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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