NCERT Solutions for Class 9 Maths | NCERT Solutions for Class 9 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.
Solution –
Given – ∠AOC + ∠BOE = 70°
∠BOD = 40°
∠AOC and ∠BOD are vertically opposite angles
∴ ∠AOC = ∠BOD = 40°
AB is a straight line,
∴ ∠AOC + ∠COE + ∠EOB = 180°
or (∠AOC + ∠BOE) + ∠COE = 180° …………. ( 1 )
put ∠AOC + ∠BOE = 70° in equation (1)
70° + ∠COE = 180°
∠COE = 180° – 70° = 110°
∴ Reflex ∠COE = 360° – 110° = 250°
∠AOC + ∠BOE = 70° …………. ( 2 )
put ∠AOC = 40° in equation ( 2)
∴ 40° + ∠BOE = 70°
∠BOE = 70° – 40° = 30°
∠BOE = 30° and reflex ∠COE = 250°
Question 2. In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.
Solution:
Given ∠POY = 90°
XY is a line , sum of linear pair is always equal to 180°
So, a +b + ∠POY = 180° ……………………… ( 1 )
Given a:b = 2:3
Suppose common ratio of a and b is x
then a = 2x and
b = 3x
Putting the value of a= 2x , b= 3x and ∠POY = 90° in equation (1) we get,
a +b + ∠POY = 180°
2x + 3x + 90° = 180°
5x = 180° – 90°
5x = 90°
x = 90° / 5
x = 18°
∴ a = 2×18° = 36°
b = 3×18° = 54°
From the diagram, b+c also forms a straight angle, so
b+c = 180°
c+54° = 180°
∴ c = 126°
Question 3. In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution –
Given – ∠PQR = ∠PRQ ………..(1 )
∠PQR + ∠PQS = 180° [Linear pair of angles ]
∠PQR = 180° – ∠PQS ………..(2 )
Similarly,
∠PRQ + ∠PRT = 180° [Linear Pair of angles ]
∠PRQ = 180° – ∠PRT ……… (3)
From (1) , (2) and (3) –
put value ∠PQR and ∠PRQ from equation (1) and (2) in equation (1)
∠PQR = ∠PRQ
180° – ∠PQS = 180° – ∠PRT
– ∠PQS = – ∠PRT
∴ ∠PQS = ∠PRT
Question 4.
In figure, if x + y = w + ⇒, then prove that AOB is a line.
Solution:
Given – (x + y) = (w + z)
Sum of all the angles at a point O = 360°
∴ x + y + w + z = 360° …………. ( 1 )
Put (x + y) = (w + z) in equation (1)
(x + y) + (w + z) = 360°
(x + y) + (x + y) = 360°
2(x + y) = 360°
or, (x + y) = 360 / 2 = 180°
∠x and ∠y are linear pair angles.
if x+y = 180° then uncommon arms of these linear pair angles form a line .
∴ AOB is a straight line.
5. In Fig. 6.17, 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:
Given (OR ⊥ PQ)
∠POQ = 180°
∠POR = 900 (OR ⊥ PQ)
∠ROQ = 900 (OR ⊥ PQ)
∠POR = ∠ROQ
∠POR written as ∠POS + ∠ROS
∠POR = ∠POS + ∠ROS = 900
∠POS + ∠ROS = 900
∠ROS = 900 – ∠POS ………………………. ( 1 )
∠ROQ written as ∠QOS – ∠ROS
∠ROQ = ∠QOS – ∠ROS = 900
∠QOS – ∠ROS = 900
– ∠ROS = 900 – ∠QOS
– ∠ROS = – ( ∠QOS – 900 )
∠ROS = ∠QOS – 900 …………………… ( 2 )
Add Equation (1) and (2)
∠ROS + ∠ROS = 900 – ∠POS + ∠QOS – 900
2∠ROS = ∠QOS – ∠POS + 900 – 900
2∠ROS = ∠QOS – ∠POS
∠ROS = ( ∠QOS – ∠POS ) / 2
∠ROS = 1/2 ( ∠QOS – ∠POS ) Hence Proved
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:
Given XYP is a straight line
∠XYZ +∠ZYP = 180°
∠XYZ + ∠ZYQ + ∠QYP = 180°………………. ( 1 ) ( ∠ZYP = ∠ZYQ + ∠QYP )
Given YQ bisects ∠ZYP
∴ ∠ZYQ = ∠QYP …………………………………….( 2 )
Putting the value of ∠XYZ = 64° and ∠ZYQ = ∠QYP in Equation (1)
∠XYZ + ∠ZYQ + ∠QYP = 180°
64° + ∠QYP + ∠QYP = 180°
2 ∠QYP = 180° – 64°
2 ∠QYP = 116°
∠QYP = 116° / 2
∠QYP = 58°
Reflex Of ∠QYP = 360° – 58° = 302°
∠XYQ = ∠XYZ + ∠ZYQ
∠XYQ = 64° + 58° = 122°
Reflex Of ∠QYP = 360° – 58° = 302°
