**NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 – Lines and Angles**

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

**Given :**

∠ AOC + ∠ BOE = 70°

∠ BOD = 40°

**To Find :**

∠ BOE = ?

Reflex ∠ COE = ?

**Solution:**

∠ AOC = ∠ BOD [ Vertically opposite angles ]

∠ AOC = ∠ BOD = 40°

∠ AOC = 40°

∠ AOC + ∠ BOE = 70° [Given]

40° + ∠ BOE = 70°

∠ BOE = 70° – 40°**∠ BOE = 30°**

∠ AOC + ∠ COE + ∠ BOE = 180° [ Angles on Straight line]

∠ AOC + ∠ BOE + ∠ COE= 180°

70° ∠ COE = 180°

∠ COE = 180° – 70°

∠ COE = 110°

Reflex ∠ COE = 360° – 110°**Reflex ∠ COE = 250°**

** **

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

**Given :**

∠ POY = 90°

a : b = 2 : 3

**To Find : **

c = ?

**Solution:**

Let, a = 2x & b = 3x

∠ XOM + ∠ MOP + ∠ POY = 180° [Angle on Straight Line]

b + c + 90° = 180°

3x + 2x + 90° = 180°

3x + 2x = 180° – 90°

5x = 90°

x = (90°)/5

x = 18°

a = 2x = 2 x 18° = 36°

b = 3x = 3 x 18° = 54°

∠ XOM + ∠ XON = 180° [Linear Pair]

54° + c = 180°

c = 180° – 54°**c = 126°**

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

**Given :**

∠ PQR = ∠ PRQ

**To Prove :**

∠ PQS = ∠ PRT

**Solution:**

∠ PQS + ∠ PQR = 180° [ Linear Pair ]

∠ PQR = 180° – ∠ PQS ——— 1

∠ PRT + ∠ PRQ = 180° [ Linear Pair ]

∠ PRQ = 180° – ∠ PRT ———- 2

∠ PQR = ∠ PRQ [ Given ]

On Equating Equation 1 and 2, we get

180° – ∠ PQS = 180° – ∠ PRT**∠ PQS = ∠ PRT**

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

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

**To Prove : **AOB is a line

**Proof:**

x + y + w + z = 360° [ Complete Angle ]

x + y = w + z [ Given ]

Since, x + y + x + y = 360°

2 ( x + y ) = 360°

( x + y ) = (360°)/2

( x + y ) = 180°**Since, x and y form a Linear Pair, AOB is a line**

**Q. 5 In given 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 = (∠ QOS – ∠ POS).**

**Given :**

OR is perpendicular to line PQ

i.e, ∠ POR = ∠ QOR = 90°

**To Prove :**

∠ ROS = (∠ QOS – ∠ POS).

**Proof:**

∠ POR = 90° [Given : OR is perpendicular to line PQ]

∠ POS + ∠ ROS = 90°

∠ ROS = 90° – ∠ POS ———– 1

∠ QOR = 90° [Given : OR is perpendicular to line PQ]

∠ QOS – ∠ ROS = 90°

∠ ROS = ∠ QOS – 90° ———– 2

∠ 2ROS = ∠ QOS – ∠POS [On adding Eq 1 and 2]

∠ ROS = (∠ QOS – ∠ POS)

**Q.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**

**Given :**

∠ XYZ = 64°

YQ bisects ∠ ZYP

i.e, ∠ ZYQ = ∠ QYP

**To Find:**

∠ XYQ = ?

Reflex ∠ QYP = ?

**Solution:**

∠ XYZ + ∠ ZYQ + QYP = 180° [Angles on a Straight line]

∠ XYZ + 2 ∠ QYP = 180° [∠ ZYQ = ∠ QYP]

64° + 2 ∠ QYP = 180°

2 ∠ QYP = 180° – 64°

2 ∠ QYP = 116°

∠ QYP = 58°

∠ ZYQ = ∠ QYP = 58°**Reflex ∠ QYP = 360° – 58° = 302°**

∠ XYQ = ∠ XYZ + ∠ ZYQ

∠ XYQ = 64° + 58°**∠ XYQ = 122°**

**NCERT Solutions for Class 9 Maths Chapter 6 Exercise 6.1 – Lines and Angles**

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