Class 9 Maths Chapter 6 Lines and Angles

Exercise 6.1

 Question 1.
In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:
Here,∠AOC and ∠BOD are vertically opposite angles.
∴  ∠AOC =  ∠BOD
⇒ ∠AOC –  40° [BOD=40° (Given)]…(i)
We have, ∠AOC + ∠BOE = 70° (Given)
40° + ∠BOE =   70° [from eq.(1)]
⇒ ∠BOE =    30°
Also, ∠AOC + ∠COE + ∠BOE = 180° (Linear pair axiom)
⇒  40° + ∠COE + 30° = 180°, ∠COE=110°
Now,  ∠COE + reflex ∠COE = 360° (Angles at a point)
110° + reflex ∠COE = 360°
⇒ Reflex ∠COE = 250°

 Question 2.
In figure lines XY and MN intersect at O. If ∠POY = 90° and a: b = 2: 3 find c.

Solution:
We have, ∠POY = 90°
⇒ ∠POY + ∠POX = 180° (Linear pair axiom)
⇒ ∠POX = 90°
⇒ a + b= 90°……….(i)
Also a : b = 2 : 3        (Given)
⇒ Let   a = 2x, b = 3x
Now, from Eq. (i), we get
2x +3x= 90° ⇒ 5x = 90° ⇒ x = 18°
∴ a = 2 x 18° = 36°
and b = 3 x 18°= 54°
Now, ∠MOX + ∠XON = 180° b + c = 180°
⇒ 54° + c = 180°
⇒ c = 126°

 Question 3.
In the given figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution:
∠PQS +∠PQR =180°   [linear Pair] …………(i)
∠PRT + ∠PRQ = 180°  [linear Pair] …………(ii)
From eqs. (i) and (ii), we have
∠PQS + ∠PQR =∠PRT + ∠PRQ
⇒  ∠PQS = ∠PRT   [∵ ∠PQR = ∠PRQ (given)]

 Question 4.
In the given figure, if x + y= w + z, then prove that AOB is a line.
Solution:

We know that, sum of all the angles around a point is 360°.
∴  x + y + z + w = 360°
⇒ (x + y) + (z + w) = 360°
⇒ (x + y) + (x + y)= 360° [ ∵ x + y = z + w, given]
⇒ 2(x + y) = 360°
⇒ (x + y)= ⇒ x + y = 180°
So, AOB is a straight line
Hence proved.

 Question 5.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
∠ROS =   (∠QOS – ∠POS)

Solution:
We have,  ∠POR = ∠ROQ = 90° (∵ Given that, OR is perpendicular to PQ)
∴  ∠POS + ∠ROS = 90°
⇒  ∠ROS = 90°- ∠POS
On adding ZROS both sides, we get
⇒ 2 ∠ROS = 90°-∠POS + ∠ROS
⇒  2∠ROS = (90°  + ∠ROS) – ∠POS
⇒ 2∠ROS = ∠QOS – ∠POS
(∵ ∠QOS = ∠ROQ + ∠ROS = 90° + ∠ROS)
⇒  ∠ROS =  (∠QOS –∠POS) Hence proved.

 Question 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.
Solution:
Here, YQ bisects ∠ZYP.

Hence ∠ZYQ  =  ∠QYP = ∠ZYP  ………..(i)
Given,    ∠XYZ =  64° …………(ii)
∵ ∠XYZ + ∠ZYQ + ∠QYP = 180°  (Linear pair axiom)
⇒ 64° + ∠ZYQ + ∠ZYQ = 180°    [From eq .(i) a and(ii)]
⇒ 2∠ZYQ =  180°-64°
⇒ ∠ZYQ =  x 116° ⇒ ∠ZYQ – 58°
∴  ∠XYQ =  ∠XYZ + ∠ZYQ
⇒ 64° + 58° = 122°
Now, ∠QYZ + reflex ∠QYP = 360°
58° + reflex ∠QYP – 360°    ⇒ reflex ∠QYP = 302

 

Exercise 6.2

 Question 1.
In the given figure, find the values of x and y and then show that  AB||CD.

Solution:
Here, l is a straight line,
So, x + 50° = 180° [by linear pair axiom]
⇒ x = 180° – 50° = 130°
Now, line l and CD intersect each other at a point.
So,          y=130°                       [vertically opposite angles]
Thus,     x = y = 130°               [alternate interior angles]
Hence,   AB || CD                     [by theorem 2] Hence proved.

 Question 2.
In the given figure, if AB || CD, CD || EF and y :z = 3: 7, then find the value of x.

Solution:
Given, AB || CD and CD || EF
∴      AB ||EF
[since, lines which are parallel to the same line, are parallel to each other]
Then,   ∠x= ∠z              [alternate interior angles] … (i)
Also, given that y : z = 3 :7
⇒ y : x =3:7                                 [from eq. (i)] …(ii)
Let ∠y = 3a and ∠x=7a
Now, AB || CD
∠x + ∠y= 180° [since, interior angles are supplementary]
⇒ 7a +3a = 180°⇒ 10a = 180° ⇒ a =  = 18°
Hence, x = 7a = (7 x 18)° = 126°

 Question 3.
In the given figure, if AB || CD, FE ⊥ CD, and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Solution:
Given, AB || CD, EF ⊥ CD and GE is a transversal.
.’.    ∠AGE = ∠GED [alternate interior angles]
∠AGE = 126° [∠AGE =126° given]
∠GEF = ∠GED – ∠FED
⇒ ∠GEF = 126° – 90° [∵  FE ⊥ CD => FED = 90°]
⇒ ∠GEF= 36°
Also, AB is a straight line and EG is a ray on it.
So, ∠FGE + ∠AGE = 180° [By linear pair axiom)
⇒  ∠FGE +126° = 180° [ ∵ ∠AGE= 126°]
⇒  ∠FGE = 180° -126° = 54°
Hence, ∠AGE = 126°, ∠GEF = 36° and ∠FGE =54°.

 Question 4.
In the given figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint: Draw a line parallel to ST through point R.]

Solution:

Given, PQ || ST, ∠PQR = 110° and ∠RST = 130°
Draw a line AB parallel to ST through R.
Now, ST || AB and SR is a transversal.
So, ∠RST + ∠SRB = 180°[since, sum of the interior angles on the same side of the transversal is 180°]
 ⇒ 130° + ∠SRB = 180° ⇒ ∠SRB = 180°-130°
⇒  ∠SRB = 50°    …(i)
Since, PQ || ST and AB || ST, so PQ || AB and then QR is a transversal.
So, ∠PQR + ∠QRA = 180° [since, sum of the interior angles on the same side of the transversal is 180°] ⇒ 110° + ∠QRA = 180° ⇒ ∠QRA = 180° -110°
⇒ ∠QRA=70°  ..(ii)
Now, ARB is a line.
∴  ∠QRA + ∠QRS + ∠SRB = 180° [by linear pair axiom]
⇒ = 70° + ∠QRS + 50° = 180° ⇒ 120° + ∠QRS = 180°
⇒ => ∠QRS = 180° -120° ⇒ ∠QRS = 60°

 Question 5.
In the given figure, if AB||CD, ∠APQ 50° and ∠PRD = 127°, find x and y.
Solution:

Given, ∠APQ = 50° and ∠PRD = 127°
In the given figure,
AB || CD and PQ is a transversal.
∴ ∠APQ + ∠PQC = 180°
[Since, pair of consecutive interior angles on the same side of the transversal is 180°]
⇒ 50° + ∠PQC = 180°  [∴ ∠APQ = 50°, given]
⇒ ∠PQC = 180° -50° = 130°
Now, CD is a straight line and QP is a ray on it
So, ∠PQC + ∠PQR = 180° [by linear pair axiom]
⇒ 130° + ∠PQR = 180°   [∴ ∠PQC = 130°]
⇒  ∠PQR = 180° -130° = 50°
∴  x = 50°    [∴ ∠PQR = x]
Also, AB || CD and PR is a transversal.
So, ∠APR = ∠PRD  [alternate interior angles]
⇒ 50° + y = 127°  [∴ ∠APR = ∠APQ + ∠QPR = 50°+y]
⇒ y = 127° – 50°  ⇒ y=77°
Hence, x = 50° and y = 77°.

 Question 6.
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.

Solution:
Draw BE ⊥ PQ and CF ⊥ L RS.
⇒ BE || CF
Also,  ∠a= ∠b      …(i)  [∵ angle of incidence = angle of reflection]
and    ∠x= ∠y   …(ii) [∵ angle of incidence = angle of reflection]

Since, BE ||CF and BC is transversal.
∴ ∠b= ∠x    [alternate interior angles]
⇒ 2∠b = 2∠x   [multiplying by 2 on both sides]
⇒ ∠b + ∠b = ∠x + ∠x
⇒ ∠a + ∠b = ∠x + ∠y                    [from eqs. (i) and (ii)]
⇒  ∠ABC = ∠DCB
which are alternate interior angles.
Hence, AB || BC Hence proved.

 

Exercise 6.3

 Question 1.
In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively.
If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:
∠RPS + ∠RPQ = 180 °
135° + ∠RPQ = 180°  [∵ ∠RPS = 135° (given)]
⇒  RPQ = 180°-135° = 45°
Now, ∠RPQ + ∠PRQ = ∠PQT  [ext. angle = sum of int. opp. angles]
⇒ 45° + ∠PRQ = 110°
⇒ ∠PRQ = 110° – 45° = 65°

 Question 2.
In the given figure, ∠X= 62°,∠XYZ= 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.
Solution:

Given, ∠X = 62° and ∠XYZ = 54°
In ΔXYZ, ∠XYZ + ∠YXZ + ∠XZY = 180° [since a sum of all the angles of a triangle is 180°]
54°+62°+∠180°⇒ 116°+∠XYZ =18O°
∠XZY=180°-116°
∴  ∠XZY = 64°
Also, given that YO and ZO are the bisectors of ∠XYZ and ∠XZY, respectively.
So, ∠OYZ =  ∠XYZ => ∠OYZ =  x 54° = 27°
and ∠OZY =  ∠XZY=  x 64°= ∠OZY=32°
Now, in ΔOYZ, ∠OYZ + ∠OZY + ∠YOZ = 180° [since, sum of all the angles of a triangle Is 180°]
27° + 32° + ∠YOZ = 180°
59°+∠YOZ = 180° ⇒ ∠YOZ = 180°—59°
∴ ∠YOZ = 121°
Hence,∠OZY = 32° and ∠YOZ = 121°

 Question 3.
In figure,If AB DE, ∠BAC= 35°and ∠CDE= 53°,find ∠DCE.

Solution:
We have, AB || DE
⇒  ∠AED = ∠ BAE (Alternate interior angles)
Now, ∠BAE = ∠BAC
⇒  ∠BAE=35° [ ∵ ∠BAC = 35°Given)]
∠AED = 35°
In Δ DCE,
∴ ∠DCE + ∠CED+ EDC = 180°, (∵ Sum of all angles of triangle is equal to 180°)
⇒ ∠DCE + 35° + 53° – 180°         (∵∠AED = ∠CED = 35°)
⇒  ∠DCE = 180 °-(35° + 53° ) ⇒  ∠DCE = 92°

 Question 4.
In the given figure, if lines PQ and RS intersect at point T such that ∠PRT = 40°, ∠RPT = 95°
and ∠TSQ = 75° then find ∠SQT.

Solution:
Given, ∠PRT = 40°, ∠RPT = 95° and
∠TSQ = 75°
In Δ PRT, ∠PRT + ∠RPT + ∠PTR = 180°          …(i) [since, sum of all the angles of a triangle is 180°]
On putting ∠PRT =40° and ∠RPT = 95° in eq. (i), we get
40° + 95° + ∠PTR = 180°
⇒ 135° + ∠PTR = 180° ⇒ ∠PTR = 180° -135° ⇒ ∠PTR = 45°
Now, ∠PTR    = ∠QTS [vertically opposite angles]
 ∴ ∠QTS – 45°
In ∠TQS, ∠QTS + ∠TSQ + ∠TQS = 180°   … (ii)
[since, sum of all the angles of a triangle is 180°]
On putting ∠QTS = 45° and ∠TSQ = 75° in eq. (ii), we get
45° + 75° + ∠TQS = 180° ⇒120° + ∠TQS = 180°
⇒  ∠TQS    – 180°-120°
 ∴  ∠TQS    =60°
Hence,∠TQS    = ∠SQT = 60°

 Question 5.
In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR – 28° and ∠QRT – 65° then find the values of x and y.

Solution:
For Δ QSR, ∠QRT is an exterior angle
∠QRT = ∠SQR + ∠QSR [∵  exterior angle = sum of interior opposite angles]
⇒ 65° = 28° + ∠QSR [∵ ∠QRT = 65° and ∠SQR = 28°, given]
∠QSR = 65° – 28° ⇒ ∠QSR = 37°
Given PQ || SR and SQ is the transversal which intersects PQ and ST at Q and S, respectively.
∴ ∠QSR = ∠PQS  [alternate interior angles]
⇒  x = 37°
Now, in ΔPQS, ∠SPQ + ∠PQS + ∠PSQ = 180°[since, sum of all the angles of a triangle is 180°]
⇒    90° + 37° + y = 180°   [∵ PQ ⊥ PS ⇒ ∠SPQ = 90°]
⇒    127° + y= 180°⇒ y = 180° -127° = 53°
Hence, x = 37° and y= 53°.

 Question 6.
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.

Solution:
In ΔTQR, ∠QTR + ∠TQR + ∠QRT = 1800 [sum of all the angles of a triangle is 180°]
∠QRT = 1800 – [∠TQR + ∠QRT]
∠PQR + ∠QRP + ∠PRT