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NCERT Solutions class 9 Maths exercise 6.3 of chapter 6-Lines and Angles

EX.6.3 MATHS CLASS 9

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.

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 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°

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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∴

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

NCERT Solutions for class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions for class 9 science 

Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
Chapter 8- MotionLast years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
Chapter 8-Trigonometry

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

NCERT Solutions for class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem Chapter 16- Probability

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NCERT solutions for class 12 maths

Chapter 1-Relations and FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
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Chapter 7- Integrals
Chapter 8-Application of Integrals

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