Lines and Angles
Chapter 6
Exercise 6.1
EX 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:
Given: Lines AB and CD intersect at 0.
∠BOD = 40°
∠AOC + ∠BOE = 70° …..(i)
∠AOC = ∠BOD [vertically opposite angles]
∴ ∠AOC = 40° [∠BOD = 40°]
from equation (i)
40°+ ∠BOE = 70°
⇒∠BOE = 70° – 40° = 30°
∠AOC + ∠BOE + ∠COE = 180° [AOB is a straight line]
⇒ 70°+∠COE = 180° [from equation (i)]
⇒∠COE = 180° – 70° = 110°
and ∠COE =360° – ∠COE= 360°- 110° = 250°
EX 6.1 QUESTION 2.
In Fig. 6.14, lines XY and MN intersect at O. If POY = 90° and a: b = 2 : 3, find c.
Solution:
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
Now from (i) and (ii), we get
⇒ 5a/2= 90°
⇒ a =(90°/5)x2 = 36°
From (ii), we get
b = 
XY and MN interstect at O,
∴ c = [a + ∠POY] [Vertically opposite angles]
or c = 36° + 90° = 126°
EX 6.1 QUESTION 3.
In Fig. 6.15, PQR = PRQ, then prove that PQS = PRT.
Solution:
ST is a straight line.
∴ ∠PQR + ∠PQS = 180° …(i) [Linear pair]
Similarly, ∠PRT + ∠PRQ = 180° …(ii) [Linear Pair]
From (i) and (ii), we have
∠PQS + ∠PQR = ∠PRT + ∠PRQ
But ∠PQR = ∠PRQ [Given]
∴ ∠PQS = ∠PRT
EX 6.1 QUESTION 4.
In Fig. 6.16, if x+y = w+z, 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) = 
∴ AOB is a straight line.
EX 6.1 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 = ½ (QOS – POS).
Solution:
Given: POQ is a straight line.
∴ ∠POS + ∠ROS + ∠ROQ = 180°
But OR ⊥ PQ
∴ ∠ROQ = 90°
⇒ ∠POS + ∠ROS + 90° = 180°
⇒ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS … (i)
Now, we have ∠ROS + ∠ROQ = ∠QOS
⇒ ∠ROS + 90° = ∠QOS
⇒ ∠ROS = ∠QOS – 90° ……(ii)
Adding (1) and (2), we have
2 ∠ROS = (∠QOS – ∠POS)
∴ ∠ROS = 1/2(∠QOS – ∠POS)
EX 6.1 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:
∠PYZ + ∠XYZ = 180° [Linear pair]
⇒∠PYZ + 64° = 180° [XYZ = 64° ]
⇒∠PYZ = 180° – 64° = 116°
But,
∠PYQ =∠ ZYQ = 1/2 ∠PYZ [∠ZYP is bisected by ray YQ}
∴∠PYQ =∠ ZYQ = 1/2 x116° = 58°
∴ ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 112°
and
∠QYP = 360 – ∠QYP = 360° – 58° = 302°
