 # Class 9 Maths NCERT Solution for Line and Angle Exercise 6.1

In this page we have NCERT book Solutions for Class 9th Maths:Line and angles for exercise 6.1 Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1
In below figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Given,
∠AOC + ∠BOE = 70° and ∠BOD = 40°
Now AOB is a straight line
∠AOC + ∠BOE +∠COE = 180°
70° +∠COE = 180°
∠COE = 110°
Also COD is a straight lne
∠COE +∠BOD + ∠BOE = 180°
110° +40° + ∠BOE = 180°
150° + ∠BOE = 180°
∠BOE = 30°

Question 2
In below figure lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c. Given,
∠POY = 90° and a : b = 2 : 3
Lt x be the common ration, then a=2x and b=3x
Now XOY is a straight line
∠POY + a + b = 180°
90° + a + b = 180°
a + b = 90°
2x + 3x = 90°
5x = 90°
x = 18°
So a = 2×18° = 36°
and b = 3×18° = 54°
Also, now angle b and Angle c forms a linear pair
b + c = 180°
54° + c = 180°
c = 126°

Question 3
In below figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT. Given
∠PQR = ∠PRQ
To prove,
∠PQS = ∠PRT
Now   ∠PQR and ∠PQS forms a linear pair
∠PQR +∠PQS = 180°
∠PQS = 180° - ∠PQR --- (a)
Also, ∠PRQ and ∠PRT forms a linear pair
∠PRQ +∠PRT = 180° ( ∠PRT = 180° - ∠PRQ
Now as (∠PQR = ∠PRQ
∠PRQ = 180° - ∠PQR --- (b)
From (a) and (b)
∠PQS = ∠PRT = 180° - ∠PQR
Therefore, ∠PQS = ∠PRT

Question 4
In below figure, if x + y = w + z, then prove that AOB is a line. Given,
x + y = w + z
To Prove,
AOB is a line or x + y = 180°
Now O is the point and we have four angles around it
x + y + w + z = 360°
(x + y) + (w + z) = 360° Now Given x + y = w + z
(x + y) + (x + y) = 360°
2(x + y) = 360°
(x + y) = 180°
Hence, x + y makes a linear pair. Therefore, AOB is a straight line.

Question 5
In below 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). Given,
OR is perpendicular to line PQ
To prove,
∠ROS = 1/2(∠QOS – ∠POS)
Proof
∠POR = ∠ROQ = 90° (Perpendicular)
From the figure, it is clear that
∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROQ --- (a)
∠POS = ∠POR - ∠ROS = 90° - ∠ROQ --- (b)
Subtracting (b) from (a)
∠QOS - ∠POS = 90° + ∠ROQ - (90° - ∠ROQ)
⇒ ∠QOS - ∠POS = 90° + ∠ROQ - 90° + ∠ROQ
⇒ ∠QOS - ∠POS = 2∠ROQ
⇒ ∠ROS = 1/2(∠QOS – ∠POS)
Hence, proved.

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. Given,
∠XYZ = 64°
YQ bisects ∠ZYP Now angle ∠XYZ and ∠ZYP forms a linear pair
∠XYZ +∠ZYP = 180°
64° +∠ZYP = 180° ∠ZYP = 116°
Now YQ bisects ∠ZYP
∠ZYQ = ∠QYP
Also ∠ZYP = ∠ZYQ + ∠QYP So ∠ZYP = 2∠ZYQ 2∠ZYQ = 116°
∠ZYQ = 58° = ∠QYP
Now,
∠XYQ = ∠XYZ + ∠ZYQ ∠XYQ = 64° + 58°
∠XYQ = 122°
Also,
reflex ∠QYP = 180° + ∠XYQ
∠QYP = 180° + 122°
=3020