm{y}+37\mathrm{°}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=180\mathrm{°}-127\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=53\mathrm{°}\end{array}$

Q.18 In Fig. 6.44, the side QR of Δ PQR is produced to a point S.
If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = (1/2) ∠QPR. Ans-

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\text{The side QR of\hspace{0.17em}}\mathrm{\Delta }\text{PQR is produced}\text{to a point S}.\text{The}\\ \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}bisectors of}\angle \text{PQR and}\mathrm{}\text{\hspace{0.17em}}\angle \text{PRS meet at point T.}\\ \text{To Prove:}\angle \text{QTR}=\frac{1}{2}\angle \text{QPR}\\ \text{Proof}:\text{\hspace{0.17em}}\mathrm{Since},\angle \mathrm{PQT}=\angle \mathrm{TQR}=\mathrm{x}\left(\mathrm{let}\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}}\mathrm{and}\angle \mathrm{PRT}=\angle \mathrm{TRS}=\mathrm{y}\left(\mathrm{let}\right)\\ \mathrm{In}\text{​\hspace{0.17em}}\mathrm{\Delta PQR},\\ \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}}\mathrm{Ext}.\text{\hspace{0.17em}}\angle \mathrm{PRS}=\angle \mathrm{PQR}\text{\hspace{0.17em}}+\angle \mathrm{QPR}\\ \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}}2\mathrm{y}=2\mathrm{x}\text{\hspace{0.17em}}+\angle \mathrm{QPR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{y}-2\mathrm{x}=\angle \mathrm{QPR}\\ \mathrm{y}-\mathrm{x}=\frac{1}{2}\angle \mathrm{QPR}...\left(\mathrm{i}\right)\\ \mathrm{In}\text{​\hspace{0.17em}}\mathrm{\Delta QTR},\\ \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}}\mathrm{Ext}.\text{\hspace{0.17em}}\angle \mathrm{TRS}=\angle \mathrm{TQR}\text{\hspace{0.17em}}+\angle \mathrm{QTR}\\ \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}}\mathrm{y}=\mathrm{x}\text{\hspace{0.17em}}+\angle \mathrm{QTR}\\ \text{\hspace{0.17em}}\mathrm{y}-\mathrm{x}=\angle \mathrm{QTR}...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{QTR}=\frac{1}{2}\angle \mathrm{QPR}\\ \mathrm{Hence}\text{Proved.}\end{array}$