Our ncert solutions for Solutions 6. Lines and Angles - Exercise 6.1 | Class 9 Mathematics - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Solutions 6. Lines and Angles - Exercise 6.1 | Class 9 Mathematics - Toppers Study

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## 6. Lines and Angles

### | Exercise 6.1 |

## Solutions 6. Lines and Angles - Exercise 6.1 | Class 9 Mathematics - Toppers Study

**Exercise 6.1**

**Q1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠ BOE = 70° and ∠BOD = 40° find ∠BOE and reflex ∠COE.**

**Solution:**

∠BOD = 40°

∠AOC = ∠BOD (Vertically opposite Angle)

∠AOC = 40°

∠AOC + ∠ BOE = 70° (Given)

∠BOE = 70°

∠BOE = 70° - 40°

∠BOE = 30°

AOB is straight line

∠AOC + ∠COE +∠BOE = 180° (linear pair)

⇒ 70° + ∠COE = 180°

⇒ ∠COE = 180° - 70°

⇒ ∠COE = 110°

Reflex ∠COE = 360 - 110°

= 250°

**Q2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.**

Solution:

∠POY=90° (given)

Let ∠a and ∠b = 2x and 3x

XOY is a straight line

∠a + ∠b + ∠POY = 180°

⇒2x + 3x + 90°= 180°

⇒5x = 180° - 90°

⇒5x = 90°

⇒x = 90°/5

⇒x = 18°

Now ∠a = 2 x 18°

= 36°

∠b =3 x 18°

= 54°

MON is a straight line

∠b + ∠c = 180°(linear pair)

∠54° + ∠c = 180°

⇒∠c = 180°- 54°

=126°

**Q3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT **

**Solution : **

**Given :** ∠PQR = ∠PRQ

**To prove : **∠PQS = ∠PRT

Proof :

∠PQS + ∠PQR = 180° .................. (1) Linear pair

∠PRT + ∠PRQ = 180° .................. (2) Linear pair

From equation (1) and (2)

∠PQS + ∠PQR = ∠PRT + ∠PRQ

Or, ∠PQS + ∠PQR = ∠PRT + ∠PQR (∠PQR = ∠PRQ given)

Or, ∠PQS = ∠PRT Proved

**Q4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line. **

**Solution: **

**Given : **x + y = w + z

**To prove :** AOB is a line.

**Proof : **

We know that;

x + y + w + z = 360^{०}

(Angle Sustained on centre)

x + y + x + y = 360^{०} (x + y = w + z given)

2x + 2y = 360^{०}

2 (x + y) = 360^{०}

x + y = 180^{०} (linear pair)

Therefore, AOB is a line

Hence, Proved

**Q5. 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**

**Solution:**

Given:

POQ is a straight line. OR ⊥ PQ and OS is

another ray lying between rays OP and OR.

To prove:

Proof: OR ⊥ PQ (given)

∴ ∠QOR = 90^{०} …………… (1)

POQ is straight line

∴ ∠POR + ∠QOR = 180^{०} (linear pair)

⇒ ∠POR + 90^{०} = 180^{०}

⇒ ∠POR = 180^{०}– 90^{०}

⇒ ∠POR = 90^{०}…………… (2)

Now, ∠ROS + ∠QOR = ∠QOS

Or, ∠ROS = ∠QOS – ∠QOR ……………. (3)

Again, ∠ROS + ∠POS = ∠POR

Or, ∠ROS = ∠POR – ∠POS ……………. (4)

Adding equation (1) and (2)

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

2 ∠ROS = ∠QOS – 90^{०}+ 90^{०}– ∠POS

2 ∠ROS = (∠QOS – ∠POS)

**Hence Proved **

**Q6. 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: **

Given: ∠ XYZ = 64°and XY is produced to point P. YQ bisects ∠ ZYP.

To Find: ∠XYQ and reflex ∠QYP.

YQ bisects ∠ZYP

∴ ∠ZYQ = ∠QYP ................. (1)

∵ XY is produced to point P.

∴ PX is a straight line.

Now, ∠ XYZ + ∠ZYQ + ∠QYP = 180° (linear pair)

Or, 64° + ∠ZYQ + ∠QYP = 180°

⇒ ∠ZYQ + ∠QYP = 180° - 64°

⇒ ∠ZYQ + ∠ZYQ = 116° [Using equation (1) ]

⇒ 2∠ZYQ = 116°

⇒ ∠ZYQ = 116°/2

⇒ ∠ZYQ = 58°

∠ZYQ = ∠QYP = 58°

∠XYQ = ∠XYZ + ∠ZYQ

= 64° + 58°

= 122°

∵ ∠QYP = 58°

∴ Reflex ∠QYP = 360° - 58°

= 302°

∠XYQ = 122°, Reflex ∠QYP = 302°

##### Other Pages of this Chapter: 6. Lines and Angles

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