NCERT Solutions for Class 9 Maths Chapter 6 – Lines and Angles
Page No: 96
Exercise 6.1
1. In Fig. 6.13, 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°
A/q,
∠AOC + ∠BOE +∠COE = 180° (Forms a straight line)
⇒ 70° +∠COE = 180°
⇒ ∠COE = 110°
also,
∠COE +∠BOD + ∠BOE = 180° (Forms a straight line)
⇒ 110° +40° + ∠BOE = 180°
⇒ 150° + ∠BOE = 180°
⇒ ∠BOE = 30°
Page No: 97
2. In Fig. 6.14, 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
A/q,
∠POY + a + b = 180°
⇒ 90° + a + b = 180°
⇒ a + b = 90°
Let a be 2x then will be 3x
2x + 3x = 90°
⇒ 5x = 90°
⇒ x = 18°
∴ a = 2×18° = 36°
and b = 3×18° = 54°
also,
b + c = 180° (Linear Pair)
⇒ 54° + c = 180°
⇒ c = 126°
3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Answer
Given,
∠PQR = ∠PRQ
To prove,
∠PQS = ∠PRT
A/q,
∠PQR +∠PQS = 180° (Linear Pair)
⇒ ∠PQS = 180° - ∠PQR --- (i)
also,
∠PRQ +∠PRT = 180° (Linear Pair)
⇒ ∠PRT = 180° - ∠PRQ
⇒ ∠PRQ = 180° - ∠PQR --- (ii) (∠PQR = ∠PRQ)
From (i) and (ii)
∠PQS = ∠PRT = 180° - ∠PQR
Therefore, ∠PQS = ∠PRT
4. In Fig. 6.16, 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° (linear pair.)
A/q,
x + y + w + z = 360° (Angles around a point.)
⇒ (x + y) + (w + z) = 360°
⇒ (x + y) + (x + y) = 360° (Given x + y = w + z)
⇒ 2(x + y) = 360°
⇒ (x + y) = 180°
Hence, x + y makes a linear pair. Therefore, AOB is a staright line.
5. 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 = 1/2(∠QOS – ∠POS).
Answer
Given,
OR is perpendicular to line PQ
To prove,
∠ROS = 1/2(∠QOS – ∠POS)
A/q,
∠POR = ∠ROQ = 90° (Perpendicular)
∠QOS = ∠ROQ+ ∠ROS = 90° + ∠ROS --- (i)
∠POS = ∠POR - ∠ROS = 90° - ∠ROS --- (ii)
Subtracting (ii) from (i)
∠QOS - ∠POS = 90° + ∠ROS - (90° - ∠ROS)
⇒ ∠QOS - ∠POS = 90° + ∠ROS - 90° + ∠ROS
⇒ ∠QOS - ∠POS = 2∠ROS
⇒ ∠ROS = 1/2(∠QOS – ∠POS)
Hence, Proved.
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
∠XYZ +∠ZYP = 180° (Linear Pair)
⇒ 64° +∠ZYP = 180°
⇒ ∠ZYP = 116°
also, ∠ZYP = ∠ZYQ + ∠QYP
∠ZYQ = ∠QYP (YQ bisects ∠ZYP)
⇒ ∠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°
⇒ ∠QYP = 302°
Page No: 103
Exercise 6.2
1. In Fig. 6.28, find the values of x and y and then show that AB || CD.
Answer
x + 50° = 180° (Linear pair)
⇒ x = 130°
also,
y = 130° (Vertically opposite)
Now,
x = y = 130° (Alternate interior angles)
Alternate interior angles are equal.
Therefore, AB || CD.
Page No: 104
2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Exercise 6.1
1. In Fig. 6.13, 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°
A/q,
∠AOC + ∠BOE +∠COE = 180° (Forms a straight line)
⇒ 70° +∠COE = 180°
⇒ ∠COE = 110°
also,
∠COE +∠BOD + ∠BOE = 180° (Forms a straight line)
⇒ 110° +40° + ∠BOE = 180°
⇒ 150° + ∠BOE = 180°
⇒ ∠BOE = 30°
Page No: 97
2. In Fig. 6.14, 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
A/q,
∠POY + a + b = 180°
⇒ 90° + a + b = 180°
⇒ a + b = 90°
Let a be 2x then will be 3x
2x + 3x = 90°
⇒ 5x = 90°
⇒ x = 18°
∴ a = 2×18° = 36°
and b = 3×18° = 54°
also,
b + c = 180° (Linear Pair)
⇒ 54° + c = 180°
⇒ c = 126°
3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Answer
Given,
∠PQR = ∠PRQ
To prove,
∠PQS = ∠PRT
A/q,
∠PQR +∠PQS = 180° (Linear Pair)
⇒ ∠PQS = 180° - ∠PQR --- (i)
also,
∠PRQ +∠PRT = 180° (Linear Pair)
⇒ ∠PRT = 180° - ∠PRQ
⇒ ∠PRQ = 180° - ∠PQR --- (ii) (∠PQR = ∠PRQ)
From (i) and (ii)
∠PQS = ∠PRT = 180° - ∠PQR
Therefore, ∠PQS = ∠PRT
4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
Given,
x + y = w + z
To Prove,
AOB is a line or x + y = 180° (linear pair.)
A/q,
x + y + w + z = 360° (Angles around a point.)
⇒ (x + y) + (w + z) = 360°
⇒ (x + y) + (x + y) = 360° (Given x + y = w + z)
⇒ 2(x + y) = 360°
⇒ (x + y) = 180°
Hence, x + y makes a linear pair. Therefore, AOB is a staright line.
5. 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 = 1/2(∠QOS – ∠POS).
Given,
OR is perpendicular to line PQ
To prove,
∠ROS = 1/2(∠QOS – ∠POS)
A/q,
∠POR = ∠ROQ = 90° (Perpendicular)
∠QOS = ∠ROQ
∠POS = ∠POR - ∠ROS = 90° - ∠ROS --- (ii)
Subtracting (ii) from (i)
∠QOS - ∠POS = 90° + ∠ROS - (90° - ∠ROS)
⇒ ∠QOS - ∠POS = 90° + ∠ROS - 90° + ∠ROS
⇒ ∠QOS - ∠POS = 2∠ROS
⇒ ∠ROS = 1/2(∠QOS – ∠POS)
Hence, Proved.
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
⇒ 64° +∠ZYP = 180°
⇒ ∠ZYP = 116°
also, ∠ZYP = ∠ZYQ + ∠QYP
∠ZYQ = ∠QYP (YQ bisects ∠ZYP)
⇒ ∠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°
⇒ ∠QYP = 302°
Page No: 103
Exercise 6.2
1. In Fig. 6.28, find the values of x and y and then show that AB || CD.
x + 50° = 180° (Linear pair)
⇒ x = 130°
also,
y = 130° (Vertically opposite)
Now,
x = y = 130° (Alternate interior angles)
Alternate interior angles are equal.
Therefore, AB || CD.
Page No: 104
2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Given,
AB || CD and CD || EF
y : z = 3 : 7
Now,
x + y = 180° (Angles on the same side of transversal.)
also,
∠O = z (Corresponding angles)
and, y
⇒ y + z = 180°
A/q,
y = 3w and z = 7w
3w + 7w = 180°
⇒ 10 w = 180°
⇒ w = 18°
∴ y = 3×18° = 54°
and, z = 7×18° = 126°
Now,
x + y = 180°
⇒ x + 54° = 180°
⇒ x = 126°
3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Answer
Given,
AB || CD
EF ⊥ CD
∠GED = 126°
A/q,
∠FED = 90° (EF ⊥ CD)
Now,
∠AGE = ∠GED (Since, AB || CD and GE is transversal. Alternate interior angles.)
∴ ∠AGE = 126°
Also, ∠GEF = ∠GED - ∠FED
⇒ ∠GEF = 126° - 90°
⇒ ∠GEF = 36°
Now,
∠FGE +∠AGE = 180° (Linear pair)
⇒ ∠FGE = 180° - 126°
⇒ ∠FGE = 54°
4. In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint : Draw a line parallel to ST through point R.]
Answer
Given,
PQ || ST, ∠PQR = 110° and ∠RST = 130°
Construction,
A line XY parallel to PQ and ST is drawn.
⇒ 110° + ∠QRX = 180°
⇒ ∠QRX = 70°
Also,
∠RST + ∠SRY = 180° (Angles on the same side of transversal.)
⇒ 130° + ∠SRY = 180°
⇒ ∠SRY = 50°
Now,
∠QRX +∠SRY
⇒ 70° + 50° + ∠QRS = 180°
⇒ ∠QRS = 60°
5. In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
Given,
AB || CD, ∠APQ = 50° and ∠PRD = 127°
A/q,
x = 50° (Alternate interior angles.)
∠PRD + ∠RPB = 180° (Angles on the same side of transversal.)
⇒ 127° + ∠RPB = 180°
⇒ ∠RPB = 53°
Now,
y + 50° + ∠RPB = 180° (AB is a straight line.)
⇒ y + 50° + 53° = 180°
⇒ y + 103° = 180°
⇒ y = 77°
6. In Fig. 6.33, 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
Let us draw BE ⟂ PQ and CF ⟂ RS.
As PQ || RSSo, BE || CF
By laws of reflection we know that,
Angle of incidence = Angle of reflection
Thus, ∠1 = ∠2 and ∠3 = ∠4 --- (i)
also, ∠2 = ∠3 (alternate interior angles because BE || CF and a transversal line BC cuts them at B and C) --- (ii)
From (i) and (ii),
∠1 + ∠2 = ∠3 + ∠4
⇒ ∠ABC = ∠DCB
⇒ AB || CD (alternate interior angles are equal)
Exercise 6.3
1. 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.
Given,
∠SPR = 135° and ∠PQT = 110°
A/q,
∠SPR +∠QPR = 180° (SQ is a straight line.)
⇒ 135° +∠QPR = 180°
⇒ ∠QPR = 45°
also,
∠PQT +∠PQR = 180° (TR is a straight line.)
⇒ 110° +∠PQR = 180°
⇒ ∠PQR = 70°
Now,
∠PQR +∠QPR + ∠PRQ = 180° (Sum of the interior angles of the triangle.)
⇒ 70° + 45° + ∠PRQ = 180°
⇒ 115° + ∠PRQ = 180°
⇒ ∠PRQ = 65°
2. In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠YOZ.
Given,
∠X = 62°, ∠XYZ = 54°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
A/q,
∠X +∠XYZ
⇒ 62° + 54° + ∠XZY = 180°
⇒ 116° + ∠XZY = 180°
⇒ ∠XZY = 64°
Now,
∠OZY = 1/2∠XZY (ZO is the bisector.)
⇒ ∠OZY = 32°
also,
∠OYZ = 1/2∠XYZ (YO is the bisector.)
⇒ ∠OYZ = 27°
Now,
∠OZY +∠OYZ
⇒ 32° + 27° + ∠O = 180°
⇒ 59° + ∠O = 180°
⇒ ∠O = 121°
3. In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Answer
Given,
AB || DE, ∠BAC = 35° and ∠ CDE = 53°
A/q,
∠BAC = ∠CED (Alternate interior angles.)
∴ ∠CED = 35°
Now,
∠DCE +∠CED + ∠CDE = 180° (Sum of the interior angles of the triangle.)
⇒ ∠DCE + 35° + 53° = 180°
⇒ ∠DCE + 88° = 180°
⇒ ∠DCE = 92°
4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Answer
Given,
∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°
A/q,
∠PRT +∠RPT + ∠PTR = 180° (Sum of the interior angles of the triangle.)
⇒ 40° + 95° + ∠PTR = 180°
⇒ 40° + 95° + ∠PTR = 180°
⇒ 135° + ∠PTR = 180°
⇒ ∠PTR = 45°
∠PTR = ∠STQ = 45° (Vertically opposite angles.)
Now,
∠TSQ +∠PTR
⇒ 75° + 45° + ∠SQT = 180°
⇒ 120° + ∠SQT = 180°
⇒ ∠SQT = 60°
Page No: 108
5. In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Given,
PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°
A/q,
x +∠SQR = ∠QRT (Alternate angles as QR is transveersal.)
⇒ x + 28° = 65°
⇒ x = 37°
also,
∠QSR = x
⇒ ∠QSR = 37°
also,
∠QRS +∠QRT = 180° (Linea pair)
⇒ ∠QRS + 65° = 180°
⇒ ∠QRS = 115°
Now,
∠P
⇒ 90° + 65° + 115° + ∠S = 360°
⇒ 270° + y + ∠QSR = 360°
⇒ 270° + y + 37° = 360°
⇒ 307° + y = 360°
⇒ y = 53°
6. 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 = 1/2∠QPR.
Answer
Given,
Bisectors of ∠PQR and ∠PRS meet at point T.
To prove,
∠QTR = 1/2∠QPR.
Proof,
∠TRS = ∠TQR
⇒ ∠QTR = ∠TRS - ∠TQR --- (i)
also,
∠SRP = ∠QPR
⇒ 2∠TRS = ∠QPR + 2∠TQR
⇒ ∠QPR = 2∠TRS - 2∠TQR
⇒ 1/2∠QPR = ∠TRS - ∠TQR --- (ii)
Equating (i) and (ii)
∠QTR - ∠TQR = 1/2∠QPR
Hence proved.
