# NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.3) Exercise 6.3

Mathematics is not only a subject but also a concept that is widely used in the daily lives of students. Therefore, students should have a strong grasp of the subject. The concepts of mathematics are used to figure out distance, prepare food, understand finances, and much more. Mathematics can not be learned, due to the conceptual nature of the subject, therefore

students must have a deep understanding of it. With proper guidance and continuous practice, concepts and calculations in the subject become easier for students. The Class 9 Mathematics NCERT Textbook does not include solutions to the questions listed in the exercises. The NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3, provided by Extramarks give a clear understanding and approach for each question of the NCERT textbook. NCERT provides the foundation for every chapter of Mathematics. Therefore, students should regularly practice the NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3.

The NCERT Class 9 Exercise 6.3 Solutions, Lines and Angles, includes the following topics: Line Segment, Collinear and Non-Collinear Points, Different Types of Angles, Linear Pairs, Vertically Opposite Angles, Complementary Angles and much more. It also includes Supplementary Angles, Adjacent Angles, Parallel Lines and a Transversal, Angle Sum Property of a triangle, and much more. There is a lot of important information covered in this chapter. Students can score good marks in their examinations with the assistance of Extramarks since it provides the essential tools to make the learning process systematic and effortless for them.

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.3) Exercise 6.3

Class 9 Maths Chapter 6 Exercise 6.3 can be downloaded in PDF format from the Extramarks website. Students can access these solutions whenever and wherever they want since they are easily accessible and also available offline.

### NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.3

NCERT is used as the primary source of information in all the schools affiliated with the CBSE. It is essential for students to thoroughly go through the NCERT curriculum, as it can be very helpful in clarifying their concepts. As the question papers of examinations are mostly based on the fundamentals of NCERT textbooks, they are the primary source of teaching in many schools. Students should therefore review the Exercise 6.3 Class 9 Maths NCERT Solutions very carefully. Exercise 6.3 Class 9 Maths NCERT Solutions help students build a firm understanding of the concepts so that they can also solve any problem that comes in front of them during the examination. Along with practising the NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3, students should also practice various other important questions and question banks in order to score well in the Class 9 Mathematics examination. The NCERT textbooks only contain a few questions that explain concepts, but it is important for students to practise more conceptual questions thoroughly for a deeper understanding of the chapter. In order to help them excel in their studies and score high grades, Extramarks provides them with learning tools such as K12 study material for boards, sample papers, past year papers, and much more. The NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3, provided by Extramarks are an example of one such learning tool. Therefore, students can join Extramarks in order to have convenient access to reliable and comprehensive study materials.

### NCERT Solutions for Class 9

The NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3, are based on several concepts such as Exterior Angle Property, Angle Sum Property of a Triangle, Linear Pair of Angles, Vertically Opposite Angles, Parallel Lines and Transversal, etc. Numerous essential concepts are discussed further in higher classes in relation to these topics. The concepts and calculations of the chapter may be difficult for some students at the beginning, but with practice and clear understanding, they can solve the question related to the chapter quickly and effortlessly. Extramarks’ NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3, provide students with an in-depth understanding of the concepts. This way, they will be able to find authentic solutions to their problems without having to look elsewhere. In addition to NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3, Extramarks also provides students with study material that clarifies their concepts and assists them with better problem-solving skills.

### CBSE Study Materials for Class 9

The subject of mathematics involves a lot of critical thinking. It helps students improve their analytical and problem-solving skills. Mathematics is a subject that requires an ample amount of practice. Since it is a conceptual subject, therefore, it cannot be learned by students. Therefore, in order to score well in the subject, it is important that students understand all the concepts of the subject very well. Scientists also assert that mathematics is more than just a subject; it is the basis for many scientific concepts.

There are many chapters in Class 9 Mathematics which are new to students, so they need to work very hard and understand the concepts thoroughly. In order to clear the fundamentals of the subject, students should learn NCERT by heart. Lines and Angles is one of the many chapters of Mathematics in Class 9. It contains a lot of difficult concepts that can be challenging for students. They can enhance their conceptual clarity with the NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3 provided by the Extramarks website. The subject experts of the learning website curate these answers and cross-check them according to the latest examination patterns. The Extramarks website provides the NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3 so that students can easily prepare and practice the chapter’s curriculum for their examinations.

### CBSE Study Materials

It is not enough for students to have access to a number of questions and question banks before the examinations; they also need detailed and step-by-step solutions as well. Most of the question banks use NCERT textbook questions as their primary source of information. The Extramarks website provides students with the NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3 to help them prepare better for their examinations. Students are more confident in their examinations when they have a good means of learning, which helps them to concentrate on their studies. Students can move ahead with their preparation by having a good understanding of NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3. These solutions help students learn the chapter effectively and perform well in their examinations.

Q.1 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.2 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.3 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.4 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.5 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.6 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}$

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Since Mathematics is a conceptual subject, it is necessary to practice every question of the NCERT Solutions For Class 9 Maths Chapter 6 Exercise 6.3. Students must practice these solutions several times in order to master the chapter. To be able to gain a deeper understanding of the concepts of the chapter, students should practice as many questions as they can. As a result, they will be able to good grades in the examinations.

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