Class 9 Lines and Angles Ex 6.1

 

ncert class 9 maths chapter 6 solution, Lines and Angles Exercise 6.1 involve complete  answers for each question in the exercise 6.1. The solutions provide students a strategic methods  to prepare for their exam. Class 9 Maths Chapter 6 Lines and Angles exercise 6.1 questions and answers helps students to perform better in exam and it will  clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems.NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles prepared by www.mathematicsandinformationtechnology.com team in very delicate, easy and creative way. 

Question 1:

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

Solution 1:

AB is a straight line, rays OC and OE stand on it.

∠AOC+∠COE+∠BOE=180º 

(∠AOC+∠BOE)+∠COE=180º

70º+∠COE=180º

∠COE=180º-70º=110º

Reflex ∠COE=360-110=250

CD is a straight line, rays OE and OB stand on it.

∠COE+∠BOE+∠BOD=180º

110º+∠BOE+40º=180º

∠BOE=180º-150º=30º

Hence, ∠BOE = 30º and Reflex ∠COE = 250º

Question 2:

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

Solution 2:

Let the common ratio between a and b be x.

∴ a = 2x, and b = 3x 

XY is a straight line, rays OM and OP stand on it.

∴ ∠XOM + ∠MOP + ∠POY = 180º

b + a + ∠POY = 180º

3x + 2x + 90º = 180º

5x = 90º

x = 18º

a = 2x = 2 × 18 = 36º

b = 3x= 3 ×18 = 54º

MN is a straight line. Ray OX stands on it.

∴ b + c = 180º (Linear Pair)

54º + c = 180º

c = 180º − 54º = 126º

∴ c = 126º

Question 3:

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

Solution 3:

In the given figure, ST is a straight line and ray QP stands on it.

∴ ∠PQS + ∠PQR = 180º (Linear Pair)

∠PQR = 180º − ∠PQS … (1)

∠PRT + ∠PRQ = 180º (Linear Pair)

∠PRQ = 180º − ∠PRT … (2)

It is given that ∠PQR = ∠PRQ.

Equating Equations (1) and (2), we obtain

180º − ∠PQS = 180 − ∠PRT

∠PQS = ∠PRT (Proved)

Question 4:

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

Solution 4:

It can be observed that,

x + y + z + w = 360º (Complete angle)

It is given that,

x + y = z + w

∴ x + y + x + y = 360º

2(x + y) = 360º

x + y = 180º

Since x and y form a linear pair, therefore, AOB is a line. (Proved)

Question 5:

In the given 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 =½(∠QOS − ∠POS)

Solution 5:

It is given that OR ⊥ PQ

∠POR = 90º

∠POS + ∠SOR = 90º

∠ROS = 90º − ∠POS … (1)

∠QOR = 90º (As OR ⊥ PQ)

∠QOS − ∠ROS = 90º

∠ROS = ∠QOS − 90º … (2)

On adding Equations (1) and (2), we obtain

2 ∠ROS = (∠QOS − ∠POS)

∠ROS =½(∠QOS − ∠POS)

Question 6:

It is given that ∠XYZ = 64o 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 6:

It is given that line YQ bisects ∠PYZ.

Hence, ∠QYP = ∠ZYQ

It can be observed that PX is a line. Rays YQ and YZ stand on it.

∠XYZ + ∠ZYQ + ∠QYP = 180º

64º + 2∠QYP = 180º

2∠QYP = 180º − 64º = 116º

∠QYP = 58º

Also, ∠ZYQ = ∠QYP = 58º

Reflex ∠QYP = 360º − 58º = 302º

∠XYQ = ∠XYZ + ∠ZYQ

= 64º + 58º = 122º

Hence, ∠XYQ = 122º, Reflex ∠QYP = 302º.