Maths Chapter 6 Lines and Angles

 

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

Ex 6.1 Class 9 Maths Question 1
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:
Since AB is a straight line,
∴ ∠AOC + ∠COE + ∠EOB = 180°
or (∠AOC + ∠BOE) + ∠COE = 180° or 70° + ∠COE = 180° [ ∵∠AOC + ∠BOE = 70° (Given)]
or ∠COE = 180° – 70° = 110°
∴ Reflex ∠COE = 360° – 110° = 250°
Also, AB and CD intersect at O.
∴∠COA = ∠BOD [Vertically opposite angles]
But ∠BOD = 40° [Given]
∴ ∠COA = 40°
Also, ∠AOC + ∠BOE = 70°
∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30°
Thus, ∠BOE = 30° and reflex ∠COE = 250°.

Ex 6.1 Class 9 Maths Question 2.
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.

Solution:
Since XOY is a straight line.
∴ b+a+∠POY= 180°
But ∠POY = 90° [Given]
∴ b + a = 180° – 90° = 90° …(i)
Also a : b = 2 : 3 ⇒ b = 3a2 …(ii)
Now from (i) and (ii), we get
3a2 + A = 90°
⇒ 5a2 = 90°
⇒ a = 90∘5×2=36∘ = 36°
From (ii), we get
b = 32 x 36° = 54°
Since XY and MN interstect at O,
∴ c = [a + ∠POY] [Vertically opposite angles]
or c = 36° + 90° = 126°
Thus, the required measure of c = 126°.

Ex 6.1 Class 9 Maths Question 3.
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

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]
∴ ∠PQS = ∠PRT

Ex 6.1 Class 9 Maths Question 4.
In figure, if x + y = w + ⇒, then prove that AOB is a line.

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.

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

Solution:
rara 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 = 12(∠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:
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°.

∴ ∠QRT = ∠RQS + ∠RSQ
[Exterior angle property of a triangle]
But ∠RQS = 28° and ∠QRT = 65°
So, 28° + ∠RSQ = 65°
⇒ ∠RSQ = 65° – 28° = 37°
Since, PQ || SR and QS is a transversal.
∴ ∠PQS = ∠RSQ = 37°
[Alternate interior angles]
⇒ x = 37°
Again, PQ ⊥ PS ⇒ AP = 90°
Now, in ∆PQS,
we have ∠P + ∠PQS + ∠PSQ = 180°
[Angle sum property of a triangle]
⇒ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
Thus, x = 37° and y = 53°

Ex 6.3 Class 9 Maths Question 6.
In figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that

Solution:
In ∆PQR, side QR is produced to S, so by exterior angle property,
∠PRS = ∠P + ∠PQR
⇒ 12∠PRS = 12∠P + 12∠PQR
⇒ ∠TRS = 12∠P + ∠TQR …(1)
[∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.]
Now, in ∆QRT, we have
∠TRS = ∠TQR + ∠T …(2)
[Exterior angle property of a triangle]
From (1) and (2),
we have ∠TQR + 12∠P = ∠TQR + ∠T
⇒ 12∠P = ∠T
⇒ 12∠QPR = ∠QTR or ∠QTR = 12∠QPR

NCERT Solutions for Class 9 Maths

• Chapter 1 Number systems
• Chapter 2 Polynomials
• Chapter 3 Coordinate Geometry
• Chapter 4 Linear Equations in Two Variables
• Chapter 5 Introduction to Euclid Geometry
• Chapter 6 Lines and Angles
• Chapter 7 Triangles
• Chapter 8 Quadrilaterals
• Chapter 9 Areas of Parallelograms and Triangles
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Heron’s Formula
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability