## 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.

Answer:

Given

∠1 + ∠3 = 70°

∠4 = 40°

Required to find

∠3

reflex ∠2

Steps

∠1 + ∠2 + ∠3 = 180° (Angles on a line)

70° + ∠2 = 180°

∠2 = 180 - 70

∠2 = 110°

Similarly, ∠2 + ∠3 + ∠4 = 180° (Angles on a line)

110° + ∠3 + 40° = 180°

150° + ∠3 = 180°

∠3 = 180 - 150

∠3 = 30°

Reflex ∠2 = 360° - ∠2

= 360° - 110° = 250°

Therefore, ∠BOE is 30° and reflex ∠COE is 250°.

Question 2: In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a:b = 2:3, find c.

Answer:

Given

∠POY = 90°

a:b = 2:3

Required to find

c

Steps

∠POY + ∠POM + ∠MOX = 180° (angles on a line)

90° + 2x + 3x = 180°

90° + 5x = 180°

5x = 180 - 90

5x = 90°

x = 90/5

x = 18°

a = 2x = 2 × 18 = 36°

b = 3x = 3 × 18 = 54°

b + c = 180° (linear pair)

54° + c = 180°

c = 180 - 54

c = 126°

Therefore, c is 126°.

Question 3: In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Answer:

Given

∠PQR = ∠PRQ

Required to prove

∠PQS = ∠PRT

Proof

If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.

Ray PQ on line ST.

Ray PR on line ST.

Therefore,

∠PQS + ∠PQR = 180° …1

∠PRT + ∠PRQ = 180° …2

Adding 1 and 2,

∠PQS + ∠PQR = ∠PRT + ∠PRQ

∠PQS = ∠PRT

Hence, proved.

Question 4: In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Answer:

Given

x + y = w + z

Required to prove

AOB is a line

Proof

Angle at O = 360°

x + y + z + w = 360°

2(x + y) = 360° (Since, x + y = w + z)

x + y = 180°

z + w = 180°

If the sum of two adjacent angles in 180°, then the non-common arms of the angles forms a line. So, AOB is a line.

Hence, proved.

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 = 1/2 (∠QOS - ∠POS).

Answer:

Given

POQ is a line

∠ROQ = 90°

Required to prove

∠ROS = 1/2 (∠QOS - ∠POS)

Proof

∠ROS = ∠SOQ - 90° …1

∠ROS = 90° - ∠POS …2

Adding 1 and 2,

∠ROS + ∠ROS = ∠SOQ - 90° + 90° - ∠POS

2∠ROS = ∠SOQ - ∠POS

∠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.

Answer:

Given

∠XYZ = 64°

Required to find

∠XYQ

reflex ∠QYP

Steps

If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.

Ray YZ stands on XP

= ∠XYZ + ∠ZXP = 180°

= 64° + ∠ZXP = 180°

= ∠ZXP = 180 - 64

= ∠ZXP = 116°

Since, ray YQ bisects ∠ZXP, ∠ZYQ and ∠QYP will be equal.

∠ZYQ + ∠QYP = 116°

x + x = 116°

2x = 116°

x = 116/2

x = 58°

∠XYQ = ∠ZYX + ∠QYZ

∠XYQ = 64° + 58°

∠XYQ = 122°

Reflex of ∠QYP = 360° - ∠QYP

= 360° - 58°

= 302°

Therefore, ∠XYQ is 122° and reflex ∠QYP is 302°.

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