These NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1
Question 1.
In Fig. 6.13 lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°. Find ∠BOE and reflex ∠COE.
Solution:
In this fig. we have given that
∠AOC + ∠BOE = 70° and ∠BOD = 40°
But, ∠BOD = ∠AOC (Vertical opposite angles)
∠AOC = 40 ……(i) (∵ ∠BOD = 40°)
Again, ∠AOC + ∠BOE = 70° (Given)
40 + ∠ZBOE = 70° (from equation (i))
∠BOE = 70 – 40 = 30°
Again, ∠AOC + ∠COE + ∠EOB = 180° (Because AB is a st. line)
40° + ∠COE + 30° = 180° (∵ ∠AOC = 40° and ∠EOB = 30°)
Therefore, ∠BOE = 30°, and Reflex ∠COE = 360° – 110° = 250°
Therefore, ∠BOE = 30°, and Reflex ∠COE = 250°.
Question 2.
In Fig. 6.14 lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Solution:
We have given
∠POY = 90°, and a : b = 2 : 3
or, \(\frac{a}{b}=\frac{2}{3}\)
∴ a = \(\frac {2}{3}\) b ……(i)
Again, ∠POY + a + b = 180° (Because OXY is a straight line)
90 + \(\frac {2}{3}\) b + b = 180°
or, \(\frac {2}{3}\) b + b = 90
b = \(\frac{90 \times 3}{5}\) = 54° ……(ii)
Again, c + b = 180° (because MN is a straight line)
c + 54° = 180° (from equation (ii))
Therefore, c = 126°
Question 3.
In Fig. 6.15, ∠PQR = ∠PRQ then prove that ∠PQS = ∠PRT.
Solution:
We have given
∠PQR = ∠PRQ
Again,
∠PQR + ∠PQS = 180° ……(i) (Linear pair)
∠PRQ = ∠PRT = 180° ……(ii) (Linear pair)
From equation (i) and (ii)
∠PQR + ∠PQS = ∠PRQ + ∠PRT
But, ∠PQR = ∠PRQ
Therefore, ∠PQS = ∠PRT
Question 4.
In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
Solution:
We have given x + y = w + z
Again, x + y + w + z = 360
or, x + y + x + y = 360 (∵ x + y = w + z)
or, 2(x + y) = 360°
or, x + y = 180°
∠AOB = 180°
which is straight line angle.
∴ AOB is a straight line.
Question 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 = \(\frac {1}{2}\) (∠QOS – ∠POS)
Solution:
We have given,
Ray OR is perpendicular to line PQ
Therefore, ∠QOR = ∠POR = 90° ……(i)
Now, ∠QOS = ∠QOR + ∠ROS
or, ∠QOS = 90° + ∠ROS ……(ii) (from equation(i))
Again, ∠POS = ∠POR – ∠ROS
or, ∠PSO = 90 – ∠ROS ……(iii) (from equation (i))
Subtract equation (iii) from equation (ii),
∠QOS – ∠POS
∠QOS – ∠POS = 90° + ∠ROS – (90° – ∠ROS)
∠QOS – ∠POS = 90° + ∠ROS – 90° + ∠ROS
∠QOS – ∠POS = 2∠ROS
∠ROS = \(\frac {1}{2}\) (∠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.
Solution:
We have given that ∠XYZ = 64°
and ∠ZYQ = ∠QYP (∵ YQ is bisector of ∠ZYP)
Now, ∠XYZ + ∠ZYQ + ∠QYP = 180° (because XYP is a straight line)
or, 64° + ∠ZYQ + ∠ZYQ = 180° (∵ ∠ZYQ + ∠QYP)
or, 2∠ZYQ = 180° – 64° = 116°
or, ∠ZYQ = 58°
Now, ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58°
∠XYQ = 122°
Again, ∠QYP = ∠ZYQ (from equation (i))
∴ ∠QYP = 58° (∵ ∠ZYQ = 58°)
∴ Reflex ∠QYP = 360° – 58° = 302°