⇒ ∠SQT = 180° – 75° – 45° = 60° In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y. If ∠POY = 90° , and a : b = 2 : 3. find c. Solution: ∴∠COA = ∠BOD [Vertically opposite angles] Also, ∠GEF + ∠FED = ∠GED ∴ ∠COA = 40° NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles are part of NCERT Solutions for Class 9 Maths. Solution: [Alternate interior angles] This proves that alternate interior angles are equal and so, AB CD. After that go through the solved examples of Lines and Angles that are given in the Class 9 NCERT Book. (ii) Interior of an angle – The interior of ∠BAC is the set of all points in its plane which lie on the same side of AB as C and also on the same side of AC as B. We, in our aim to help students, have devised detailed chapter wise solutions for them to understand the concepts easily. ∴ AB || CD. This topic introduces you to the basic Geometry primarily focusing on the properties of the angles formed i) when two lines intersect each other and ii) when a line intersects two or more parallel lines at distinct points. [Vertically opposite angles] 1. ∴ b + a = 180° – 90° = 90° …(i) Now, in ∆CDE, we have ∠CDE + ∠DEC + ∠DCE = 180° In Fig. Ex 6.2 Class 9 Maths Question 5. If AOC +BOE = 70° and BOD = 40°, find BOE and reflex COE. ⇒ \(\frac { 1 }{ 2 }\)∠PRS = \(\frac { 1 }{ 2 }\)∠P + \(\frac { 1 }{ 2 }\)∠PQR ∠YOZ + ∠OYZ + ∠OZY = 180° Thus, the required measure of c = 126°. 6.43, if PQ ⊥ PS, PQ SR, SQR = 28° and QRT = 65°, then find the values of x and y. x +SQR = QRT (As they are alternate angles since QR is transversal). Again, AB || CD So, GED = AGE = 126° (As they are alternate interior angles). If a side of a triangle is produced, the exterior angle so formed is equal to the … Solution: But ∠RQS = 28° and ∠QRT = 65° Ex 6.1 Class 9 Maths Question 1. ∴ b+a+∠POY= 180° Again, AB || CD and PR is a transversal. Adding (1) and (2), we get Before starting to solve the exercise problems, you must first read the theory part and get to know the basic terms, definitions and theorems. All the exercise questions of Maths Class 9 Chapters are solved and it will be a great help for the students in their exam preparation and revision. Again, PQ ⊥ PS ⇒ AP = 90° [Vertically opposite angles] we have ∠TQR + \(\frac { 1 }{ 2 }\)∠P = ∠TQR + ∠T 2. ⇒ z + y = 180° … (2) [By (1)] So, PRS = QPR+PQR (According to triangle property). So, you can easily score marks if you have a thorough understanding of this topic. In ΔABC, ∠A = 50° and the external bisectors of ∠B and ∠C meet at O as shown in figure. In NCERT Solutions for Class 9 Maths Chapter 6, you will learn to solve the questions related to all the concepts of Lines and Angles. These solutions help students prepare for their upcoming Board Exams by covering the whole syllabus, in accordance with the NCERT guidelines. Here, BE ⊥ CF and the transversal line BC cuts them at B and C, So, 2 = 3 (As they are alternate interior angles), So, AB CD alternate interior angles are equal). 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Chapter 6, Chapter 4 linear equation in two variables, Chapter 5 introduction to euclids geometry, Chapter 9 areas of parallelograms and triangle. ⇒ ∠ROS = 90° – ∠POS … (1) ∴ AB || EF The architecture uses lines and angles to design the structure of a building. In figure, lines XY and MN intersect at 0. But ∠GED = 126° [Given] Mathematics NCERT Grade 9, Chapter 6: Lines and Angles: In this chapter students will study the properties of the angle formed when two lines intersect each other and properties of the angle formed when a line intersects two or more parallel lines at distinct points.The chapter starts from zero level, the first topic of the chapter being Basic Terms and Definitions. You can download the complete solution pdf of NCERT Chapter 6 Line and Angles of Class 9 by clicking on the link below: List of Exercises in class 9 Maths Chapter 6, Exercise 6.1 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question)Exercise 6.2 Solutions 6 Questions (3 Short Answer Questions, 3 Long Answer Question)Exercise 6.3 Solutions 6 Questions (5 Short Answer Questions, 1 Long Answer Question). Question 1. Solution: Since XY and MN interstect at O, These NCERT Solutions … Solution: LINES AND ANGLES 91 An acute angle measures between 0° and 90°, whereas a right angle is exactly equal to 90°. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP. [Angle sum property of a triangle] Solution: Here, ∠ AOC and ∠ BOD are vertically opposite angles. 2(x + y) = 360° KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 are part of KSEEB Solutions for Class 9 Maths. So. It is also known that alternate interior angles are same and so, QRS +QRT = 180° (As they are a Linear pair). ∴ AOB is a straight line. ∴ ∠APQ = ∠PQR Ex 6.1 Class 9 Maths Question 1 As you can see that it constitutes approximately 27% of weightage. 6.14, lines XY and MN intersect at O. 4. 2. ⇒ \(\frac { 5a }{ 2 }\) = 90° (1) ⇒ 10z = 7 x 180° Prove that AB || CD. ∴ 75° + 45° + ∠SQT = 180° [ ∵ ∠TSQ = 75° and ∠STQ = 45°] Question 1. We know that the angles on the same side of transversal is equal to 180°. The chapter deals with lines and angles, its different types and formulas etc. Lines and Angles Class 9 MCQs Questions with Answers. 6.41, if AB DE, BAC = 35° and CDE = 53°, find DCE. ⇒ x = 180° – 50° = 130° …(2) In Fig. Solution: ∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360° Putting the value of POY = 90° (as given in the question) we get, Similarly, b can be calculated and the value will be. ⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP] The RS Aggarwal Solutions for Class 9 Chapter-7 Lines and Angles Solutions Maths have been provided here for the benefit of the CBSE Class 9 students. Get clarity on concepts like linear pairs, vertically opposite angles, co-interior angles, alternate interior angles etc. Refer to the NCERT Solutions of Class 9 provided by our Experts below. [Angle sum property of a triangle] ⇒ ∠SRF = 180° – 130° = 50° [∵ ∠XYZ = 64° (given)] In Fig 3.13, lines AB and CD intersect at O. [∵ BL || PQ and CM || RS] Stay tuned for further updates on CBSE and other competitive exams. Lines and Angles (Mathematics) Class 9 - NCERT Questions. 5. Now from (i) and (ii), we get We know that AE is a transversal since AB DE. If ∠POY = and ... Read more . ∵ PQ || RS ⇒ BL || CM ∴ Its complement = 90° – x. So, ∠BAC = ∠AED But ∠BOD = 40° [Given] Since ∠PQR =∠PRQ (as given in the question). NCERT Solutions for Class 9 Maths Chapter 6 are created by the BYJU’S expert faculty to help students in the preparation of their examinations. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ. But PQ and RS intersect at T. Ex 6.1 Class 9 Maths Question 1. Ex 6.2 Class 9 Maths Question 3. We have, ∠TQP + ∠PQR = 180° In the figure, we have CD and PQ intersect at F. Ex 6.2 Class 9 Maths Question 1. It is given that XYZ = 64° and XY is produced to point P. Draw a figure from the given information. ⇒ ∠PQR = ∠QRF [Alternate interior angles] But ∠PQR = 110° [Given] If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. and EF || ST [Construction] Question 1. Angle of incidence = Angle of reflection (By the law of reflection), We also know that alternate interior angles are equal. In figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair] RD Sharma Solutions for Class 9 Mathematics CBSE, 10 Lines and Angles. CBSETuts.com provides you Free PDF download of NCERT Exemplar of Class 9 Maths Chapter 6 Lines And Angles solved by expert teachers as per NCERT (CBSE) Book guidelines. 1. Ex 6.1 Class 9 Maths Question 2. ⇒ 2∠QYP = 180° – 64° = 116° ⇒ \(\frac { 1 }{ 2 }\)∠P = ∠T An angle greater than 90° but less than 180° is called an obtuse angle. ∠PQS + ∠PQR = ∠PRT + ∠PRQ Now, you must be wondering why we are studying Lines and Angles. All questions and answers from the Rs Aggarwal 2018 Book of Class 9 Math Chapter 7 are provided here for you for free. Q 1. Ex 6.1 Class 9 Maths Question 3. ⇒ ∠APQ + ∠QPR = 127° RD Sharma Solutions contains all in one solution for the different problem sets along with solved examples for ease of understanding. 2. lines which are parallel to a given lines are parallel to each other. ⇒ 50° + y = 127° [ ∵ ∠APQ = 50° (given)] We know that a linear pair is equal to 180°. 6.29, if AB CD, CD EF and y : z = 3 : 7, find x. Solution: \(\frac { 3a }{ 2 }\) + A = 90° All the solutions of Lines and Angles - Mathematics explained in detail by experts to help students prepare for their CBSE exams. ⇒ ∠TRS = \(\frac { 1 }{ 2 }\)∠P + ∠TQR …(1) NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1. All the chapter wise questions with solutions to help you to revise the complete CBSE syllabus and score more marks in Your board examinations. The answer is that lines and angles are everywhere around us. ∠GEF = 126° -90° = 36° In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT. An angle which is greater than 180° but less than 360° is called a reflex angle.Further, two angles whose sum is 90° are ∴ ∠ROQ = 90° Draw ray BL ⊥PQ and CM ⊥ RS Required fields are marked *. Then you can start solving the exercise problems with the help of NCERT Solutions. So, 28° + ∠RSQ = 65° ∠TRS = ∠TQR + ∠T …(2) ∴ ∠PTR = ∠QTS Solution: Thus, ∠OZY = 32° and ∠YOZ = 121°, Ex 6.3 Class 9 Maths Question 3. Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.1. 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. To access interactive Maths and Science Videos download BYJU’S App and subscribe to YouTube Channel. 6.32, if AB CD, APQ = 50° and PRD = 127°, find x and y. Also a : b = 2 : 3 ⇒ b = \(\frac { 3a }{ 2 }\) …(ii) In the question, it is given that (OR ⊥ PQ) and POQ = 180°, Now, POS+ROS = 180°- 90° (Since POR = ROQ = 90°), As POS + ROS = 90° and QOS – ROS = 90°, we get. Solution: MCQ Questions for Class 9 Maths Chapter 6 Lines and Angles with Answers MCQs from Class 9 Maths Chapter 6 – Lines and Angles are provided here to help students prepare for their upcoming Maths exam. Now, in ∆OYZ, we have Prove that AB CD. Solution: For proving AOB is a straight line, we will have to prove x+y is a linear pair. ⇒ ∠YOZ + 27° + 32° = 180° Cuemath experts provide Maths NCERT solutions with detailed explanations class 9. From the diagram, b+c also forms a straight angle so. OS is another ray lying between rays OP and OR. Get NCERT Solutions of all exercise questions and examples of Chapter 6 Class 9 Lines and Angles free at teachoo. Adding (1) and (2), we have In figure, find the values of x and y and then show that AB || CD. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Students can now freely access RD Sharma Class 9 Maths solutions for chapter 8 here. ∴ 54° + ∠YZX + 62° = 180° Consider the ΔPQR. x = 126°. We hope the NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 help you. Again ST || EF and RS is a transversal MCQs from CBSE Class 9 Maths Chapter 6: Lines and Angles 1. ∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30° When you stop at a signal and then move on when the signal light is green, then you either take a left angle turn or right-angle turn or move in a straight line. In the given figure, lines AB and CD intersect at O. ⇒ ∠ROS = ∠QOS – 90° ……(2) Thus, the values of x and y are calculated as: 6. 6.13, lines AB and CD intersect at O. From (1) and (2), [Angle sum property of a triangle] ∴ c = [a + ∠POY] [Vertically opposite angles] In ∆PRT, we have ∠P + ∠R + ∠PTR = 180° ⇒ 95° + 40° + ∠PTR =180° Solution: Since AB is a straight line, ∴ ∠AOC + ∠COE + ∠EOB = 180°. We know that the angles around a point are 360° so. or ∠FGE + 126° = 180° In Fig. Now, we have ∠ROS + ∠ROQ = ∠QOS ∴ 53° + 35° + ∠DCE =180° Here, the side QP is extended to S and so, SPR forms the exterior angle. (ii) Interior of an angle: The interior of ∠AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A. Answer : Q2 : In the given figure, lines XY and MN intersect at O. Since ∠XYQ = ∠XYZ + ∠ZYQ 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. But ∠XYZ = 54° and ∠ZXY = 62° Also, ∠AOC + ∠BOE = 70° [YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ] PDF download free. What are the real-life applications of it? Since AB is a straight line, By putting the value of XYZ = 64° and ZYQ = 58° we get. In figure, if x + y = w + ⇒, then prove that AOB is a line. Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 9 so that you can refer them as and when required. Now, in ∆PQS, NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1. ∴ x = z [Alternate interior angles] …. Exercise 4A. In figure, ∠X = 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ. ⇒ x + y = 180° [Co-interior angles] Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 6. RD Sharma Solution for Class 9 Chapter 8 includes several exercises of Lines and Angles to help the students practice the concepts more effectively. ∴ y = 130° …(1) Now, for the linear pairs on the line XY-. NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles deals with the questions and answers related to the chapter Lines and Angles. Draw a line EF parallel to ST through R. z = \(\frac { 7 }{ 3 }\) y = \(\frac { 7 }{ 3 }\)(180°- z) [By (2)] Telangana SCERT Class 9 Math Chapter 4 Lines and Angles Exercise 4.3 Math Problems and Solution Here in this Post. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions. If ∠POY = 90° and a : b = 2 : 3, find c. 3. ⇒ 110° + ∠PQR = 180° Putting the values as given in the question we get. In figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. In Fig. ⇒ a = \(\frac { { 90 }^{ \circ } }{ 5 } \times 2\quad =\quad { 36 }^{ \circ }\) = 36° 6.40, X = 62°, XYZ = 54°. Extra Questions for Class 9 Maths Ex 6.1 Class 9 Maths Question 6. or 50° + x = 180° It is given the TQR is a straight line and so, the linear pairs (i.e. ∴ ∠FGE + ∠GED = 180° [Co-interior angles] If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. 3. 6.31, if PQ ST, PQR = 110° and RST = 130°, find QRS. Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair] Refer to NCERT Solutions for CBSE Class 9 Mathematics Chapter 6 Lines and Angles at TopperLearning for thorough Maths learning. UP board high school students also use these solutions as UP Board Solutions updated for academic session 2020-2021. In figure, lines AB and CD intersect at 0. In ∆XYZ, we have ∠XYZ + ∠YZX + ∠ZXY = 180° ∴ ∠PQS = ∠PRT. Lines and Angles Class 9 Solutions are prepared by highly qualified and professional teachers at Vedantu. Since PQ || ST [Given] ⇒ y = 180° – 90° – 37° = 53° First, construct a line XY parallel to PQ. This topic introduces you to the basic Geometry primarily focusing on the properties of the angles formed i) when two lines intersect each other and ii) when a line intersects two or more parallel lines at distinct points. We computed that the value of XYQ = 122°. Now, as the sum of the interior angles of the triangle. In Fig. If YO and ZO are the bisectors of XYZ and XZY respectively of Δ XYZ, find OZY and YOZ. AB || CD and GE is a transversal. Thus, ∠SQT = 60°, Ex 6.3 Class 9 Maths Question 5. Intersecting lines cut each other at: a) […] Your email address will not be published. In two parallel lines, the alternate interior angles are equal. Since XOY is a straight line. If POY = 90° and a : b = 2 : 3, find c. We know that the sum of linear pair are always equal to 180°. [Exterior angle property of a triangle] After solving the Line and Angles chapter of Class 9 Maths, you will get to know the following points: We hope this information on “NCERT Solution for Class 9 Maths Chapter 6 Lines and Angles” is useful for students. ⇒ 64° + 2∠QYP = 180° NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1, Exercise 6.2 and Exercise 6.3 in English Medium as well as Hindi Medium updated for new academic session 2020-2021 based on latest NCERT Books. Since, the side QP of ∆PQR is produced to S. In figure, PQ and RS are two mirrors placed parallel to each other. Extra Questions for Class 9 Maths Chapter 6 Lines and Angles. 6.28, find the values of x and y and then show that AB CD. Skip to content. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = ½ QPR. Since, angle of incidence = Angle of reflection If an angle is half of its complementary angle, then find its degree measure. Lines and Angles Class 9 Exercise 6.1 : Solutions of Questions on Page Number : 96 Q1 : In the given figure, lines AB and CD intersect at O. But OR ⊥ PQ Now, putting the value of TQP = 110° we get. 6.42, if lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75°, find SQT. Now PTR will be equal to STQ as they are vertically opposite angles. ⇒ ∠DCE = 180° – 53° – 35° = 92° RD Sharma Class 11 Solutions Free PDF Download, NCERT Solutions for Class 12 Computer Science (Python), NCERT Solutions for Class 12 Computer Science (C++), NCERT Solutions for Class 12 Business Studies, NCERT Solutions for Class 12 Micro Economics, NCERT Solutions for Class 12 Macro Economics, NCERT Solutions for Class 12 Entrepreneurship, NCERT Solutions for Class 12 Political Science, NCERT Solutions for Class 11 Computer Science (Python), NCERT Solutions for Class 11 Business Studies, NCERT Solutions for Class 11 Entrepreneurship, NCERT Solutions for Class 11 Political Science, NCERT Solutions for Class 11 Indian Economic Development, NCERT Solutions for Class 10 Social Science, NCERT Solutions For Class 10 Hindi Sanchayan, NCERT Solutions For Class 10 Hindi Sparsh, NCERT Solutions For Class 10 Hindi Kshitiz, NCERT Solutions For Class 10 Hindi Kritika, NCERT Solutions for Class 10 Foundation of Information Technology, NCERT Solutions for Class 9 Social Science, NCERT Solutions for Class 9 Foundation of IT, PS Verma and VK Agarwal Biology Class 9 Solutions, Chapter 4 Linear Equations in Two Variables, Chapter 5 Introduction to Euclid Geometry, Chapter 9 Areas of Parallelograms and Triangles, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, Periodic Classification of Elements Class 10, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, CBSE Previous Year Question Papers Class 12, CBSE Previous Year Question Papers Class 10. ⇒ ∠YZX = 180° – 54° – 62° = 64° These solutions are designed by subject matter experts who have assembled model questions covering all the exercise questions from the textbook. 6.15, PQR = PRQ, then prove that PQS = PRT. ∴ (x + y) + (x + y) = 360° or, ⇒ 70° + ∠PRQ = 135° [∠PQR = 70°] Ray OR is perpendicular to line PQ. 5. AB || DE and AE is a transversal. We know that QT and RT bisect PQR and PRS respectively. ∴ ∠QRT = ∠RQS + ∠RSQ Solution: In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x. These solutions for Lines And Angles are extremely popular among Class 9 students for Math Lines And Angles Solutions come handy for quickly completing your homework and preparing for exams. Thus, these are some questions for the different chapters starting from Class 9 Chapter 8 Introduction to Lines and Angles. The sum of the three angles of a triangle is 180 degree. Lines and Angles NCERT solution. If SPR = 135° and PQT = 110°, find PRQ. In figure, lines AB and CD intersect at 0. 6.14, lines XY and MN intersect at O. ∴∠AGE = 126° (Triangle property). 6. Now, by putting the values of AOC+BOE = 70° and BOD = 40° we get. ⇒ 50° = x [ ∵ ∠APQ = 50° (given)] If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1, drop a comment below and we will get back to you at the earliest. ⇒ ∠PTR = 180° – 95° – 40° = 45° Ex 6.1 Class 9 Maths Question 5. In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE. ∴ ∠ROS = \(\frac { 1 }{ 2 } (\angle QOS-\angle POS)\). Lines and Angles Class 9 Extra Questions Very Short Answer Type. It will make your concepts more clear. ∴ ∠APR = ∠PRD [Alternate interior angles] In Fig. We have AB || CD and PQ is a transversal. NCERT Solutions for Class 9 Maths Chapter 6 are useful for students as it helps them to score well in the class exams. 3. If ray YQ bisects ZYP, find XYQ and reflex QYP. In Fig 3.13, Lines AB and CD intersect at O you can start solving the sums this. For Class 9 Math Chapter 4 Lines and Angles ( reflex, complementary supplementary. 6.40, x = ( 90° – x ) ⇒ 2x = 90° – x help of textbooks. 180° is called an obtuse angle for Chapter 8 here measures between 0° and 90°, a! Concepts easily that AB || CD, CD || EF and y are calculated as:.... Book of Class 9 Maths Chapter 3 Lines and Angles Class 9 Maths Chapter 6 Lines and Angles 6.1! As up Board Solutions updated for academic session 2020-2021 a line: Q2: in the question we get exercise! ⇒ 2x = 90° and a: b = 2: 3. find 3... = AGE = 126° ( as they are alternate interior Angles are equal +BOD +BOE ) a! Aggarwal 2018 Book of Class 9 Maths Chapter 6: Lines and Angles 6.1... Ab DE 180° is called an obtuse angle a given Lines are parallel to other! Of interior opposite Angles, co-interior Angles, its different types and formulas etc cut each other at: )! 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