Class 9 Maths - Lines and Angles NCERT Solutions

Chapter 6: Lines and Angles (NCERT Solutions)

Exercise 6.1

Q1. In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Since AB is a straight line, ∠AOC + ∠COE + ∠BOE = 180° (angles on a straight line).
Given: ∠AOC + ∠BOE = 70°
So: 70° + ∠COE = 180°
∠COE = 110°.
Reflex ∠COE = 360° - 110° = 250°.

Since AB and CD intersect at O, vertically opposite angles are equal.
∠AOC = ∠BOD = 40°.
∠BOE = 70° - ∠AOC = 70° - 40° = 30°.

Q2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a:b = 2:3, find c.

Since XOY is a straight line: ∠POY + ∠POX = 180°
∠POX = 180° - 90° = 90°.
Now a + b = 90° (since PO lies between them and ∠POX = 90°).
Given a:b = 2:3. Let a = 2k, b = 3k.
2k + 3k = 90° ⇒ 5k = 90° ⇒ k = 18°.
b = 3 × 18 = 54°.
c = b (vertically opposite angles) = 54°.

Q3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Proof:
ST is a straight line, so ∠PQR + ∠PQS = 180° (linear pair) ...(1)
Also ∠PRQ + ∠PRT = 180° (linear pair) ...(2)
From (1) and (2): ∠PQR + ∠PQS = ∠PRQ + ∠PRT.
Given ∠PQR = ∠PRQ.
Subtracting equal angles from both sides:
∠PQS = ∠PRT. (Proved)

Q4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.

Proof:
All angles around point O sum to 360°.
So x + y + w + z = 360° ...(1)
Given: x + y = w + z ...(2)
From (1) and (2): 2(x + y) = 360°
x + y = 180°.
Since ∠AOC + ∠COB = x + y = 180°, they form a linear pair.
Therefore, AOB is a straight line. (Proved)

Q5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = ½(∠QOS - ∠POS).

Proof:
OR ⊥ PQ, so ∠POR = ∠QOR = 90°.
∠QOS = ∠QOR - ∠ROS = 90° - ∠ROS ...(1)
∠POS = ∠POR - ∠ROS = 90° - ∠ROS is incorrect. OS lies between OP and OR.
∠POS = ∠POR - ∠ROS = 90° - ∠ROS is also incorrect. Let us write:
∠POS + ∠ROS = ∠POR = 90° ⇒ ∠POS = 90° - ∠ROS ...(2)
∠QOS = ∠QOR + ∠ROS = 90° + ∠ROS ...(3) [OS is between OR and OQ side]
Wait - OS lies between OP and OR, so ∠QOS = 90° + ∠ROS and ∠POS = 90° - ∠ROS.
∠QOS - ∠POS = (90° + ∠ROS) - (90° - ∠ROS) = 2∠ROS.
½(∠QOS - ∠POS) = ∠ROS.
Therefore ∠ROS = ½(∠QOS - ∠POS). (Proved)

Q6. 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.

XYP is a straight line, so ∠XYZ + ∠ZYP = 180° (linear pair).
64° + ∠ZYP = 180° ⇒ ∠ZYP = 116°.
YQ bisects ∠ZYP, so ∠ZYQ = ∠QYP = 116°/2 = 58°.
∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°.
Reflex ∠QYP = 360° - ∠QYP = 360° - 58° = 302°.

Exercise 6.2

Q1. In Fig. 6.28, find the values of x and y and then show that AB ∥ CD.

x + 50° = 180° (linear pair) ⇒ x = 130°.
y = 130° (vertically opposite angles).
Now x = 130° and y = 130°. Since they are alternate interior angles and equal, AB ∥ CD.

Q2. In Fig. 6.29, if AB ∥ CD, CD ∥ EF, and y:z = 3:7, find x.

AB ∥ CD and CD ∥ EF implies AB ∥ EF.
Co-interior angles (same side of transversal): x + y = 180° ...(1)
Alternate interior angles: x = z ...(2) [AB ∥ EF]
y + z = 180° (co-interior angles between AB and EF).
y:z = 3:7 ⇒ y = 3k, z = 7k. y + z = 10k = 180° ⇒ k = 18.
z = 126°. Therefore x = z = 126°.

Q3. In Fig. 6.30, if AB ∥ CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

AB ∥ CD, GE is a transversal.
∠AGE = ∠GED = 126° (alternate interior angles, AB ∥ CD).
∠GEF = ∠GED - ∠FED = 126° - 90° = 36° (since EF ⊥ CD, ∠FED = 90°).
∠FGE = 90° - ∠GEF = 90° - 36° = 54°. [Co-interior angles sum to 180° between AB and CD, ∠AGE + ∠FGE + 90° = 180°].
Actually: ∠AGE + ∠GEF complement inside triangle: ∠FGE = 180° - 90° - 36° = 54°.

Q4. In Fig. 6.31, if PQ ∥ ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

Draw XY through R parallel to PQ and ST.
PQ ∥ XY ⇒ ∠PQR + ∠QRX = 180° (co-interior angles)
∠QRX = 180° - 110° = 70°.
XY ∥ ST ⇒ ∠XRS + ∠RST = 180° (co-interior angles)
∠XRS = 180° - 130° = 50°.
∠QRS = ∠QRX + ∠XRS = 70° + 50° = 120°.

Q5. In Fig. 6.32, if AB ∥ CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

∠APQ = ∠PQR = x (alternate interior angles with AB ∥ CD).
Therefore x = 50°.
∠APR = ∠PRD = 127° (alternate interior angles).
∠APR = ∠APQ + ∠QPR = 50° + y = 127°.
y = 127° - 50° = 77°.

Q6. In Fig. 6.33, PQ and RS are two mirrors. If AB ∥ CD, find ∠CEF.

Draw EG ∥ AB ∥ CD through E.
∠AEG = ∠EAB = 30° (alternate interior angles, since AB ∥ EG).
∠GEC = ∠ECD = 65° (alternate interior angles, since EG ∥ CD).
∠AEC = ∠AEG + ∠GEC = 30° + 65° = 95°.
∠CEF = 180° - ∠AEC = 180° - 95° = 85°.

Exercise 6.3

Q1. In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

∠PQR + ∠PQT = 180° (linear pair)
∠PQR = 180° - 110° = 70°.
∠SPR + ∠QPR = 180° (linear pair)
∠QPR = 180° - 135° = 45°.
In ΔPQR: ∠QPR + ∠PQR + ∠PRQ = 180°
45° + 70° + ∠PRQ = 180°
∠PRQ = 180° - 115° = 65°.

Q2. In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.

In ΔXYZ: ∠X + ∠XYZ + ∠XZY = 180°
62° + 54° + ∠XZY = 180°
∠XZY = 64°.
ZO bisects ∠XZY: ∠OZY = ∠XZY / 2 = 64° / 2 = 32°.
YO bisects ∠XYZ: ∠OYZ = ∠XYZ / 2 = 54° / 2 = 27°.
In ΔOYZ: ∠OZY + ∠OYZ + ∠YOZ = 180°
32° + 27° + ∠YOZ = 180°
∠YOZ = 121°.

Q3. In Fig. 6.41, if AB ∥ DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

AB ∥ DE, AC is a transversal.
∠BAC = ∠DCA = 35° (alternate interior angles).
In ΔDCE: ∠DCE + ∠CDE + ∠DEC = 180°.
Wait – ∠ACD = ∠DCE = 35° (vertical / alternate angles).
∠DCE + ∠CDE = 35° + 53° = 88°
∠DEC = 180° - 88° = 92°.
∠DCE = 180° - 53° - 92° = 35°.

Q4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

In ΔPRT: ∠PTR = 180° - ∠PRT - ∠RPT = 180° - 40° - 95° = 45°.
∠QTS = ∠PTR = 45° (vertically opposite angles).
In ΔQST: ∠SQT + ∠TSQ + ∠QTS = 180°
∠SQT + 75° + 45° = 180°
∠SQT = 60°.

Q5. In Fig. 6.43, if PQ ⊥ PS, PQ ∥ SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

PQ ∥ SR, QR is a transversal.
∠PQR = ∠QRT = 65° (alternate interior angles).
x = ∠PQR - ∠SQR = 65° - 28° = 37°.
In ΔSQR: PQ ⊥ PS ⇒ ∠P = 90°.
∠PSQ + ∠PQS = 90° (angle sum with ∠P=90°). Wait – for finding y:
In ΔPQS: ∠P + x + y = 180°
90° + 37° + y = 180°
y = 53°.

Q6. In Fig. 6.44, 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 = ½∠QPR.

Proof:
In ΔPQR, exterior angle ∠PRS = ∠QPR + ∠PQR ...(1) [Exterior angle property]
In ΔTQR, exterior angle ∠TRS = ∠TQR + ∠QTR ...(2) [Exterior angle property]
TQ bisects ∠PQR ⇒ ∠TQR = ½∠PQR.
TR bisects ∠PRS ⇒ ∠TRS = ½∠PRS = ½(∠QPR + ∠PQR) [from (1)].
From (2): ½(∠QPR + ∠PQR) = ½∠PQR + ∠QTR.
½∠QPR + ½∠PQR = ½∠PQR + ∠QTR.
∠QTR = ½∠QPR. (Proved)

Class 9 Maths - Lines and Angles Practice

Chapter 6: Lines and Angles (Practice Questions)

RD Sharma / Extra Practice

Q1. Two supplementary angles are in the ratio 4:5. Find the angles.

Let the angles be 4x and 5x. They are supplementary, so 4x + 5x = 180°.
9x = 180° ⇒ x = 20°.
Angles are 4 × 20 = 80° and 5 × 20 = 100°.

Q2. Find the complement of each of the following: (i) 35° (ii) 47° (iii) 65° (iv) 90°.

Complement = 90° - given angle.
(i) 90° - 35° = 55°.
(ii) 90° - 47° = 43°.
(iii) 90° - 65° = 25°.
(iv) 90° - 90° = 0°.

Q3. Two lines AB and CD intersect at O. If ∠AOC = 50°, find all the four angles.

∠AOC = 50°.
∠BOD = ∠AOC = 50° (vertically opposite).
∠AOD = 180° - ∠AOC = 130° (linear pair).
∠BOC = 130° (vertically opposite to ∠AOD).
Angles: ∠AOC = 50°, ∠BOD = 50°, ∠AOD = 130°, ∠BOC = 130°.

Q4. In the figure, if l ∥ m and n is a transversal making ∠1 = 75°, find all remaining angles.

∠1 = 75°.
∠2 = 180° - 75° = 105° (linear pair with ∠1).
∠3 = ∠1 = 75° (corresponding angles, l ∥ m).
∠4 = ∠2 = 105° (corresponding angles).
∠5 = ∠1 = 75° (alternate interior angles).
∠6 = ∠2 = 105° (alternate interior angles).
∠7 = ∠1 = 75° (vertically opposite to ∠3).
∠8 = ∠2 = 105° (vertically opposite to ∠4).

Q5. Prove that if two parallel lines are cut by a transversal, then the bisectors of the alternate interior angles are parallel.

Let lines AB ∥ CD be cut by transversal EF at P and Q respectively. Let PG and QH bisect the alternate interior angles ∠BPQ and ∠PQD respectively.
∠BPQ = ∠PQD (alternate interior angles).
½∠BPQ = ½∠PQD ⇒ ∠GPQ = ∠PQH.
Since ∠GPQ and ∠PQH are alternate interior angles and equal, PG ∥ QH. (Proved)

Q6. An angle is 4 times its complement. Find the angle.

Let the angle be x. Its complement = 90° - x.
x = 4(90° - x)
x = 360° - 4x
5x = 360° ⇒ x = 72°.

Q7. If the angles of a triangle are in the ratio 2:3:4, find all the angles of the triangle.

Let angles be 2x, 3x, 4x. By angle sum property: 2x + 3x + 4x = 180°.
9x = 180° ⇒ x = 20°.
Angles: 40°, 60°, 80°.

Q8. In a triangle, one angle is 60° and the other two angles are equal. Find them.

Let the two equal angles be x each.
60° + x + x = 180°
2x = 120° ⇒ x = 60°.
All three angles are 60° (equilateral triangle).

Q9. The exterior angle of a triangle is 105°. One of the two interior opposite angles is 60°. Find the other interior opposite angle.

By the Exterior Angle Property: Exterior angle = Sum of two non-adjacent interior angles.
105° = 60° + other angle.
Other angle = 105° - 60° = 45°.

Q10. In ΔABC, ∠B = 45°, ∠C = 55° and bisector of ∠A meets BC at D. Find ∠ADB and ∠ADC.

∠A = 180° - 45° - 55° = 80°.
AD bisects ∠A ⇒ ∠BAD = ∠CAD = 40°.
In ΔABD: ∠ADB = 180° - ∠B - ∠BAD = 180° - 45° - 40° = 95°.
∠ADC = 180° - ∠ADB = 180° - 95° = 85°.

Q11. Two angles are supplementary. One angle is 20° more than three times the other. Find both angles.

Let one angle = x. Then the other = 3x + 20°.
x + 3x + 20° = 180°
4x = 160° ⇒ x = 40°.
Angles: 40° and 140°.

Q12. AB ∥ CD. A transversal EF cuts AB at G and CD at H. If ∠AGF = 3x - 15° and ∠GHD = 2x + 10°, find x and both angles.

∠AGF = ∠GHD (alternate exterior angles with AB ∥ CD).
3x - 15 = 2x + 10
x = 25.
∠AGF = 3(25) - 15 = 60°. ∠GHD = 2(25) + 10 = 60°.

Q13. What is the angle formed between two adjacent supplementary angles at their vertex called?

The two supplementary adjacent angles together form a linear pair. The combined angle formed is a straight angle = 180°

Q14. The angles of a quadrilateral are in the ratio 1:2:3:4. Find all four angles.

Sum of angles of a quadrilateral = 360°.
Let angles be x, 2x, 3x, 4x.
x + 2x + 3x + 4x = 360° ⇒ 10x = 360° ⇒ x = 36°.
Angles: 36°, 72°, 108°, 144°.

Q15. In ΔABC, if ∠A = 2∠B = 6∠C, find all three angles.

Let ∠A = 6k, ∠B = 3k, ∠C = k (since 2∠B = 6k and 6∠C = 6k both satisfy ∠A = 6k).
6k + 3k + k = 180° ⇒ 10k = 180° ⇒ k = 18°.
∠A = 108°, ∠B = 54°, ∠C = 18°.

Class 9 Maths - Lines and Angles Summary

Chapter 6: Lines and Angles (Concepts & Formulas)

1. Types of Angles

Acute Angle: Less than 90°
Right Angle: Exactly 90°
Obtuse Angle: Between 90° and 180°
Straight Angle: Exactly 180°
Reflex Angle: Between 180° and 360°
Complementary Angles: Two angles whose sum = 90°
Supplementary Angles: Two angles whose sum = 180°
Linear Pair: Adjacent supplementary angles on a straight line

2. Key Axioms

Linear Pair Axiom: If a ray stands on a line, the two adjacent angles form a linear pair summing to 180°.
Converse: If two adjacent angles sum to 180°, the outer rays form a straight line.
Corresponding Angles Axiom: If a transversal cuts two parallel lines, corresponding angles are equal.
Converse: If corresponding angles are equal, the lines are parallel.

3. Intersecting Lines Theorems

Vertically Opposite Angles: When two lines intersect, vertically opposite angles are equal.

4. Transversal and Parallel Lines

When a transversal cuts two parallel lines:

Corresponding angles are equal.
Alternate interior angles are equal.
Alternate exterior angles are equal.
Co-interior (same side interior) angles are supplementary (sum = 180°).

Converses of all the above also hold true.

5. Triangle Angle Properties

Angle Sum Property: Sum of all interior angles of a triangle = 180°.
∠A + ∠B + ∠C = 180°
Exterior Angle Property: An exterior angle of a triangle equals the sum of the two non-adjacent interior angles.
Exterior ∠ = ∠B + ∠C (if exterior is at A)

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