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
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 = 3a/2 …(ii)
Now from (i) and (ii), we get
3a/2 + A = 90°
⇒ 5a/2 = 90°
⇒ a = 90∘/5×2=36∘ = 36°
From (ii), we get
b = 3/2 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 = 1/2(∠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°.
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
Lines and Angles Class 9 Extra Questions Maths Chapter 6
Extra Questions for Class 9 Maths Chapter 6 Lines and Angles
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