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NCERT Solutions for Class 9 Maths: Chapter 6 Lines and Angles

Lines and Angles Class 9

Exercise 6.1 : Solutions of Questions on Page Number : 96

Q1 : In the given figure, lines AB and CD intersect at O. If and find ∠BOE and reflex ∠COE.  Q2 : In the given figure, lines XY and MN intersect at O. If ∠POY = and a:b = 2 : 3, find c. Let the common ratio between a and b be x.
∴ a = 2x, and b = 3x
XY is a straight line, rays OM and OP stand on it.
∴ ∠ XOM + ∠ MOP + ∠ POY = 180º
b + a + ∠ POY = 180º
3x + 2x + 90º = 180º
5x = 90º
x = 18º
a = 2x = 2 x 18 = 36º
b = 3x= 3 x 18 = 54º
MN is a straight line. Ray OX stands on it.
∴ b + c = 180º (Linear Pair)
54º + c = 180º
c = 180º – 54º = 126º
∴ c = 126º

Q3 : In the given figure, ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT. In the given figure, ST is a straight line and ray QP stands on it.
∴ ∠PQS + ∠PQR = 180 º (Linear Pair)
∠PQR = 180 º – ∠PQS (1)
∠PRT + ∠PRQ = 180 º (Linear Pair)
∠PRQ = 180 º – ∠PRT (2)
It is given that ∠PQR = ∠PRQ.
Equating equations (1) and (2), we obtain
180 º – ∠PQS = 180 – ∠PRT
∠PQS = ∠PRT

Q4 : In the given figure, if x + y = W + Z then prove that AOB is a line. It can be observed that,
x + y + z + w = 360º (Complete angle)
It is given that,
x + y = z + w
∴ x + y + x + y = 360º
2(x + y) = 360º
x + y = 180º
Since x and y form a linear pair, AOB is a line.

Q5 : In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that  It is given that OR ⊥ PQ
∴ ∠POR = 90 º
⇒ ∠POS + ∠SOR = 90 º
∠ROS = 90 º – ∠POS … (1)
∠QOR = 90 º (As OR ⊥ PQ)
∠QOS – ∠ROS = 90 º
∠ROS = ∠QOS – 90 º … (2)
On adding equations (1) and (2), we obtain
2 ∠ROS = ∠QOS – ∠POS 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. It is given that line YQ bisects ∠ PYZ.
Hence, ∠ QYP = ∠ ZYQ
It can be observed that PX is a line. Rays YQ and YZ stand on it.
∴ ∠ XYZ + ∠ ZYQ + ∠ QYP = 180º
⇒ 64º + 2∠ QYP = 180º
⇒ 2∠ QYP = 180º – 64º = 116º
⇒ ∠ QYP = 58º
Also, ∠ ZYQ = ∠ QYP = 58º
Reflex ∠ QYP = 360º – 58º = 302º
∠ XYQ = ∠ XYZ + ∠ ZYQ
= 64º + 58º = 122º

Exercise 6.2 : Solutions of Questions on Page Number : 103

Q1 : In the given figure, find the values of x and y and then show that AB || CD. It can be observed that,
50º + x = 180º (Linear pair)
x = 130º … (1)
Also, y = 130º (Vertically opposite angles)
As x and y are alternate interior angles for lines AB and CD and also measures of these angles are equal to each other, therefore, line AB || CD.

Q2 : In the given figure, if AB || CD, CD || EF and y: z = 3: 7, find x. It is given that AB || CD and CD || EF
∴ AB || CD || EF (Lines parallel to the same line are parallel to each other)
It can be observed that
x = z (Alternate interior angles) … (1)
It is given that y: z = 3: 7
Let the common ratio between y and z be a.
∴ y = 3a and z = 7a
Also, x + y = 180º (Co-interior angles on the same side of the transversal)
z + y = 180º [Using equation (1)]
7a + 3a = 180º
10a = 180º
a = 18º
∴ x = 7a = 7 x 18º = 126º

Q3 : In the given figure, If AB || CD, EF ⊥ CD and ∠ GED = 126º, find ∠ AGE, ∠ GEF and ∠ FGE. It is given that,
AB || CD
EF ⊥ CD
∠ GED = 126º
⇒ ∠ GEF + ∠ FED = 126º
⇒ ∠ GEF + 90º = 126º
⇒ ∠ GEF = 36º
∠ AGE and ∠ GED are alternate interior angles.
⇒ ∠ AGE = ∠ GED = 126º
However, ∠ AGE + ∠ FGE = 180º (Linear pair)
⇒ 126º + ∠ FGE = 180º
⇒ ∠ FGE = 180º – 126º = 54º
∴ ∠ AGE = 126º, ∠ GEF = 36º, ∠ FGE = 54º

Q4 : In the given figure, if PQ || ST, ∠ PQR = 110º and ∠ RST = 130º, find ∠ QRS.
[Hint: Draw a line parallel to ST through point R.]  Let us draw a line XY parallel to ST and passing through point R.
∠ PQR + ∠ QRX = 180º (Co-interior angles on the same side of transversal QR)
⇒ 110º + ∠ QRX = 180º
⇒ ∠ QRX = 70º
Also,
∠ RST + ∠ SRY = 180º (Co-interior angles on the same side of transversal SR)
130º + ∠ SRY = 180º
∠ SRY = 50º
XY is a straight line. RQ and RS stand on it.
∴ ∠ QRX + ∠ QRS + ∠ SRY = 180º
70º + ∠ QRS + 50º = 180º
∠ QRS = 180º – 120º = 60º

Q5 : In the given figure, if AB || CD, ∠ APQ = 50º and ∠ PRD = 127º, find x and y. ∠ APR = ∠ PRD (Alternate interior angles)
50º + y = 127º
y = 127º – 50º
y = 77º
Also, ∠ APQ = ∠ PQR (Alternate interior angles)
50º = x
∴ x = 50º and y = 77º

Q6 : In the given 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.  Let us draw BM ⊥ PQ and CN ⊥ RS.
As PQ || RS,
Therefore, BM || CN
Thus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.
∴∠ 2 = ∠ 3 (Alternate interior angles)
However, ∠ 1 = ∠ 2 and ∠ 3 = ∠ 4 (By laws of reflection)
∴ ∠ 1 = ∠ 2 = ∠ 3 = ∠ 4
Also, ∠ 1 + ∠ 2 = ∠ 3 + ∠ 4
∠ ABC = ∠ DCB
However, these are alternate interior angles.
∴ AB || CD

Exercise 6.3 : Solutions of Questions on Page Number : 107

Q1 : In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠ SPR = 135º and ∠ PQT = 110º, find ∠ PRQ. It is given that,
∠ SPR = 135º and ∠ PQT = 110º
∠ SPR + ∠ QPR = 180º (Linear pair angles)
⇒ 135º + ∠ QPR = 180º
⇒ ∠ QPR = 45º
Also, ∠ PQT + ∠ PQR = 180º (Linear pair angles)
⇒ 110º + ∠ PQR = 180º
⇒ ∠ PQR = 70º
As the sum of all interior angles of a triangle is 180º, therefore, for ΔPQR,
∠ QPR + ∠ PQR + ∠ PRQ = 180º
⇒ 45º + 70º + ∠ PRQ = 180º
⇒ ∠ PRQ = 180º – 115º
⇒ ∠ PRQ = 65º

Q2 : In the given figure, ∠ X = 62º, ∠ XYZ = 54º. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of ΔXYZ, find ∠ OZY and ∠ YOZ. As the sum of all interior angles of a triangle is 180 º, therefore, for ΔXYZ,
∠X + ∠XYZ + ∠XZY = 180 º
62 º + 54 º + ∠XZY = 180 º
∠XZY = 180 º – 116 º
∠XZY = 64 º
∠OZY = = 32 º (OZ is the angle bisector of ∠XZY)
Similarly, ∠OYZ = = 27 º
Using angle sum property for ΔOYZ, we obtain
∠OYZ + ∠YOZ + ∠OZY = 180 º
27 º + ∠YOZ + 32 º = 180 º
∠YOZ = 180 º – 59 º
∠YOZ = 121 º

Q3 : In the given figure, if AB || DE, ∠ BAC = 35º and ∠ CDE = 53º, find ∠ DCE. AB || DE and AE is a transversal.
∠ BAC = ∠ CED (Alternate interior angles)
∴ ∠ CED = 35º
In ΔCDE,
∠ CDE + ∠ CED + ∠ DCE = 180º (Angle sum property of a triangle)
53º + 35º + ∠ DCE = 180º
∠ DCE = 180º – 88º
∠ DCE = 92º

Q4 : In the given figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40º, ∠ RPT = 95º and ∠ TSQ = 75º, find ∠ SQT. Using angle sum property for ΔPRT, we obtain
∠ PRT + ∠ RPT + ∠ PTR = 180º
40º + 95º + ∠ PTR = 180º
∠ PTR = 180º – 135º
∠ PTR = 45º
∠ STQ = ∠ PTR = 45º (Vertically opposite angles)
∠ STQ = 45º
By using angle sum property for ΔSTQ, we obtain
∠ STQ + ∠ SQT + ∠ QST = 180º
45º + ∠ SQT + 75º = 180º
∠ SQT = 180º – 120º
∠ SQT = 60º

Q5 : In the given figure, if PQ ⊥ PS, PQ || SR, ∠ SQR = 28º and ∠ QRT = 65º, then find the values of x and y. It is given that PQ || SR and QR is a transversal line.
∠PQR = ∠QRT (Alternate interior angles)
x + 28 º = 65 º
x = 65 º – 28 º
x = 37 º
By using the angle sum property for ΔSPQ, we obtain
∠SPQ + x + y = 180 º
90 º + 37 º + y = 180 º
y = 180 º – 127 º
y = 53 º
x = 37 º and y = 53 º

Q6 : In the given figure, 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= ∠QPR. Answer :
In ΔQTR, ∠TRS is an exterior angle.
∠QTR + ∠TQR = ∠TRS
∠QTR = ∠TRS – ∠TQR (1)
For ΔPQR, ∠PRS is an external angle.
∠QPR + ∠PQR = ∠PRS
∠QPR + 2∠TQR = 2∠TRS (As QT and RT are angle bisectors)
∠QPR = 2(∠TRS – ∠TQR)
∠QPR = 2∠QTR [By using equation (1)]
∠QTR = ∠QPR