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#### Class Nine Maths NCERT Solution Chapter 6

Chapter 6: Lines and Angles

TEXTBOOK’S EXERCISE – 6.1

1. In the, figure, lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 70° and ∠ BOD = 40°, find ∠ BOE and reflex ∠ COE.

Sol. Lines AB and CD intersect at O.

∠AOC + ∠BOE = 70° …(1) [Given]

∠BOD = 40° …(2) [Given]

Since, ∠AOC = ∠BOD = 40°

[Vertically opposite angles]

∴ 40° + ∠BOE = 70° [From (1)]

⇒ ∠BOE = 70° – 40° = 30°

Also, ∠AOC + ∠BOE + ∠COE = 180°

[Since AOB is a straight line]

⇒ 70° + ∠COE = 180° [From (1)]

⇒ ∠COE = 180° – 70° = 110°

Now, reflex ∠COE = 360° – 110° = 250° Hence, ∠BOE = 30° and reflex ∠COE = 250°

2. In the given figure, lines XY and MN intersect at O. If ∠POY = 90°

and a : b = 2 : 3, find c.

Sol. In the figure, lines XY and MN intersect at O and ∠ POY = 90°.

Also, given a : b = 2 : 3

Let a = 2x and b = 3x.

Since, ∠XOM + ∠POM + ∠POY = 180° [Linear pair axiom]

⇒ 3x + 2x + 90° = 180°

⇒ 5x = 180° – 90° ⇒ x = 90/5° = 18°

∴ ∠XOM = b = 3x = 3 × 18° = 54°

and ∠POM = a = 2x = 2 × 18° = 36°

Now, ∠XON = c = ∠MOY = ∠POM + ∠POY [Vertically opposite angles]

= 36° + 90° = 126°

Hence, c = 126°

3. In the given figure, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.

Sol. ∠PQS + ∠PQR = 180° …(1) [Linear pair axiom]

∠PRQ + ∠PRT = 180° …(2) [Linear pair axiom]

But, ∠PQR = ∠PRQ [Given]

∴ From (1) and (2),

∠PQS = ∠PRT Proved.

4. In the figure, if x + y = w + z, then prove that AOB is a line.

Sol. Assume AOB is a line.

Therefore, x + y = 180° …(1) [Linear pair axiom]

w + z = 180° …(2) [Linear pair axiom]

Now, from (1) and (2)

x + y = w + z

Hence, our assumption is correct, AOB is a line. Proved.

5. In the 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 = 12 (∠QOS – ∠ POS)

Sol. ∠ROS = ∠ROP – ∠POS …(1)

and ∠ROS = ∠QOS – ∠QOR …(2)

∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠ROP – ∠POS

⇒ 2∠ROS = ∠QOS – ∠POS

(Since ∠QOR = ∠ROP = 90°)

⇒ ∠ROS = 12 (∠QOS – ∠POS) Proved.

6. It is given that ∠ XYZ = 64° and XY is produced to a point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠ QYP.

Sol. From figure,

∠XYZ = 64° [Given]

Now, ∠ZYP + ∠XYZ = 180° [Linear pair axiom]

⇒ ∠ZYP + 64° = 180°

⇒ ∠ZYP = 180° – 64° = 116°

Also, given that ray YQ bisects ∠ZYP.

But, ∠ZYP = ∠QYP + ∠QYZ = 116°

Therefore, ∠QYP = 58° and ∠QYZ = 58°

Also, ∠XYQ = ∠XYZ + ∠QYZ

⇒ ∠XYQ = 64° + 58° = 122°

and reflex ∠QYP = 360° – ∠QYP = 360° – 58° = 302° [since ∠QYP = 58°]

Hence, ∠XYQ = 122° and reflex ∠QYP = 302°.