## MP Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.1

**Class 9 Chapter 6 Maths Exercise 6.1 Question 1.**

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

Solution:

Given

∠AOC + ∠BOE = 70°

∠BOD = 40°

To find:

∠BOE and Reflex ∠COE

Calculation:

∠BOD

= ∠AOC = 40° (VOA’s)

∠AOC + ∠BOE = 70°

40° + ∠BOE = 70°

∴ ∠BOE = 30°

∠AOC + ∠COE + ∠BOE = 180°

(∴ Angles on the same line)

70° + ∠COE = 180°

∴ ∠COE – 110°

Reflex ∠COE = 360° – ∠COE

= 360° – 110° = 250°

**Class 9th Maths Chapter 6 Exercise 6.1 Question 2.**

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

Solution:

Given

∠POY= 90°

a : b = 2 : 3

To find. ∠C

Calculation:

a = 2x

b = 3x

b + a + 90° = 180° (∴ Angles on the same line)

3x + 2x + 90° = 180°

5x = 90°

x = 18°

a = 2 x 18° = 36°

b = 3 x 18° = 54°

b + c = 180 (LPA’s)

c = 180° – 54° = 126°

**9th Class Maths Chapter 6 Exercise 6.1 Question 3.**

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

Solution:

Given

∠PQR = ∠PRQ

To prove:

∠PQS = ∠PRT

Proof:

∠PQS + ∠PQR = 180° (LPA’s) …..(1)

∠PRT + ∠PRQ = 180° (LPA’s)

=> ∠PRT + ∠PQR = 180° (∴ ∠PRQ = ∠PQR) …(2)

From (1) and (2), we get

∠PQS + ∠PQR = ∠PRT + ∠PQR

∠PQS = ∠PRT Proved.

**Exercise 6.1 Class 9 Question 6 Question 4.**

In Fig. below, 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:

x + y + z + w = 360° (Complete angle)

x + y + x + y = 360° (∴ w + z = x + y)

2(x + y) = 360°

x + y = 180° (LPA’s)

AOB is a line. Proved

**Math Class 9 Chapter 6 Exercise 6.1 Question 5.**

In Fig. below. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS).

Solution:

Given

POQ is a line, OR ⊥ PQ

To prove:

∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS).

Proof:

∠QOS = ∠QOR + ∠ROS

∠QOS = 90° + ∠ROS ….(1)

∠POS = ∠POR – ∠ROS

∠POS = 90° – ∠ROS …..(2)

Subtracting (1) and (2), we get

∠QOS – ∠POS = 90° + ∠ROS – (90° – ∠ROS)

= 90° + ∠ROS – 90° + ∠ROS

∠QOS – ∠POS = 2 ∠ROS

∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)

**Class 9 Maths Exercise 6.1 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.

Solution:

Given

∠XYZ = 64°

To find:

∠XYQ and Reflex ∠QYP

Calculation:

∠XYZ + ∠ZYQ + ∠QYP

= 180° (∴ Angles on the same line)

64° + x + x = 180°

2x = 180° – 64° = 116°

x = 58°

∠XYQ = ∠XYZ + ∠ZYQ

= 64° + 58° = 122°

Reflex ∠QYP = 360° – 58° = 302°

Transversal:

A line which intersects two or more lines at distinct points is called a transversal.

Angles and their Name formed by Transversal:

Line intersects line m and n at point P and Q respectively. So line l is a transversal ftk lines m and n. Observe that four angles are formed at each of the points P and Q. Name these angles ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8. There are two type of angles: exterior angles and interior angles as shown in Fig. above.

(a) ∠1, ∠2, ∠7 and ∠8 are Exterior angles and

(b) ∠3, ∠4, ∠5 and ∠6 are Interior angles

Same pairs of angles are formed when a transversal intersects two lines. These are given below:

(a) Corresponding angles:

- ∠1 and ∠5
- ∠2 and ∠6
- ∠4 and ∠8
- ∠3 and ∠7

(b) Alternate interior angles:

- ∠4 and ∠6
- ∠3 and ∠5

(c) Alternate exterior angles:

- ∠1 and ∠7
- ∠2 and ∠8

(d) Interior angles on the same side of the transversal:

- ∠4 and ∠5
- ∠3 and ∠6

Axiom 3:

If a transversal intersects two parallel lines, then each pair of corresponding angles is equal.

If AB ∥ CD and l is the transversal.

∠1 = ∠5; ∠4 = ∠8

(Corresponding angles)

∠2 = ∠6; ∠3 = ∠7

Axiom 4:

If a transversal intersects two lines such that a pair of correspond-ing angles are equal, then the two lines are parallel to each other.

In Fig., AB and CD are two lines, intersected by transversal ‘l’ such that ∠1 = ∠2 (corresponding angles,) then AB ∥ CD.

Theorem 1.

If a transversal intersects two parallel lines, then each pair of alternate interior angles are equal.

Given. AB ∥ CD and l is the transversal.

To prove:

- ∠3 = ∠5
- ∠4 = ∠6

Proof:

∠1 = ∠3 (V.O.A’s) ……(i)

∠1 = ∠5 (Corresponding angle’s) …(ii)

From (i) and (ii), we get

∠3 = ∠5

∠4 = ∠2 (V.O.A’s)

∠2 = ∠6 (Corresponding angle’s)

From (iii) and (iv), we get

∠4 = ∠6. Proved.

Theorem 3.

if a transversal intersect two lines, such that a pair of alternate interior angle is equal, then the two lines are parallel.

Given

AB and CD are two lines, and l is the transversal.

∠2 = ∠3

To prove:

AB ∥ CD

Proof:

∠2 = ∠3 (V.O.A’s) …..(i)

∠1 = ∠2 (Corresponding angle’s) …(ii)

From (i) and (ii), we get

∠1 = ∠2

∠1 and ∠3 are corresponding angles and are equal.

AB ∥ CD. Proved.

Theorem 4.

If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.

Given

AB ∥ CD and l is the transversal.

To prove:

∠4 + ∠5 = 180° and ∠3 + ∠6 = 180°

Proof

∠3 + ∠4 = 180° (LPA’s) …..(i)

∠3 = ∠5 (AIA’s) …..(ii)

∠4 = ∠6 (AIA’s) …..(iii)

From (i) and (ii) we get

∠5 + ∠4 = 180°

From (i) and (iii), we get

∠3 + ∠6 = 180s.

Theorem 5.

If a transversal intersects two lines, such that a pair of interior angles on the same side of transversal is supplementary, then the two lines are parallel.

Given

AB and CD are two lines, l is the transversal.

∠4 + ∠5 = 180°

To prove

AB ∥ CD

Proof

∠4 + ∠5 = 180° (Given) …..(i)

∠1 + ∠4 = 180° (LPA’s) …..(ii)

From (i) and (ii), we get

∠4 + ∠5 = ∠1 + ∠4

∠5 = ∠1

∠1 and ∠5 are corresponding angles and are equal.

AB ∥ CD. Proved.

Theorem 6.

Lines which are parallel to the same line are parallel to each other.

Given, p, q and r are three lines

P ∥ r and q ∥ r

To prove

p ∥ q

Proof

p ∥ r and l is the transversal.

∠1 = ∠3 (Corresponding angles) …..(i)

q ∥ r and l is the transversal.

∠2 = ∠3 (Corresponding angles) …..(ii)

From (i) and (ii), we get

∠1 = ∠2

∠1 and ∠2 are corresponding angles and are equal.

p ∥ q. Proved.