Mathematics NCERT Grade 9, Chapter 6: Lines and Angles: In this chapter students will study the properties of the angle formed when two lines intersect each other and properties of the angle formed when a line intersects two or more parallel lines at distinct points. The chapter starts from zero level, the first topic of the chapter being Basic Terms and Definitions.
• Acute angle: Measures between 0$°$ and 90$°$.
• Right angle: Exactly equal to 90$°$.
• Obtuse angle: Angle greater than 90$°$ but less than 360$°$.
• Reflex angle: Angle greater than 180$°$ but less than 360$°$.
• Two angles whose sum is 90$°$ are called complementary angles.
• Two angles whose sum is 180$°$ are called supplementary angles.
• Adjacent angles, linear pair of angles and vertically opposite angles are other important topics in this chapter.
Section 6.3 is about the topic- Intersecting Lines and Non-Intersecting Lines.
• If two lines intersect each other, then the vertically opposite angles are equal
After that, the topic- Pair of Angles is discussed. 2 axioms, 1 theorem, and some solved examples are also given in this section.
• Axiom 6.1 states about linear pair of angles.
• Linear pair axiom is discussed.
Exercise 6.1 contains various questions based on all the concepts discussed in the section.
• Lines that are parallel to a given line are parallel to each other.
• Corresponding angles axiom is also discussed.
• Theorem on alternate interior angles is also given in this chapter.
Moving on, students will find a detailed description of Lines parallel to the Same Line. Exercise 6.2 contains 6 questions.
Lines parallel to the same line is discussed with the help of corresponding angles axiom
Then chapter will also lay emphasis on Angle Sum Property.
• The sum of the three angles of a triangle is 180$°$
• Exterior angle of a triangle is greater than the sum of the interior opposite angles.
Theorem and examples are given in this section. After that, another unsolved exercise i.e 6.3 is given.
​In the end, a summary of the chapter is provided.

#### Question 1:

In the given figure, lines AB and CD intersect at O. If and find ∠BOE and reflex ∠COE.

##### Video Solution for lines and angles (Page: 96 , Q.No.: 1)

NCERT Solution for Class 9 maths - lines and angles 96 , Question 1

#### Question 2:

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

Let the common ratio between a and b be x.

a = 2x, and b = 3x

XY is a straight line, rays OM and OP stand on it.

∴ ∠XOM + ∠MOP + ∠POY = 180º

b + a + ∠POY = 180º

3x + 2x + 90º = 180º

5x = 90º

x = 18º

a = 2x = 2 × 18 = 36º

b = 3x= 3 ×18 = 54º

MN is a straight line. Ray OX stands on it.

b + c = 180º (Linear Pair)

54º + c = 180º

c = 180º − 54º = 126º

c = 126º

##### Video Solution for lines and angles (Page: 97 , Q.No.: 2)

NCERT Solution for Class 9 maths - lines and angles 97 , Question 2

#### Question 3:

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

In the given figure, ST is a straight line and ray QP stands on it.

∴ ∠PQS + ∠PQR = 180º (Linear Pair)

∠PQR = 180º − ∠PQS (1)

∠PRT + ∠PRQ = 180º (Linear Pair)

∠PRQ = 180º − ∠PRT (2)

It is given that ∠PQR = ∠PRQ.

Equating equations (1) and (2), we obtain

180º − ∠PQS = 180 − ∠PRT

∠PQS = ∠PRT

##### Video Solution for lines and angles (Page: 97 , Q.No.: 3)

NCERT Solution for Class 9 maths - lines and angles 97 , Question 3

#### Question 4:

In the given figure, if then prove that AOB is a line.

It can be observed that,

x + y + z + w = 360º (Complete angle)

It is given that,

x + y = z + w

x + y + x + y = 360º

2(x + y) = 360º

x + y = 180º

Since x and y form a linear pair, AOB is a line.

##### Video Solution for lines and angles (Page: 97 , Q.No.: 4)

NCERT Solution for Class 9 maths - lines and angles 97 , Question 4

#### Question 5:

In the 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

It is given that OR ⊥ PQ

∴ ∠POR = 90º

⇒ ∠POS + SOR = 90º

∠ROS = 90º − ∠POS … (1)

∠QOR = 90º (As OR ⊥ PQ)

∠QOS − ∠ROS = 90º

∠ROS = ∠QOS − 90º … (2)

On adding equations (1) and (2), we obtain

2 ∠ROS = ∠QOS − ∠POS

∠ROS = (∠QOS − ∠POS)

##### Video Solution for lines and angles (Page: 97 , Q.No.: 5)

NCERT Solution for Class 9 maths - lines and angles 97 , Question 5

#### Question 6:

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

It is given that line YQ bisects ∠PYZ.

Hence, ∠QYP = ∠ZYQ

It can be observed that PX is a line. Rays YQ and YZ stand on it.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180º

⇒ 64º + 2∠QYP = 180º

⇒ 2∠QYP = 180º − 64º = 116º

⇒ ∠QYP = 58º

Also, ∠ZYQ = ∠QYP = 58º

Reflex QYP = 360º − 58º = 302º

XYQ = XYZ + ZYQ

= 64º + 58º = 122º

##### Video Solution for lines and angles (Page: 97 , Q.No.: 6)

NCERT Solution for Class 9 maths - lines and angles 97 , Question 6

#### Question 1:

In the given figure, find the values of x and y and then show that AB || CD.

It can be observed that,

50º + x = 180º (Linear pair)

x = 130º … (1)

Also, y = 130º (Vertically opposite angles)

As x and y are alternate interior angles for lines AB and CD and also measures of these angles are equal to each other, therefore, line AB || CD.

##### Video Solution for lines and angles (Page: 103 , Q.No.: 1)

NCERT Solution for Class 9 maths - lines and angles 103 , Question 1

#### Question 2:

In the given figure, if AB || CD, CD || EF and y: z = 3: 7, find x.

It is given that AB || CD and CD || EF

∴ AB || CD || EF (Lines parallel to the same line are parallel to each other)

It can be observed that

x = z (Alternate interior angles) … (1)

It is given that y: z = 3: 7

Let the common ratio between y and z be a.

y = 3a and z = 7a

Also, x + y = 180º (Co-interior angles on the same side of the transversal)

z + y = 180º [Using equation (1)]

7a + 3a = 180º

10a = 180º

a = 18º

x = 7a = 7 × 18º = 126º

##### Video Solution for lines and angles (Page: 104 , Q.No.: 2)

NCERT Solution for Class 9 maths - lines and angles 104 , Question 2

#### Question 3:

In the given figure, If AB || CD, EF ⊥ CD and ∠GED = 126º, find ∠AGE, ∠GEF and ∠FGE.

It is given that,

AB || CD

EF ⊥ CD

∠GED = 126º

⇒ ∠GEF + ∠FED = 126º

⇒ ∠GEF + 90º = 126º

⇒ ∠GEF = 36º

∠AGE and ∠GED are alternate interior angles.

⇒ ∠AGE = ∠GED = 126º

However, ∠AGE + ∠FGE = 180º (Linear pair)

⇒ 126º + ∠FGE = 180º

⇒ ∠FGE = 180º − 126º = 54º

∴ ∠AGE = 126º, ∠GEF = 36º, ∠FGE = 54º

##### Video Solution for lines and angles (Page: 104 , Q.No.: 3)

NCERT Solution for Class 9 maths - lines and angles 104 , Question 3

#### Question 4:

In the given figure, if PQ || ST, ∠PQR = 110º and ∠RST = 130º, find ∠QRS.

[Hint: Draw a line parallel to ST through point R.]

Let us draw a line XY parallel to ST and passing through point R.

∠PQR + ∠QRX = 180º (Co-interior angles on the same side of transversal QR)

⇒ 110º + ∠QRX = 180º

⇒ ∠QRX = 70º

Also,

∠RST + ∠SRY = 180º (Co-interior angles on the same side of transversal SR)

130º + ∠SRY = 180º

∠SRY = 50º

XY is a straight line. RQ and RS stand on it.

∴ ∠QRX + ∠QRS + ∠SRY = 180º

70º + ∠QRS + 50º = 180º

∠QRS = 180º − 120º = 60º

##### Video Solution for lines and angles (Page: 104 , Q.No.: 4)

NCERT Solution for Class 9 maths - lines and angles 104 , Question 4

#### Question 5:

In the given figure, if AB || CD, ∠APQ = 50º and ∠PRD = 127º, find x and y.

∠APR = ∠PRD (Alternate interior angles)

50º + y = 127º

y = 127º − 50º

y = 77º

Also, ∠APQ = ∠PQR (Alternate interior angles)

50º = x

x = 50º and y = 77º

##### Video Solution for lines and angles (Page: 104 , Q.No.: 5)

NCERT Solution for Class 9 maths - lines and angles 104 , Question 5

#### Question 6:

In the given figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Let us draw BM ⊥ PQ and CN ⊥ RS.

As PQ || RS,

Therefore, BM || CN

Thus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.

∴∠2 = ∠3 (Alternate interior angles)

However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)

∴ ∠1 = ∠2 = ∠3 = ∠4

Also, ∠1 + ∠2 = ∠3 + ∠4

∠ABC = ∠DCB

However, these are alternate interior angles.

∴ AB || CD

##### Video Solution for lines and angles (Page: 104 , Q.No.: 6)

NCERT Solution for Class 9 maths - lines and angles 104 , Question 6

#### Question 1:

In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135º and ∠PQT = 110º, find ∠PRQ.

It is given that,

∠SPR = 135º and ∠PQT = 110º

∠SPR + ∠QPR = 180º (Linear pair angles)

⇒ 135º + ∠QPR = 180º

⇒ ∠QPR = 45º

Also, ∠PQT + ∠PQR = 180º (Linear pair angles)

⇒ 110º + ∠PQR = 180º

⇒ ∠PQR = 70º

As the sum of all interior angles of a triangle is 180º, therefore, for ΔPQR,

∠QPR + ∠PQR + ∠PRQ = 180º

⇒ 45º + 70º + ∠PRQ = 180º

⇒ ∠PRQ = 180º − 115º

⇒ ∠PRQ = 65º

##### Video Solution for lines and angles (Page: 107 , Q.No.: 1)

NCERT Solution for Class 9 maths - lines and angles 107 , Question 1

#### Question 2:

In the given figure, ∠X = 62º, ∠XYZ = 54º. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.

As the sum of all interior angles of a triangle is 180º, therefore, for ΔXYZ,

∠X + ∠XYZ + ∠XZY = 180º

62º + 54º + ∠XZY = 180º

∠XZY = 180º − 116º

∠XZY = 64º

∠OZY = = 32º (OZ is the angle bisector of ∠XZY)

Similarly, ∠OYZ = = 27º

Using angle sum property for ΔOYZ, we obtain

∠OYZ + ∠YOZ + ∠OZY = 180º

27º + ∠YOZ + 32º = 180º

∠YOZ = 180º − 59º

∠YOZ = 121º

##### Video Solution for lines and angles (Page: 107 , Q.No.: 2)

NCERT Solution for Class 9 maths - lines and angles 107 , Question 2

#### Question 3:

In the given figure, if AB || DE, ∠BAC = 35º and ∠CDE = 53º, find ∠DCE.

AB || DE and AE is a transversal.

∠BAC = ∠CED (Alternate interior angles)

∴ ∠CED = 35º

In ΔCDE,

∠CDE + ∠CED + ∠DCE = 180º (Angle sum property of a triangle)

53º + 35º + ∠DCE = 180º

∠DCE = 180º − 88º

∠DCE = 92º

##### Video Solution for lines and angles (Page: 107 , Q.No.: 3)

NCERT Solution for Class 9 maths - lines and angles 107 , Question 3

#### Question 4:

In the given figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40º, ∠RPT = 95º and ∠TSQ = 75º, find ∠SQT.

Using angle sum property for ΔPRT, we obtain

∠PRT + ∠RPT + ∠PTR = 180º

40º + 95º + ∠PTR = 180º

∠PTR = 180º − 135º

∠PTR = 45º

∠STQ = ∠PTR = 45º (Vertically opposite angles)

∠STQ = 45º

By using angle sum property for ΔSTQ, we obtain

∠STQ + ∠SQT + ∠QST = 180º

45º + ∠SQT + 75º = 180º

∠SQT = 180º − 120º

∠SQT = 60º

##### Video Solution for lines and angles (Page: 107 , Q.No.: 4)

NCERT Solution for Class 9 maths - lines and angles 107 , Question 4

#### Question 5:

In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28º and ∠QRT = 65º, then find the values of x and y.

It is given that PQ || SR and QR is a transversal line.

∠PQR = ∠QRT (Alternate interior angles)

x + 28º = 65º

x = 65º − 28º

x = 37º

By using the angle sum property for ΔSPQ, we obtain

∠SPQ + x + y = 180º

90º + 37º + y = 180º

y = 180º − 127º

y = 53º

x = 37º and y = 53º

##### Video Solution for lines and angles (Page: 108 , Q.No.: 5)

NCERT Solution for Class 9 maths - lines and angles 108 , Question 5

#### Question 6:

In the given figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR=∠QPR.

In ΔQTR, ∠TRS is an exterior angle.

∠QTR + ∠TQR = ∠TRS

∠QTR = ∠TRS − ∠TQR (1)

For ΔPQR, ∠PRS is an external angle.

∠QPR + ∠PQR = ∠PRS

∠QPR + 2∠TQR = 2∠TRS (As QT and RT are angle bisectors)

∠QPR = 2(∠TRS − ∠TQR)

∠QPR = 2∠QTR [By using equation (1)]

∠QTR = ∠QPR

##### Video Solution for lines and angles (Page: 108 , Q.No.: 6)

NCERT Solution for Class 9 maths - lines and angles 108 , Question 6

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