# Lines and Angles (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 …(ii)
Now from (i) and (ii), we get
3a/2 + A = 90°
⇒ 5a/2= 90°
⇒ a =(90°/5)x2 = 36°
From (ii), we get
b = 3/2 x 36° = 54°
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) =  = 180°
∴ 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°

error: Content is protected !!