**NCERT
Solutions of Class 9 Maths Chapter 6 Exercise 6.1**

In this post, you
can find a complete explanation of exercise 6.1 class 9 maths which belongs to
Chapter 6 Lines and angles. Before you go through maths exercise 6.1 class 9,
you should read and study the theorems of chapter 6.

**NCERT Maths
Class 9 Chapter 6 Exercise 6.1**

Here you will find
complete solutions to class
9 maths chapter 6 exercise 6.1
along with the video. You can watch videos related to ex
6.1 class 9 which will help
you to understand the topic and questions of class 9 ex 6.1 lines and
angles exercise 6.1. Questions of maths
class 9 chapter 6 exercise 6.1are
asked in the examination based on theorems of chapter 6 Lines and angles.

**Class 9 Maths Chapter 6 Exercise 6.1**

**Let’s
start solving questions of ex 6.1 class 9**

#### Class 9 Maths Chapter 6 Exercise 6.1 **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 **Question 1** ex 6.1 class
9

Therefore, ∠AOC=40° …………..(i)

Since ∠BOD=40° (given)

Now, ∠AOC+∠BOE=70° (given)

So, , 40°+∠BOE=70° (from (i))

Therefore, , ∠BOE=70°-40°=30°…………..(ii)

Now, ∠AOC+∠COB=180° (leaner pair)

So, ∠AOC+∠COE+∠BOE=180°

Or 40°+∠COE+30°=180° [from (i) and (ii)]

So, ∠COE=180°-40°-30°=110°

Therefore, Reflex ∠COE=360°-110°=250°

#### Class 9 Maths Chapter 6 Exercise 6.1 **Question 2**

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

Solution **Question 2 **ex 6.1 class
9

Let the angles be 2x and 3x

a=2x,b=3x

∠POX=a+b

90°=2x+3x

5x=90°

x=`frac{90°}{5}`=18°

a=2x=2×18°=36°

b=3x=3×18°=54°

Also, MN is a line

Since ray OX stands on MN, therefore

∠MOX+∠XON=180° (leaner pair)

b+c=180°

⟹c+54°=180°

⟹180°-54°=126°

Hence c= 126°

#### Class 9 Maths Chapter 6 Exercise 6.1 **Question
3**

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

Solution **Question 3 **ex 6.1 class
9

SRT is a line

Since QP stands on the line SRT

Therefore, ∠PQS+∠PQR=180° (linear pair)……….(i)

Since RP stands on the line SRT,

Therefore, ∠PRQ+ ∠PRT=180° (linear pair)……(ii)

From (i) and (ii) , we have:

∠PQS+∠PQR=∠PRQ+∠PRT ………..(iii)

Also, ∠PQR=∠PRQ (given)………..(iv)

Subtracting (iv) from (iii), we have

∠PQS=∠PRT

Hence proved

#### Class 9 Maths Chapter 6 Exercise 6.1 **Question
4**

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

Solution **Question 4 **ex 6.1 class
9

Since the sum of all the angles around a point is equal to 360°

Therefore

(∠BOC+∠COA)+(∠BOD+∠AOD)=360°

⟹ (x+y)+(w+z)=360°

But x+y=w+z (given)

∴ x+y=w+z=`frac{360°}{2}`=180°

Thus, ∠BOC and ∠COA as well as ∠BOD and ∠AOD form linear pairs. Consequently, OA and OB are two rays. Therefore AOB is a straight line.

#### Class 9 Maths Chapter 6 Exercise 6.1 **Question
5**

**In the figure, PQR 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).**

Solution **Question
5 **ex 6.1 class 9

Therefore

∠POR=∠ROQ

Or ∠POS+∠ROS=∠QOS-∠ROS

⇒2∠ROS=∠QOS-∠POS

⇒∠ROS= `frac{1}{2}` (∠QOS-∠POS)

Hence proved

#### Class 9 Maths Chapter 6 Exercise 6.1 **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 ∠XYP and reflex ∠QYP.**

Solution **Question 6 **ex 6.1 class
9

Since XY is produced to point P, therefore Xp is a straight line.

Now ray YZ stands XP

∴ ∠XYZ+ ∠ZYP=180° (linear pair)

⟹ 64°+∠ZYP=180° (∵∠XYZ=64°)

⟹ ∠ZYP=180°- 64°=116°

Since ray YQ bisects ∠ZYP,

Therefore,

∠QYP=∠ZYQ=(116°)/2=58°

Now, ∠XYQ=∠XYZ+∠ZYQ

⇒ ∠XYQ=64°+58°=122°

And reflex ∠QYP=180°+∠XYQ

∠QYP=180°+122°=302°

Now you have understood the all questions of chapter 6 maths class 9 exercise 6.1, To better understand 6.1 maths class 9, **watch the video given below**.

Related Topics for you

3. Watch the Video of Exercise 6.1