NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Lines and Angles |

Exercise |
Ex 6.1 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1

**Ex 6.1 Class 9 Maths Question 1.**

In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

**Solution:
**Here,∠AOC and ∠BOD are vertically opposite angles.

∴ ∠AOC = ∠BOD

⇒ ∠AOC – 40° [BOD=40° (Given)]…(i)

We have, ∠AOC + ∠BOE = 70° (Given)

40° + ∠BOE = 70° [from eq.(1)]

⇒ ∠BOE = 30°

Also, ∠AOC + ∠COE + ∠BOE – 180° (Linear pair axiom)

⇒ 40° + ∠COE + 30° = 180° ∠COE=110°

Now, ∠COE + reflex ∠COE = 360° (Angles at a point)

^{ }110° + reflex ∠COE = 360°

⇒ Reflex ∠COE = 250°

**Ex 6.1 Class 9 Maths Question 2.**

In figure lines XY and MN intersect at O. If ∠POY = 90° and a: b = 2: 3 find c.

**Solution:**

We have, ∠POY = 90°

⇒ ∠POY + ∠POX = 180° (Linear pair axiom)

⇒ ∠POX = 90°

⇒ a + b= 90°……….(i)

Also a : b = 2 : 3 (Given)

⇒ Let a = 2k, b = 3k

Now, from Eq. (i), we get

2k +3k- 90° ⇒ 5k = 90° ⇒ k = 18°

∴ a = 2 x 18° = 36°

and b = 3 x 18°= 54°

Now, ∠MOX + ∠XON = 180° b + c = 180°

⇒ 54° + c = 180°

⇒ c = 126°

**Ex 6.1 Class 9 Maths Question 3.**

In the given figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

**Solution:**

∠PQS +∠PQR =180° [linear Pair] …………(i)

∠PRT + ∠PRQ = 180° [linear Pair] …………(ii)

From eqs. (i) and (ii), we have

∠PQS + ∠PQR =∠PRT + ∠PRQ

⇒ ∠PQS = ∠PRT [∵ ∠PQR = ∠PRQ (given)]

**Ex 6.1 Class 9 Maths Question 4.**

In the given figure, if x + y= w + z, then prove that AOB is a line.

**Solution:
**

We know that, sum of all the angles around a point is 360°.

∴ x + y + z + w = 360°

⇒ (x + y) + (z + w) = 360°

⇒ (x + y) + (x + y)= 360° [ ∵ x + y = z + w, given]

⇒ 2(x + y) = 360°

⇒ (x + y)=\(\cfrac { 360 }{ 2 } \) ⇒ x + y = 180°

So, AOB is a straight line

**Hence proved.**

**Ex 6.1 Class 9 Maths Question 5.**

In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

∠ROS = \(\cfrac { 1 }{ 2 }\) (∠QOS – ∠POS)

**Solution:**

We have, ∠POR = ∠ROQ = 90° (∵ Given that, OR is perpendicular to PQ)

∴ ∠POS + ∠ROS = 90°

⇒ ∠ROS = 90°- ∠POS

On adding ZROS both sides, we get

⇒ 2 ∠ROS = 90°-∠POS + ∠ROS

⇒ 2∠ROS = (90° + ∠ROS) – ∠POS

⇒ 2∠ROS = ∠QOS – ∠POS

(∵ ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS)

⇒ ∠ROS = \(\cfrac { 1 }{ 2 }\) (∠QOS –∠POS) **Hence proved.**

**Ex 6.1 Class 9 Maths Question 6.**

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.

**Solution:**

Here, YQ bisects ∠ZYP.

Hence ∠ZYQ = \(\cfrac { 1 }{ 2 }\) ∠QYP = ∠ZYP ………..(i)

Given, ∠XYZ = 64° …………(ii)

∵ ∠XYZ + ∠ZYQ + ∠QYP = 180° (Linear pair axiom)

⇒ 64° + ∠ZYQ + ∠ZYQ = 180° [From eq .(i) a and(ii)]

⇒ 2∠ZYQ = 180°-64°

⇒ ∠ZYQ = \(\cfrac { 1 }{ 2 }\) x 116° ⇒ ∠ZYQ – 58°

∴ ∠XYQ = ∠XYZ + ∠ZYQ

⇒ 64° + 58° = 122°

Now, ∠QYZ + reflex ∠QYP = 360°

58° + reflex ∠QYP – 360° ⇒ reflex ∠QYP = 302

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