**NCERT solutions class 9 maths exercise 6.2-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.6.28, find the values of x and y and then show that AB ll CD.**

Ans. It can be observed from the figure that 50° and x forms linear pair

So, x + 50° =180°

x = 180° – 50° = 130°

130° and y forms vertically opposite angles

So, y = 130°

x = y = 130°, implies that interior alternative angles are equal

Since alternative angles are equal, therefore AB ll CD, Hence proved

**Q2.In fig. 6.29,if AB ll CD, CD ll EF and y: z = 3 : 7, find x.**

Ans.If AB ll CD and CD ll EF

Then AB ll EF

Therefore x = z (alternate angle)

We are given

y: z = 3 : 7

Putting z = x

y : x = 3 : 7

Let y = 3a and x = 7a

y and x are co-interior angles ,so the sum of both x and y is 180°

y + x = 180°

3a + 7a = 180°

10 a = 180°

a = 18

Therefore y = 3**×**18 = 54° and y =7× 18 = 126°

Hence the value of x is 126°

**Q3.In fig.6.30 if AB ll CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE,∠GEF and ∠FGE.**

Ans. We are given that ∠GED = 126°

EF ⊥ CD, so ∠DEF = 90°

∠AGE = ∠GED (alternate angle)

∠AGE =126°

From fig.we have

∠GEF = ∠GED -∠DEF

Putting the value of ∠GED =126° and ∠DEF= 90°

∠GEF = 126° – 90°= 36°

∠AGE and ∠FGE are linear pair

So,∠AGE + ∠FGE = 180°

Putting the value ∠AGE = 126°

126°+ ∠FGE = 180°

∠FGE = 180° – 126° = 54°

Hence the value of ∠AGE= 126°,∠GEF=36° and ∠FGE= 54°

**Q4. In fig. 6.31 ,if PQ ll ST,∠PQR = 110° and ∠RST = 130°, find ∠QRS.(Hint: draw a line parallel to ST through point R).**

Ans.In the fig. we are given

∠PQR = 110° and ∠RST = 130°

PQ ll ST

Drawing a line XY ll PQ ll ST

Since XY ll PQ and QR is the transversal

∠PQR and ∠QRX are co-interior angles

So, ∠PQR + ∠QRX = 180°

Putting the value of ∠PQR = 110°

110° +∠QRX = 180°

∠QRX = 180°- 110° = 70°

From the fig. we have

XY ll ST and SR is the transversal

So,∠RST and ∠SRY are the co-interior angles

∠RST + ∠SRY = 180°

Putting the value of ∠RST = 130°

130° + ∠SRY = 180°

∠SRY = 180° – 130° = 50°

It can be observed in the figure that ∠SRX and ∠SRY are the linear pair

∴ ∠SRX + ∠SRY = 180°

∠QRX + ∠QRS + ∠SRY = 180°

Putting the value of ∠QRX = 70° and ∠SRY = 50°

∠QRS + 70° + 50° = 180°

∠QRS + 120° = 180°

∠QRS = 180° – 120° = 60°

Therefore the value of ∠QRS is 60°

**Q5.In fig.6.32 , if AB ll CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.**

Ans.We are given in the figure

AB ll CD , ∠APQ = 50° and ∠PRD = 127°

AB ll CD and SR is the transversal, so ∠APR and ∠PRD are the alternates angle

∠APR = ∠PRD

∠APQ + y = 127°

50° + y = 127°

y = 127°- 50° = 77°

AB ll CD and PQ is the transversal, so ∠APQ and ∠PQR are the alternates angle

∠APQ = ∠PQR

x = 50°

Hence the value of x is 50° and of y is 77°

**Q6. In fig. 6.33, PQ and RS are the two mirrors placed parallel to each other. 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. Prove that AB ll CD.**

Ans. We are given the plane mirrors PQ ll RS, the ray AB is incident on PQ and ray CD incident on RS.

Let BM is normal to the plane PQ and CN is normal to the plane RS

Let ∠ABM = i_{1}, is the incident angle on PQ, r_{1} is the reflected angle on PQ, i_{2} is the incident angle on RS and r_{2} is the reflected angle on RS

PQ and RS can be supposed as parallel lines and BC as transversal then ∠ABC and ∠BCD are the alternate angles

∴ ∠ABC = ∠BCD

i_{1}= r_{1}and i_{2}= r_{2 }(incident angle = reflected angle)

BM ⊥PQ and CN ⊥ RS

So, BM ll CN (Perpendiculars on parallel lines are parallel) and BC transversal

Therefore r_{1}= i_{2 }(alternate angles)

2r_{1}= 2i_{2}

_{i1}+r_{1} = i_{2} + r_{2}

∠ABC = ∠BCD (alternate angles)

∴AB ll CD

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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 |

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Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

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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 |

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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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