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Updated on 24 Feb 2026, 16:17 IST
Understanding how lines and angles interact is one of the most important foundations of geometry. NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles help students learn core geometric concepts such as linear pairs, vertically opposite angles, transversals, and angle relationships in a clear and structured way.
Prepared according to the latest CBSE Class 9 Maths syllabus and NCERT guidelines, these solutions by Infinity Learn provide step-by-step explanations for every exercise question. Each NCERT Solution is designed to strengthen conceptual understanding, improve logical reasoning, and help students confidently solve exam-level problems.
In this chapter, students explore how parallel lines and transversals create predictable angle relationships — concepts that are essential not only for Class 9 examinations but also for higher classes and competitive exams like JEE Foundation and Olympiads.
The NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles PDF provide a complete guide to mastering angle properties and geometric reasoning. Students can access detailed explanations for all exercises and practice questions prepared by experienced mathematics educators at Infinity Learn.
Question 1: In the figure: lines AB and CD intersect at O. If ∠ AOC + ∠ BOE = 700 and ∠ BOD = 400, find ∠ BOE and reflex ∠ COE.
Answer:
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Lines AB and CD intersect at O.
∠AOC + ∠BOE = 700………………………………. (1)
(Given)
∠BOD = 400 ………………………………………… (2)
(Given)
Since, ∠AOC = ∠BOD
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(Vertically opposite angles)
Therefore, ∠AOC = 400
[From equation 2]
and 400 + ∠BOE = 700
[From equation 1]
=> ∠BOE = 700 – 400 = 300
Also, ∠AOC + ∠BOE + ∠COE = 1800
[Since AOB is a straight line]
=> 700 + ∠COE = 1800
[Form equation (1)]
=> ∠COE = 1800 – 700 = 1100
Now, reflex ∠COE = 3600 – 1100 = 2500
Hence, ∠BOE = 300 and reflex ∠COE = 2500
Question 2: In the figure, lines XY and MN intersect at O. If ∠POY = 900 and a: b = 2: 3, find c.
Answer:
In the figure, lines XY and MN intersect at O and ∠ POY = 900.
Also, given a: b = 2: 3
Let a = 2x and b = 3x.
Since, ∠XOM + ∠POM + ∠POY = 1800
[Linear pair axiom]
=> 3x + 2x + 900 = 1800
=> 5x = 1800 – 900
=> x = 900/50 = 180
So, ∠XOM = b = 3x = 3 × 180 = 540
and ∠POM = a = 2x = 2 × 180 = 360
Now, ∠XON = c = ∠MOY = ∠POM + ∠POY
[Vertically opposite angles]
=> c = 360 + 900 = 1260
Hence, c = 1260
Question 3: In the figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠ PRT.
Answer:
∠PQS + ∠PQR = 180° …………… (1)
(Linear pair axiom)
∠PRQ + ∠PRT = 180° …………… (2)
(Linear pair axiom)
But, ∠PQR = ∠PRQ
(Given)
From equation 1 and 2, we get
∠PQS = ∠PRT
Question 4: In the figure, if x + y = w + z, then prove that AOB is a line.
Answer:
Assume AOB is a line.
Therefore, x + y = 180°……………. (1)
[Linear pair axiom]
and w + z = 180° ....................... (2)
[Linear pair axiom]
Now, from equation (1) and (2), we get
x + y = w + z
Question 5: In the 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) / 2
Answer:
From the figure,
∠ROS = ∠ROP – ∠POS ……………... (1)
and ∠ROS = ∠QOS – ∠QOR ....... (2)
Adding equation (1) and (2), we get
∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠ROP – ∠POS
=> 2∠ROS = ∠QOS – ∠POS
[Since ∠QOR = ∠ROP = 900]
=> ∠ROS = (∠QOS – ∠ POS) / 2
Question 6: It is given that ∠ XYZ = 640 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:
From the figure,
∠XYZ = 640
(Given)
Now, ∠ZYP + ∠XYZ = 1800
[Linear pair axiom]
=> ∠ZYP + 640 = 1800
=> ∠ZYP = 1800 – 640 – 1160
Also, given that ray YQ bisects ∠ZYP.
But, ∠ZYP = ∠QYP = ∠QYZ = 1160
Therefore, ∠QYP = 580 and ∠QYZ = 580
Also, ∠XYQ = ∠XYZ + ∠QYZ
=> ∠XYQ = 640 + 580 = 1220
and reflex ∠QYP = 3600 – ∠QYP = 3600 – 580 = 3020
[Since ∠QYP = 580]
Hence, ∠XYQ = 1220 and reflex ∠QYP = 3020
Question 1: In the figure, find the values of x and y and then show that AB || CD.
Answer:
In the given figure, a transversal intersects two lines AB and CD such that
x + 500 = 1800
[Linear pair axiom]
=> x = 1800 – 500 = 1300
and y = 1300
[Vertically opposite angles]
Therefore, ∠x = ∠y = 1300
[Alternate angles]
Hence, AB || CD
[Converse of alternate angles axiom]
Question 2: In the figure, if AB || CD, CD || EF and y: z = 3: 7, find x.
Answer:
In the given figure, AB || CD, CD || EF and y: z = 3: 7.
Let y = 3a and z = 7a
∠DHI = y
[vertically opposite angles]
∠DHI + ∠FIH = 1800
[Interior angles on the same side of the transversal]
=> y + z = 1800
=> 3a + 7a = 1800
=> 10a = 1800
=> a = 180
So, y = 3 × 18° = 540
and z = 180 × 7 = 1260
Also, x + y = 1800
=> x + 540 = 1800
=> x = 1800 – 540 = 1260
Hence, x = 1260
Question 3: In the figure, if AB || CD, EF ⊥ CD and ∠GED = 126°. Find ∠ AGE, ∠GEF and ∠FGE.
Answer:
In the given figure, AB || CD, EF ⊥ CD and ∠GED = 1260
∠AGE = ∠LGE
[Alternate angle]
So, ∠AGE = 1260
Now, ∠GEF = ∠GED – ∠DEF
= 1260 – 900 = 360
[Since ∠DEF = 900]
Also, ∠AGE + ∠FGE = 1800
[Linear pair axiom]
=> 1260 + FGE = 1800
=> ∠FGE = 1800 – 1260 = 540
Question 4: In the figure, if PQ || ST, ∠PQR = 1100 and ∠ RST = 1300, find ∠QRS.
Answer:
Extend PQ to Y and draw LM || ST through R.
∠TSX = ∠QXS
[Alternate angles]
=> ∠QXS = 1300
∠QXS + ∠RXQ = 1800
[Linear pair axiom]
=> ∠RXQ = 1800 – 1300 = 500...…………... (1)
∠PQR = ∠QRM
[Alternate angles]
=> ∠QRM = 1100……………… (2)
∠RXQ = ∠XRM
[Alternate angles]
=> ∠XRM = 500
[from equation (1)]
∠QRS = ∠QRM – ∠XRM
= 1100 – 500 = 600
Question 5: In the figure, if AB || CD, ∠APQ = 500 and ∠PRD = 127°, find x and y.
Answer:
In the given figure, AB || CD, ∠APQ = 500 and ∠PRD = 1270
∠APQ + ∠PQC = 1800
[Pair of consecutive interior angles are supplementary]
=> 500 + ∠PQC = 1800
=> ∠PQC = 1800 – 500 = 1300
Now, ∠PQC + ∠PQR = 1800
[Linear pair axiom]
=> 1300 + x = 1800
=> x = 1800 – 1300 = 500
Also, x + y = 1270
[Exterior angle of a triangle is equal to the sum of the two-interior opposite angles]
=> 500 + y = 1270
=> y = 1270 – 500 = 770
Hence, x = 500 and y = 770
Question 6: In the 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.
Answer:
At point B, draw BE ⊥ PQ and at point C, draw CF ⊥ RS.
∠1 = ∠2 …………... (1)
[Angle of incidence is equal to angle of reflection]
∠3 = ∠4 …………... (2)
[Angle of incidence is equal to angle of reflection]
Also, ∠2 = ∠3 ......(3)
[Alternate angles]
=> ∠1 = ∠4
[From equation 1, 2, and 3]
=> 2∠1 = 2∠4
=> ∠1 + ∠1 = ∠4 + ∠4
=> ∠1 + ∠2 = ∠3 + ∠4
[From (1) and (2)]
=> ∠BCD = ∠ABC
Hence, AB || CD.
[Alternate angles are equal]
Hence, Proved.
Question 1: In the figure, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠SPR = 1350 and ∠PQT = 1100, find ∠PRQ.
Answer:
In the given figure, ∠SPR = 1350 and ∠PQT = 1100.
∠PQT + ∠PQR = 1800
[Linear pair axiom]
=> 1100 + ∠PQR = 1800
=> ∠PQR = 1800 – 1100 = 700
Also, ∠SPR + ∠QPR = 1800
[Linear pair axiom]
=> 1350 + ∠QPR = 1800
=> ∠QPS = 1800 – 1350 = 450
Now, in the triangle PQR
∠PQR + ∠PRQ + ∠QPR = 1800
[Angle sum property of a triangle]
=> 700 + ∠PRQ + 450 = 1800
=> ∠PRQ + 1150 = 1800
=> ∠PRQ = 1800 – 1150 = 650
Hence, ∠ PRQ = 650
Question 2: In the figure, ∠ X = 620, ∠ XYZ = 540. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of ∆ XYZ, find ∠OZY and ∠YOZ.
Answer:
In the given figure,
∠X = 620 and ∠XYZ = 540.
∠XYZ + ∠XZY + ∠YXZ = 1800 ……… (1)
[Angle sum property of a triangle]
=> 540 + ∠XZY + 620 = 1800
=> ∠XZY + 1160 = 1800
=> ∠XZY = 1800 – 1160 = 640
Now, ∠OZY = 1/2 × ∠XZY
[Since ZO is bisector of ∠XZY]
= 1/2 × 640 = 320
Similarly, ∠OYZ = 1/2 × 540 = 270
Now, in ∆OYZ, we have
∠OYZ + ∠OZY + ∠YOZ = 1800
[Angle sum property of a triangle]
=> 270 + 320 + ∠YOZ = 1800
=> ∠YOZ = 1800 – 590 = 1210
Hence, ∠OZY = 320 and ∠YOZ = 1210
Question 3: In the figure, if AB || DE, ∠ BAC = 350and ∠CDE = 530, find ∠ DCE.
Answer:
In the given figure,
∠BAC = ∠CED
[Alternate angles]
=> ∠CED = 350
In ∆CDE,
∠CDE + ∠DCE + ∠CED = 1800
[Angle sum property of a triangle]
=> 530 + ∠DCE + 350 = 1800
=> ∠DCE + 880 = 1800
=> ∠DCE = 1800 – 880 = 920
Hence, ∠DCE = 920
Question 4: In the figure, if lines PQ and RS intersection point T, such that ∠ PRT = 400, ∠RPT = 950 and ∠TSQ = 750, find ∠SQT.
Answer:
In the given figure, lines PQ and RS intersect at point T, such that ∠PRT = 400,
∠RPT = 950 and ∠TSQ = 750.
In ∆PRT
∠PRT + ∠RPT + ∠PTR = 1800
[Angle sum property of a triangle]
=> 400 + 950 + ∠PTR = 1800
=> 1350 + ∠PTR = 1800
=> ∠PTR = 1800 – 1350 = 450
Also, ∠PTR = ∠STQ
[Vertical opposite angles]
So, ∠STQ = 450
Now, in ∆STQ,
∠STQ + ∠TSQ + ∠SQT = 180°
[Angle sum property of a triangle]
=> 450 + 750 + ∠SQT = 1800
=> 1200 + ∠SQT = 1800
=> ∠SQT = 1800 – 1200 = 600
Hence, ∠SQT = 600
Question 5: In the figure, if PT ⊥ PS, PQ || SR, ∠SQR = 280 and ∠QRT = 650, then find the values of x and y.
Answer:
In the given figure, lines PQ ⊥ PS, PQ || SR,
∠SQR = 280 and ∠QRT = 650
∠PQR = ∠QRT
[Alternate angles]
=> x + 280 = 650
=> x = 650 – 280 = 370
In ∆PQS,
∠SPQ + ∠PQS + ∠QSP = 1800
[Angle sum property of a triangle]
=> 900 + 370 + y = 1800
[Since PQ ⊥ PS, ∠PQS = x = 370 and ∠QSP = y)
=> 1270 + y = 1800
=> y = 1800 – 1270 = 530
Hence, x = 370 and y = 530
Question 6: In the figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠ PRS meet at point T, then prove that ∠QTR = ∠QPR / 2.
Answer:
Exterior ∠PRS = ∠PQR + ∠QPR
[Exterior angle property]
Therefore, ∠PRS/2 = ∠PQR/2 + ∠QPR/2
=> ∠TRS = ∠TQR + ∠QPR/2 …………… (1)
But in ∆QTR,
Exterior ∠TRS = ∠TQR + ∠QTR ………. (2)
[Exterior angles property]
Therefore, from equation (1) and (2), we get
∠TQR + ∠QTR = ∠TQR + ∠QPR/2
=> ∠QTR = ∠QPR/2
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In NCERT Class 9 Maths Chapter 6 – Lines and Angles, students learn how different types of lines intersect to form angles and predictable geometric relationships. The chapter introduces key concepts such as linear pairs, vertically opposite angles, transversals, and angle relationships formed by parallel lines. These concepts form the foundation of geometry in the CBSE Class 9 Maths syllabus and are essential for solving theorem-based and diagram questions in examinations as well as higher mathematics.
A transversal is a line that intersects two or more lines at different points. When a transversal cuts two parallel lines, several angle pairs are formed, including corresponding angles, alternate interior angles, and co-interior angles. For example, railway tracks crossed by a road create a transversal situation where equal angle relationships can be observed. Understanding transversals helps students apply angle rules correctly in NCERT exercise problems.
Vertically opposite angles are formed when two straight lines intersect each other. These angles are equal because each pair forms supplementary angles with adjacent angles on a straight line. Since the total angle on a straight line is 180°, mathematical reasoning proves that vertically opposite angles must have the same measure. This theorem is frequently used in NCERT Solutions Class 9 Maths Chapter 6 to find unknown angle values.
A linear pair consists of two adjacent angles whose non-common sides form a straight line. The sum of a linear pair is always 180°, making them supplementary angles. This concept is one of the most important rules in Chapter 6 because many NCERT questions require identifying linear pairs before calculating missing angles.
When a transversal intersects two parallel lines, special angle relationships are created:
These properties help students prove lines are parallel and solve geometry problems step by step in CBSE exams.
Two lines can be proven parallel if any one of the following conditions is satisfied:
In NCERT Class 9 Maths Chapter 6, students apply these angle properties to justify geometric proofs and solve theorem-based questions.
Students, download the best NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles PDF from Infinity Learn to access step-by-step explanations aligned with the latest CBSE syllabus. The PDF helps in revision, homework practice, and quick concept clarification before exams.