- Introduction
- |
- What is angle
- |
- Congruence Angle
- |
- Adjacent Angles
- |
- Complimentary Angles
- |
- Supplementary Angles
- |
- Linear Pair Axioms
- |
- Transversal across the parallel Lines
- |
- Angle sum property of Triangles

In this page we have ** NCERT book Solutions for Class 9th Maths:Line and angles** for
exercise 1 Hope you like them and do not forget to like , social share
and comment at the end of the page.

**Question 1 **

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

**Answer**

Given,

∠AOC + ∠BOE = 70° and ∠BOD = 40°

Now AOB is a straight line

∠AOC + ∠BOE +∠COE = 180°

70° +∠COE = 180°

∠COE = 110°

Also COD is a straight lne

∠COE +∠BOD + ∠BOE = 180°

110° +40° + ∠BOE = 180°

150° + ∠BOE = 180°

∠BOE = 30°

**Question 2 **

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

**Answer**

Given,

∠POY = 90° and a : b = 2 : 3

Lt x be the common ration, then a=2x and b=3x

Now XOY is a straight line

∠POY + a + b = 180°

90° + a + b = 180°

a + b = 90°

2x + 3x = 90°

5x = 90°

x = 18°

So a = 2×18° = 36°

and b = 3×18° = 54°

Also, now angle b and Angle c forms a linear pair

b + c = 180°

54° + c = 180°

c = 126°

**Question 3**

In below figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

**Answer**

Given

∠PQR = ∠PRQ

To prove,

∠PQS = ∠PRT

Now ∠PQR and ∠PQS forms a linear pair

∠PQR +∠PQS = 180°

∠PQS = 180° - ∠PQR --- (a)

Also, ∠PRQ and ∠PRT forms a linear pair

∠PRQ +∠PRT = 180° (

∠PRT = 180° - ∠PRQ

Now as (∠PQR = ∠PRQ

∠PRQ = 180° - ∠PQR --- (b)

From (a) and (b)

∠PQS = ∠PRT = 180° - ∠PQR

Therefore, ∠PQS = ∠PRT

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

**Answer**

Given,

x + y = w + z

To Prove,

AOB is a line or x + y = 180°

Now O is the point and we have four angles around it

x + y + w + z = 360°

(x + y) + (w + z) = 360°

Now Given x + y = w + z

(x + y) + (x + y) = 360°

2(x + y) = 360°

(x + y) = 180°

Hence, x + y makes a linear pair. Therefore, AOB is a straight line.

**Question 5**

In below 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 = 1/2(∠QOS – ∠POS).

**Answer**

Given,

OR is perpendicular to line PQ

To prove,

∠ROS = 1/2(∠QOS – ∠POS)

**Proof**

∠POR = ∠ROQ = 90° (Perpendicular)

From the figure, it is clear that

∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROQ --- (a)

∠POS = ∠POR - ∠ROS = 90° - ∠ROQ --- (b)

Subtracting (b) from (a)

∠QOS - ∠POS = 90° + ∠ROQ - (90° - ∠ROQ)

⇒ ∠QOS - ∠POS = 90° + ∠ROQ - 90° + ∠ROQ

⇒ ∠QOS - ∠POS = 2∠ROQ

⇒ ∠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°

YQ bisects ∠ZYP

Now angle ∠XYZ and ∠ZYP forms a linear pair

∠XYZ +∠ZYP = 180°

64° +∠ZYP = 180°

∠ZYP = 116°

Now YQ bisects ∠ZYP

∠ZYQ = ∠QYP

Also ∠ZYP = ∠ZYQ + ∠QYP

So ∠ZYP = 2∠ZYQ

2∠ZYQ = 116°

∠ZYQ = 58° = ∠QYP

Now,

∠XYQ = ∠XYZ + ∠ZYQ

∠XYQ = 64° + 58°

∠XYQ = 122°

Also,

reflex ∠QYP = 180° + ∠XYQ

∠QYP = 180° + 122°

=302^{0}

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