NCERT Solutions for Class 9th: Ch 6 Lines and Angles Maths

NCERT Solutions for Class 9th: Ch 6 Lines and Angles Maths

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° + ∠ROQ --- (i)
∠POS = ∠POR - ∠ROS = 90° - ∠ROQ --- (ii)
Subtracting (ii) from (i)
∠QOS - ∠POS = 90° + ∠ROQ - (90° - ∠ROQ)
⇒ ∠QOS - ∠POS = 90° + ∠ROQ - 90° + ∠ROQ
⇒ ∠QOS - ∠POS = 2∠ROQ
⇒ ∠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.


Answer

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 + ∠O = 180° (Linear pair)
⇒ 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.]
∠PQR + ∠QRX = 180° (Angles on the same side of transversal.)
⇒ 110° + ∠QRX = 180°
⇒ ∠QRX = 70°
Also,
∠RST + ∠SRY = 180° (Angles on the same side of transversal.)
⇒ 130° + ∠SRY = 180°
⇒ ∠SRY = 50°
Now,
∠QRX +∠SRY + ∠QRS = 180°
⇒ 70° + 50° + ∠QRS = 180°
⇒ ∠QRS = 60°

5. In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
 Answer

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°

Page No: 107

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.
Answer

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.
Answer

Given,
∠X = 62°, ∠XYZ = 54°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
A/q,
∠X +∠XYZ + ∠XZY = 180° (Sum of the interior angles of the triangle.)
⇒ 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 + ∠O = 180° (Sum of the interior angles of the triangle.)
⇒ 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 + ∠SQT = 180° (Sum of the interior angles of the triangle.)
⇒ 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.
Answer

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 + ∠Q+ ∠R +∠S = 360° (Sum of the angles in a quadrilateral.)
⇒ 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 (Exterior angle of a triangle equals to the sum of the two interior angles.)
⇒ ∠QTR = ∠TRS - ∠TQR --- (i)
also,
∠SRP = ∠QPR + ∠PQR
⇒ 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.
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