**NCERT Solutions class 9 maths exercise 6.1-Lines and Angles**

NCERT solutions class 9 maths exercise 6.1 of chapter 6-Lines and Angles is an easy chapter since the students of 9 class already would have studied Lines and Angles in their previous classes. In this chapter 6- Lines and Angles, the questions are based on lines and angles. All unsolved questions of exercise 6.1 are solved by an expert of maths as per the CBSE norms by a step by step method with a suitable diagram.

**NCERT Solutions class 9 maths Chapter 6-Lines and Angles**

**NCERT Solutions of 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 |

**Q1.In fig.6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.**

Ans. We are given

∠AOC + ∠BOE = 70°

∠BOD = 40°

∠BOD = ∠AOC (Vertically opposite angle)

∴ ∠AOC = 40°

Putting values of ∠AOC in the given equation ∠AOC + ∠BOE = 70°

40 + ∠BOE = 70

∠BOE = 30°

∠AOE + ∠BOE = 180° (Linear pair)

∠AOC + ∠COE + ∠BOE = 180°

40° + ∠COE + 30° = 180°

∠COE = 180° – 70° = 110°

Reflex ∠COE = 360° – ∠COE

Reflex ∠COE = 360° – 110° =250°

Therefore ∠BOE =30° and reflex ∠COE =250°

**Q2.In fig.6.4 ,lines XY and MN intersect at O .If ∠POY =90° and a: b =2: 3,find c.**

Ans.We are given

∠POY =90° and a: b =2: 3

∠POX = 180° – 90° = 90°(Linear pair)

Let a = 2x and b = 3x

∠POX = a + b = 90

2x + 3x = 90

5x = 90°

x = 18°

Therefore a = 2×18 = 36° and b = 3× 18 = 54°

Since MON is a line therefore b and c are linear pair

∴b + c =180°

54° + c = 180°

c = 180° – 54° = 126°

Hence the value of c is 126°

Q3. In fig. 6.15 ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT

Ans.

**Given: **∠PQR = ∠PRQ

**To Prove:** ∠PQS = ∠PRT

**Proof: **In the given fig.

∠PQS + ∠PQR = 180°(Linear pair)…..(i)

∠PRT + ∠PRQ = 180°(Linear pair)…..(ii)

From equations (i) and (ii)

∠PQS + ∠PQR = ∠PRT + ∠PRQ

∠PQR = ∠PRQ (given)

∠PQS + ∠PRQ = ∠PRT + ∠PRQ

∠PQS = ∠PRT

Hence Proved

**Q4. In fig.6.16 if x +y = w +z, then prove that AOB is a line.**

Ans. We have In the given figure

**Given:** x +y = w +z

**To Prove: **AOB is a line

**Proof: **From the figure, we can observe that all the given angles form a complete angle of 360°

Therefore we can write

x +y + w +z = 360° (complete angle)

w + z + w + z = 360°

2w + 2z = 360°

2(w + z) = 360°

w + z = 180

The sum of w and z implies that both angle w and z forms a linear pair, so AOB is a line.

**Q5. 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.

**Given:** POQ is a line

OR ⊥ PQ

∴ ∠POR = 90° and ∠QOR = 90°

**To Prove:** ∠ROS = 1/2(∠QOS -∠POS)

Proof: POQ is a line (given)

Therefore ∠POR and ∠QOR are the linear pair

∠POR + ∠QOR = 180°

∠POS + ∠ROS + 90° = 180°

∠POS + ∠ROS = 90°…..(i)

From the fig. we have

∠QOS = ∠QOR + ∠ROS

∠QOS = 90° + ∠ROS

∠QOS – ∠ROS = 90°……(ii)

From (i) and (ii)

∠POS + ∠ROS = ∠QOS – ∠ROS

∠ROS + ∠ROS = ∠QOS – ∠POS

2∠ROS = ∠QOS – ∠POS

∠ROS = 1/2(∠QOS – ∠POS)

Hence proved

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**Q6.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.

We are given

∠XYZ = 64°

Let ∠PYZ = 2x

∠QYZ = ∠QYP = x ( YQ is the bisector of ∠PYZ)

Since PYX is a line,so

∠PYZ +∠XYZ = 180°

∠QYZ + ∠QYP + 64° = 180°

x + x = 180° – 64° = 116°

2x = 116°

x = 58°,so ∠QYZ = ∠QYP = 58°

∴∠XYQ = ∠XYZ + ∠QYZ = 64° + 58° = 122°

Since ∠QYP = 58°, therefore reflex ∠QYP = 360°-58° = 302°

Hence ∠XYQ = 122° and reflex ∠QYP = 302°

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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**

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**NCERT solutions of 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 |

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