NCERT Solutions Class 9th Maths Chapter – 6 Lines and Angles Exercise – 6.1

NCERT Solutions Class 9th Maths Chapter – 6 Lines and Angles

TextbookNCERT
Class 9th
Subject Mathematics
Chapter6th
Chapter NameLines and Angles
CategoryClass 9th Mathematics
Medium English
SourceLast Doubt

Class 9th Maths Chapter – 6 Lines and Angles Exercise – 6.1 In This Chapter We Will Learn About Lines and Angles, Line-segment, Collinear points, Angle, Arms, Vertex, Ray, Acute, Right angle, Obtuse angle, Straight angle, Reflex angle, Complementry angle, Supplementry angle, Adjacent, Vertically opposite angle,  with Class 9th Maths Chapter – 6 Lines and Angles Exercise – 6.1.

NCERT Solutions Class 9th 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.
CH 6 6.1
Solution – From the diagram, we have
(∠AOC +∠BOE +∠COE) and (∠COE +∠BOD +∠BOE) forms a straight line.
So, ∠AOC+∠BOE +∠COE = ∠COE +∠BOD+∠BOE = 180°
Now, by putting the values of ∠AOC + ∠BOE = 70° and ∠BOD = 40° 
we get, ∠COE = 110° and ∠BOE = 30°
So, reflex ∠COE = 360o – 110o = 250o
Question 2. In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.
CH 6 6.1
Solution – Since XOY is a straight line.
= a + b + ∠POY = 180°
= a + b + 90° = 180°
= a + b = 180° – 90°
= a + b = 90°
Now, it is given that a : b = 2 : 3 so,
Let a be 2x and b be 3x
2x + 3x = 90°
5x = 90°
x = 90/5
x = 18
Now, 2x = 2 × 18 = 36°
3x = 3 × 18 = 54°
From the diagram, b+c also forms a straight angle so,
b + c = 180° (linear pair)
c + 54° = 180°
So, c = 126°

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

CH 6 6.1
Solution – ST is a straight line.
∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair]
Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair]
From (1) and (2), we have
∠PQS + ∠PQR = ∠PRT + ∠PRQ
But ∠PQR = ∠PRQ [Given]
So, ∠PQS = ∠PRT

Question 4. In figure, if x + y = w + ⇒, then prove that AOB is a line.
CH 6 6.1
Solution – Sum of all the angles at a point = 360°
∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360°
But (x + y) = (⇒ + w) [Given]
∴ (x + y) + (x + y) = 360° or,
2(x + y) = 360°
or, (x + y) = 360∘2 = 180°
∴ AOB is a straight line.

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 = 1/2(∠QOS−∠POS)

CH 6 6.1
Solution – 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 have
2 ∠ROS = (∠QOS – ∠POS)
∴ ∠ROS = 1/2(∠QOS−∠POS)

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.
CH 6 6.1
Solution – 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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