Gujarat Board Textbook Solutions Class 9 Maths Chapter 6 Lines and Angles
GSEB Solutions Class 9 Maths Chapter 6 Lines and Angles
Ex 6.1
Question 1.
In the given figure AB and CD intersect at O. If ∠AOC + ∠BOE = 70°, and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
Solution:
It is given that
∠AOC +∠BOE = 70° ………..(1)
∠AOC = ∠BOD
⇒ ∠AOC = 40°
Putting in eqn. (1),
40° + ∠BOE = 70°
∠BOE = 70° – 40°
⇒ ∠BOE = 30°
∠AOE + ∠BOE = 180° (Linear pair)
∠AOC + ∠COE + ∠BOE = 180°
⇒ ∠AOC + ∠BOE + ∠COE = 1800
⇒ 70° +∠COE = 180°
⇒ ∠COE = 180° – 70°
⇒ ∠COE = 110°
Now, Reflex ∠COE = 360° – 110° = 250°
Question 2.
In figure lines XY and MN intersect at O. If ∠POY = 90°and a : b = 2 : 3, find c.
Solution:
Let a = 2 x and b = 3x
∠POX + ∠POY = 180° (Linear pair)
∠POX + 90° = 180°
∠POX = 180° – 90° = 90°
⇒ a + b = 90°
⇒ 2x + 3x = 90°
⇒ 5x = 90°
⇒ x = 90°/5 = 18°
∴ a = 2x = 2 x 18°= 36°
⇒ b = 3x = 3 x 18°= 54°
Now, ∠XON + ∠YON = 180° (Linear pair)
⇒ c + 54° = 180° (∠XOM = b and ∠XON = c)
⇒ c = 180° – 54°
⇒ c = 126°
Question 3.
In figure ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Solution:
Given: ∠PQR = ∠PRQ
To Prove: ∠PQS = ∠PRT
Proof: ∠PQS + ∠PQR = 180° ………. (1) (Linear pair)
∠PRQ +∠PRT = 180° ………(2) (Linear pair)
But ∠PQR =∠PRQ
Putting value ∠PQR from eqn. (3) in eqn. (1)
∠PQS + ∠PRQ = 180°
Now from eqn. (2) and (4), we have
∠PQS + ∠PRQ = ∠PRQ + ∠PRT
∠PQS = ∠PRT
Question 4.
In the given figure, if x + y = w + z, then prove that AOB is a line.
Solution:
Given: x + y = w + z
To Prove: AOB is a line.
Proof: ∠BOC + ∠AOC +∠BOD + ∠AOD = 360°
(Sum of all angles round a point is equal to 360°)
x + y + w + z = 360° ……..(1)
But x + y = w + z ……..(2)
From eqn. (1) and (2), we have
w + z + w + z = 360°
⇒ 2w + 2z = 360°
⇒2(w + z) = 360°
⇒ w + z = 360°/2
⇒ w + z = 180°
w + z = 180° (Linear pair)
Hence,
∠BOD + ∠AOD = 180°
Therefore AOB is a line.
Question 5.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between ray OP and OR. Prove that
∠ROS = 1/2(∠QOS – ∠POS).
Solution:
Given: Ray OR ⊥ PQ and ray OS is another ray lying between OP and OR.
To Prove: ∠ROS = (∠QOS – ∠POS)
Proof: Ray OR ⊥ PQ
∴ ∠POR = ∠ROQ = 90°
Now, ∠POS = ∠POR – ∠ROS
∠POS = 90° – ∠ROS ………..(1)
∠QOS = ∠QOR + ∠ROS
∠QOS = 90° + ∠ROS ……..(2)
Subtracting eqn (2) from eqn (1),
∠QOS – ∠POS = 90° + ∠ROS – (90°- ∠ROS)
∠QOS – ∠POS = 90° + ∠ROS – 90° + ∠ROS
∠QOS – ∠POS = 2∠ROS
2 ∠ROS = ∠QOS – ∠POS
∠ROS = 1/2(∠QOS – ∠POS)
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.
Solution:
∠XYZ = 64°
Ray YZ stands on PX.
∠XYZ + ∠PYZ = 180° (Linear pair)
64° + ∠PYZ = 180°
⇒ ∠PYZ = 180°- 64°
⇒ ∠PYZ = 116°
⇒ ∠QYZ = ∠PYZ/2
(YQ is the bisector of ∠PYZ)
∠QYZ = 116°/2
∠QYZ = 58°
∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58°
∠XYQ = 122°
∠QYP = ∠QYZ
(YQ is the bisector of ∠PYZ)
∠QYP = 58°
Reflex ∠QYP = 360° – 58°
Reflex ∠QYP = 302°
Ex 6.2
Question 1.
In figure, find the values of x and y and then show that AB || CD.
Solution:
x + 50° = 180° (Linear pair)
⇒ x = 180° – 50°
⇒ x = 130°
y = 130° (Vertically opposite angles)
∴ x = y (Alternate interior angles)
If alternate interior angles are equal then lines are parallel.
∴ AB || CD
Question 2.
In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.
Solution:
Let y = 3a
and z = 7a
b = y (Vertically opposite angles)
CD || EF
∴ b+z = 180°
(Interior angles of same side of transversal)
⇒ y + z = 180°
⇒ 3a + 7a = 180°
⇒ 10a = 180°
a = 180°/10
a = 18°
∴ y = 3a = 3 x 18°
⇒ y = 54°
z = 7a = 7 x 18°
2 = 126°
Now AB || CD and CD || EF
∴ AB || EF
(Lines parallel to same line are parallel to one another.)
⇒ x = z
(Alternate interior angles) x = 126°
Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Solution:
We have AB || CD, EF ⊥ CD
and ∠GED = 126°
∠AGE = ∠GED (Alternate interior angles)
⇒∠AGE = 126°
Again, ∠GED = ∠GEF + ∠FED
⇒ 126° = ∠GEF + 90°
⇒ ∠GEF = 126° – 90°
⇒ ∠GEF = 36°
∠FGE + ∠GED = 180° (co-interior angles)
⇒ ∠FGE + 126° = 180°
⇒ ∠FGE = 180° – 126°
∠FGE = 54°
Question 4.
In figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
(Hint: Draw a line parallel to ST through point P).
Solution:
We draw a line AB || ST through R
∠TSR + ∠SRB = 180° (Interior angles of same side of transversal)
⇒ 130° + ∠SRB = 180°
⇒ ∠SRB = 180° – 130°
⇒ ∠SRB = 50° ……….(1)
PQ || AB (Lines parallel to same line are parallel to each other)
∠PQR + ∠ARQ = 180° (Interior angles of same side of transversal)
⇒ 110° + ∠APQ = 180°
⇒ ∠ARQ = 180° – 110°
⇒ ∠ARQ = 70° ……….(2)
Now, ∠ARQ + ∠QRB = 180°
(Linear pair)
⇒ ∠ARQ + ∠QRS + ∠SRB = 180°
⇒ 70° + ∠QRS + 50° = 180°
[Putting values from eqn. (1) and (2)]
⇒ ∠QRS + 120° = 180°
∠QRS = 180° – 120° = 60°
Question 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD 127°, find x and y.
Solution:
AB || CD
then ∠APQ = ∠PQR (Alternate interior angles)
⇒ 50° = x
⇒ x = 50°
and ∠APR – ∠PRD
⇒ ∠APQ + ∠QPR = 127°
⇒ 50° + y = 127°
⇒ y= 127° – 50°
⇒ y = 77°
Question 6.
In 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.
Solution:
Given: PQ || RS, incident ray AB which strikes the mirror PQ at B, and reflected ray BC which strikes the mirror RS at C and again reflects back along CD.
To Prove: AB || CD
Construction: Draw BM ⊥ PQ and CN ⊥ RS
Proof: We have BM ⊥ PQ and CN ⊥ RS
∠2 = ∠3 ………..(1)
(Alternate interior angles)
∠1 = ∠2 ……….(2)
(Angle of incidence is equal to the angle of reflection)
∠3 = ∠4 ……….(3)
(By law of reflection, angle of incidence is equal to the angle of reflection)
From eqn. (1), (2) and (3)
∠1 = ∠4 ……….(4)
Adding eqn. (1) and (4)
∠1 + ∠2 = ∠3 + ∠4
∠ABC = ∠BCD (Alternate interior angles)
∴ AB || CD
(If. alternate interior angles are equal then lines are parallel).
Ex 6.3
Question 1.
In figure sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Solution:
∠PQT + ∠PQR = 180° (Linear pair)
⇒ 110° + ∠PQR = 180°
⇒ ∠PQR = 180° – 110°
⇒ ∠PQR = 70°
∠RPS = ∠PQR + ∠PRQ (Exterior angle property)
⇒ 135° – 70° + ∠PRQ
⇒ ∠PRQ = 135° – 70°
∠PRQ = 65°
Question 2.
In figure ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.
Solution:
In ΔXYZ,
∠YXZ + ∠XYZ + ∠XZY = 180° (Angle sum property)
⇒ 62° + 54° + ∠XZY = 180°
⇒ 116° + ∠XZY = 180°
⇒ ∠XZY = 180° – 116°
∠XZY = 64°
In ΔOYZ,
∠OYZ + ∠OZY + ∠YOZ = 180°
(Angle sum property)
⇒ 27° + 32° + ∠YOZ = 180°
⇒ 59° + ∠YOZ = 180°
⇒ ∠YOZ = 180° – 59°
⇒ ∠YOZ = 121°
Question 3.
In figure AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Solution:
AB || DE
∠DEC = ∠BAC
(Alternate interior angles)
⇒ ∠DEC = 35°
Now in ACDE,
∠DCE + ∠CDE + ∠DEC = 180° (Angle sum property)
∠DCE + 53° + 35° = 180°
∠DCE = 180° – 88° ∠DCE = 92°
Question 4.
In figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Solution:
In ΔPRT,
∠PRT + ∠RPT + ∠PTR = 180° (Angle sum property)
40° + 95° + ∠PTR = 180°
135° + ∠PTR = 180°
PTR = 180° – 135° ∠PTR = 45°
∠STQ = ∠PTR
(Vertically opposite angles)
∠STQ = 45°
Now in ΔSQT,
∠STQ + ∠SQT ∠QST = 180° (Angle sum property)
⇒ 45° + ∠SQT + 75° = 180°
⇒ ∠SQT + 120° – 180°
⇒ ∠SQT = 180° – 120°
⇒ ∠SQT = 60°
Question 5.
In figure, if PQ ⊥PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Solution:
We have PQ || SR
∴ PQ || ST
∴ ∠PQR = ∠QTR (Alternate interior angles)
⇒ x + 28° = 65°
⇒ x = 65° – 28°
⇒ x = 37°
In ΔPQS,
∠PQS + ∠PSQ + ∠SPQ = 180° (Angle sum property)
⇒ x + y + 90° = 180°
37° + 90° + y = 180°
127° + y = 180°
⇒ y= 180° – 127°
⇒ y = 53°
Question 6.
In figure, the side QR of APQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.
Solution:
Given: ΔPQR in which side QR produced to S and bisectors of ∠QPRPQR and ∠QPRPRS meet at T.
To Prove: ∠QPR = 1/2 ∠QPR
Proof: ∠TRS is the exterior angle of ΔTQR.
∴ ∠TRS = ∠1 + ∠2 …….(1)
(Exterior angle of triangle is equal to sum of two interior opposite angles)
∠PRS is an exterior angle of ΔPQR.
∴ ∠PRS = ∠PQR + ∠QPR
(Exterior angle of a triangle is equal to the sum of its two interior opposite angles)
⇒ 2 ∠TRS = 2∠1 + ∠QPR
2(∠1 + ∠2) = 2∠1 + ∠QPR [From eqn(Δ)]
⇒ 2∠1 + 2∠2 = 2∠1 + ∠QPR
⇒ 2∠2 = ∠QPR
⇒ ∠2 = 1/2 ∠QPR
⇒ ∠QTR = 1/2 ∠QPR