Future Study Point

NCERT Solutions Class 9 Maths exercise 6.3 -Lines and Angles

EX.6.3 MATHS CLASS 9

NCERT Solutions Class 9 Maths exercise 6.3 of the chapter 6-Lines and Angles are the solutions of the unsolved questions of exercise 6.3 -Lines and angles of class 9 maths. The questions of exercise 6.3 are easy to understand because the concept of geometry is based on the maths of previous classes as an example angle sum property of the triangles, the relationship between exterior angle and its opposite interior angles of a triangle, linear pair etc. Here you can study NCERT solutions of science and maths from class 9 to class 12, our articles on science and maths, sample papers, previous year question papers, tips for government and other entrance exams, and career in online jobs.

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

 

You can also study NCERT Solutions

 Exercise 6.1-Lines and Angles

Exercise 6.2-Lines and Angles

NCERT Solutions of class 9 maths

Q1.In fig. 6.39, side QP and RQ of ΔPQR are produced to points S and T respectively, if ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Fig.6.39 exercise 6.3 class 9 maths

 

Ans.We are given in figure ∠SPR = 135° and ∠PQT = 110°

∠PQT and ∠PQR are the linear pair(TQR is a line)

So, ∠PQT + ∠PQR  = 180°

110°+ ∠PQR = 180°

∠PQR = 180°- 110° = 70°

∠QPR and  ∠SPR are the linear pair (QPS is a line)

∠QPR +  ∠SPR = 180°

∠QPR + 135° = 180°

∠QPR = 180° – 135° = 45°

∠PQR +∠QPR +∠PRQ = 180°(angle sum property of triangle)

∠PRQ + 70° + 45° = 180°

∠PRQ + 115° = 180°

∠PRQ = 180° – 115° = 65°

Q2. In fig.6.40 , ∠X = 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY  and ∠YOZ.

Fig.6.40 class 9 maths ex.6.3

Ans.We are given ∠X = 62°, ∠XYZ = 54°

YO is the bisector of ∠XYZ  and  ZO is the bisector of ∠XZY

∴∠OYZ = 54°/2 = 27° and ∠OZY = ∠XZY/2

∠X + ∠XYZ + ∠XZY = 180° (angle sum property of triangle)

62° + 54° + ∠XZY = 180°

116 + ∠XZY = 180°

∠XZY = 180° – 116 = 64°

Now ,In ΔYOZ we have

∠OZY = ∠XZY/2 = 64°/2 = 32°

Applying the angle sum property of triangle in ΔYOZ

∠OYZ + ∠OZY  + ∠YOZ = 180°

27° + 32° + ∠YOZ = 180°

59 + ∠YOZ = 180°

∠YOZ = 180° – 59  = 121°

Hence the value of ∠OZY  is 32° and  of ∠YOZ= 121°

Q3.In fig.6.41 ,if AB ll DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

fig.6.41 exercise 6.3 class 9 maths

Ans.We are given AB ll DE, ∠BAC = 35° and ∠CDE = 53°

Since AB ll DE and AE is transversal

So, ∠BAC = ∠DEC (alternate angle)

∠DEC = 35°

Applying the angle sum property of triangle in ΔDEC

∠DEC + ∠DCE + ∠CDE  = 180°

35° + 53° + ∠DCE = 180°

88 + ∠DCE = 180°

∠DCE = 180° – 88 = 92°

Hence the value of the ∠DCE is 92°

Click-For online purchasing of Computer and Accessories

Q4.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. We are given in the fig.

∠PRT =40°,∠RPT =95° and ∠TSQ = 75°

Applying the angle sum property of the triangle in ΔPRT

∠RPT +∠PRT+ ∠PTR = 180°

95° + 40° + ∠PTR = 180°

135° + ∠PTR  = 180°

∠PTR = 180°- 135° = 45°

Since PQ and RS are the lines

So, ∠PTR = ∠QTS (Vertically opposite angle)

∠QTS = 45°

Applying the angle sum property of triangle in ΔQTS

∠QTS + ∠TSQ + ∠SQT = 180°

45° + 75° + ∠SQT  = 180°

120° + ∠SQT = 180°

∠SQT = 180°- 120° = 60°

Hence the value of ∠SQT  is 60°

Q5. In fig.6.43, if PQ ⊥ PS, PQ ll SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

class 9 maths ex.6.3 fig.6.43

Ans.In the fig. we are given PQ ⊥ PS ⇒∠QPS =90°, PQ ll SR, ∠SQR = 28° and ∠QRT = 65°

Since PQ ll SR and QR is the transversal,so∠PQR (x + 28) and  ∠QRT(65°) are the alternate angles.

x + 28° = 65°

x = 65° – 28° = 37°

Applying the angle sum property of triangle in ΔQPS

∠QPS +y+ x = 180°

90° + y + x = 180°

x + y = 180° – 90° = 90°

Putting the value of x=37°

y + 37° = 90°

y = 90°- 37° = 53°

Therefore the value of x is 37° and of y is 53°

Q6. In fig.6.44 ,the side QR of ΔPQR is produced to a point S.If the bisector of ∠PQR and ∠PRS meet at point T,then prove that ∠QTR = 1/2 ∠QPR.

Fig 6.44 exercise 6.3

Ans. In the fig. we are given QT is the bisector of ∠PQR and TR is the bisector of ∠PRS

∴∠TRS = 1/2 ∠PRS and ∠TQR = 1/2 ∠PQR

In ΔPQR , ∠PRS is the exterior angles of ΔPQR and ∠PQR and ∠QPR are the opposite interior angles.

We know the exterior angle of a triangle is the sum of opposite interior angles.

∠PRS = ∠QPR + ∠PQR……..(i)

In ΔQTR, ∠TRS is the exterior angles of ΔQTR and ∠TQR and ∠QTR are the opposite interior angles.

We know the exterior angle of a triangle is the sum of opposite interior angles

∠TRS = ∠TQR  + ∠QTR

Since, we are given that ∠TRS = 1/2 ∠PRS and ∠TQR = 1/2 ∠PQR

∠PRS/2 = ∠PQR/2 + ∠QTR

∠PRS = ∠PQR + 2∠QTR…….(ii)

From the equation (i) and (ii), we have

∠QPR + ∠PQR = ∠PQR + 2∠QTR

∠QTR = 1/2 ∠QPR, Hence proved∴

See the video of each solution of execrcise  6.3

You can compensate us

Paytm number 9891436286

The money collected by us will be used for the education of poor students who leaves their study because of a lack of money.

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial Chapter 10-Circles
Chapter 3- Coordinate Geometry Chapter 11-Construction
Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles Chapter 14-Statistics
Chapter 7-Triangles Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science 

Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure? Chapter 10- Gravitation
Chapter3- Atoms and Molecules Chapter 11- Work and Energy
Chapter 4-Structure of the Atom Chapter 12- Sound
Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ?
Chapter 6- Tissues Chapter 14- Natural Resources
Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources
Chapter 8- Motion Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial Chapter 10-Circles
Chapter 3-Linear equations Chapter 11- Construction
Chapter 4- Quadratic equations Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume
Chapter 6-Triangle Chapter 14-Statistics
Chapter 7- Co-ordinate geometry Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds Chapter 12- Electricity
Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process Chapter 14-Sources of Energy
Chapter 7-Control and Coordination Chapter 15-Environment
Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets Chapter 9-Sequences and Series
Chapter 2- Relations and functions Chapter 10- Straight Lines
Chapter 3- Trigonometry Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations Chapter 15- Statistics
Chapter 8- Binomial Theorem  Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra
Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability Chapter 13-Probability
Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution