NCERT Solutions for Class 9 Maths | NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

Chapter 6 Lines and Angles 

Exercise – 6.1

Question 1 In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

 

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Q1

 

 

Solution – 

Given –  ∠AOC + ∠BOE = 70°

∠BOD = 40°

∠AOC and ∠BOD are vertically opposite angles

∴ ∠AOC = ∠BOD = 40°

AB is a straight line,
∴ ∠AOC + ∠COE + ∠EOB = 180°
or (∠AOC + ∠BOE) + ∠COE = 180° …………. ( 1 )

put ∠AOC + ∠BOE = 70°  in equation (1)

70° + ∠COE = 180°
∠COE = 180° – 70° = 110°
∴ Reflex ∠COE = 360° – 110° = 250°

∠AOC + ∠BOE = 70°      …………. ( 2 )

put  ∠AOC = 40° in equation ( 2)
∴ 40° + ∠BOE = 70°

∠BOE = 70° – 40° = 30°
∠BOE = 30° and reflex ∠COE = 250° 

 

 

 

Question 2. In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Q2

 

Solution:

Given   ∠POY = 90°

XY is a line ,  sum of linear pair is always equal to 180°

So,   a +b + ∠POY  = 180° ……………………… ( 1 )

Given                  a:b = 2:3

Suppose common ratio of a and b is x

then a = 2x and

b = 3x

Putting the value of  a= 2x , b= 3x and  ∠POY = 90° in equation (1)  we get,

a +b + ∠POY  = 180°

2x + 3x + 90°  = 180°

5x = 180° – 90°

5x = 90°

x = 90° / 5

x = 18°

∴ a = 2×18° = 36°

b = 3×18° = 54°

From the diagram, b+c also forms a straight angle, so

b+c = 180°

c+54° = 180°

∴ c = 126°

 

Question 3. In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Q3

Solution –

Given – ∠PQR = ∠PRQ ………..(1 )

∠PQR + ∠PQS = 180° [Linear pair of angles ]

∠PQR = 180° – ∠PQS  ………..(2 )
Similarly,

∠PRQ + ∠PRT = 180°  [Linear Pair of angles ]

∠PRQ =  180° –  ∠PRT ……… (3)

From (1) ,  (2) and (3)  –

put value ∠PQR and ∠PRQ from equation (1) and (2) in equation (1)

∠PQR = ∠PRQ

180° – ∠PQS  = 180° –  ∠PRT
– ∠PQS = –  ∠PRT

∴ ∠PQS = ∠PRT

 

Question 4.
In figure, if x + y = w + ⇒, then prove that AOB is a line.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Q4
Solution:

Given  –    (x + y) = (w + z)
Sum of all the angles at a point O = 360°

∴ x + y +  w + z  = 360° …………. ( 1 )

Put (x + y) = (w + z)  in equation (1)
(x + y) + (w + z)   = 360°
(x + y) + (x + y) = 360°
2(x + y) = 360°
or, (x + y) = 360 / 2 = 180°

∠x  and  ∠y are linear pair angles.

if  x+y = 180° then uncommon arms of these linear pair angles form a line .
∴ AOB is a straight line.

 

5. 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 = ½ (∠QOS – ∠POS).

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Solution:

Given  (OR ⊥ PQ)

∠POQ = 180°

∠POR =  900           (OR ⊥ PQ)

∠ROQ = 900           (OR ⊥ PQ)

∠POR = ∠ROQ

∠POR written as ∠POS + ∠ROS

∠POR = ∠POS + ∠ROS  = 900

∠POS + ∠ROS  = 900

∠ROS = 90– ∠POS ………………………. ( 1 )

∠ROQ written  as ∠QOS – ∠ROS

∠ROQ = ∠QOS – ∠ROS = 900

∠QOS – ∠ROS = 900

– ∠ROS = 900  – ∠QOS

– ∠ROS = – ( ∠QOS – 90)

∠ROS =   ∠QOS – 90…………………… ( 2 )

Add Equation (1) and (2)

∠ROS + ∠ROS = 90– ∠POS + ∠QOS – 900

2∠ROS  =  ∠QOS – ∠POS + 900 – 900

2∠ROS = ∠QOS – ∠POS

∠ROS = ( ∠QOS – ∠POS ) / 2

∠ROS = 1/2 ( ∠QOS – ∠POS )         Hence Proved 

 

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.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Q6

Solution:

Given  XYP is a straight line

∠XYZ +∠ZYP = 180°

∠XYZ +  ∠ZYQ +  ∠QYP = 180°………………. ( 1 )            ( ∠ZYP   = ∠ZYQ +  ∠QYP )

Given YQ bisects ∠ZYP

∴ ∠ZYQ = ∠QYP …………………………………….( 2 )

Putting the value of ∠XYZ = 64° and ∠ZYQ = ∠QYP in Equation (1)

∠XYZ +  ∠ZYQ +  ∠QYP = 180°

64° + ∠QYP + ∠QYP  = 180°

2 ∠QYP = 180° – 64°

2 ∠QYP   = 116°

∠QYP = 116° / 2

∠QYP = 58°

Reflex Of ∠QYP  = 360° – 58°  = 302° 

∠XYQ = ∠XYZ + ∠ZYQ

∠XYQ = 64°  +  58° = 122°

Reflex Of ∠QYP  = 360° – 58°  = 302° 

 

 

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