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

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

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 6 Lines and Angles Ex 6.1. 

• NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1
• NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2
• NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

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.(i)]
⇒ ∠BOE = 30°
Also, ∠AOC + ∠COE + ∠BOE = 180°        (Linear pair of angles)
⇒  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 of angles)
⇒ ∠POX = 90°
⇒ a + b = 90°      ……….(i)
Also, a : b = 2 : 3        (Given)
⇒ Let a = 2k and b = 3k
Now, from Eq. (i), we get
2k + 3k = 90° 

⇒ 5k = 90° 

⇒ k = 18°
∴ a = 2 × 18° = 36° and b = 3 × 18° = 54°
Now, ∠MOX + ∠XON = 180°           (Linear pair of angles)

⇒ 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 of angles] …………(i)
∠PRT + ∠PRQ = 180°      [Linear pair of angles] …………(ii)
From equations. (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 the sum of all the angles at a point = 360°
∴ x + y + z + w = 360°

or, (x + y) + (z + w) = 360°
But (x + y) = (z + w)             [Given]
∴ (x + y) + (x + y) = 360°
⇒ 2(x + y) = 360°
or, (x + y) = 360°/2 = 180°
∴ AOB is a straight line.

 

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 = ½ (∠QOS – ∠POS)

Solution:
We have, POQ is a straight line.     [Given]
∴ ∠POS + ∠ROS + ∠ROQ = 180°
But OR ⊥ PQ,
∴ ∠ROQ = 90°
⇒ ∠POS + ∠ROS + 90° = 180°
⇒ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS      … (1)
Now, we have ∠ROS + ∠ROQ = ∠QOS
⇒ ∠ROS + 90° = ∠QOS
⇒ ∠ROS = ∠QOS – 90°      …(2)

Adding (1) and (2), we get
2∠ROS = (∠QOS – ∠POS)
∴ ∠ROS = ½ (∠QOS − ∠POS)

 

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:
We have, XYP is a straight line.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180°
⇒ 64° + ∠ZYQ + ∠QYP = 180°         [∵ ∠XYZ = 64°  (given)]
⇒ 64° + 2∠QYP = 180°                     [YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ]

⇒ 2∠QYP = 180° – 64° = 116°
⇒ ∠QYP = 116°/2 = 58°
∴ Reflex ∠QYP = 360° – 58° = 302°
Since ∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + ∠QYP                     [∵ ∠XYZ = 64° (Given) and ∠ZYQ = ∠QYP]
⇒ ∠XYQ = 64° + 58° = 122°             [∠QYP = 58°]
Thus, ∠XYQ = 122° and reflex ∠QYP = 302°.

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