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# Ex 6.3 NCERT Solutions- Lines and Angles Class 9 Notes | EduRev

## Class 9 Mathematics by Full Circle

Created by: Full Circle

## Class 9 : Ex 6.3 NCERT Solutions- Lines and Angles Class 9 Notes | EduRev

The document Ex 6.3 NCERT Solutions- Lines and Angles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.
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Question 1. In the adjoining figure, sides QP and RQ of Î”PQR are produced to points S and T respectively. If âˆ  SPR = 135Âº and âˆ  PQT = 110Âº, find âˆ  PRQ.
Solution:
âˆµ TQR is a straight line,

âˆ´ âˆ  TQP + âˆ  PQR = 180Âº                             [Linear pair]
â‡’ 110Âº + âˆ  PQR = 180Âº
â‡’ âˆ  PQR = 180Âº - 110Âº = 70Âº
Since, the side QP of Î”PQR is produced to S.
âˆ´ Exterior angle so formed is equal to the sum of interior opposite angles.
âˆ´ âˆ  PQR + âˆ  PRQ = 135Âº
â‡’ 70Âº + âˆ  PRQ = 135Âº                    [âˆµ âˆ  PQR = 70Âº]
â‡’ âˆ  PRQ = 135 - 70Âº
â‡’ âˆ  PRQ = 65Âº

Question 2. In the adjoining figure, âˆ  X = 62Âº, âˆ  XYZ = 54Âº. If YO and ZO are the bisectors of âˆ  XYZ and âˆ  XZY respectively of Î”XYZ, find âˆ  OZY and âˆ  YOZ.
Solution:
In âˆ  XYZ, âˆ  XYZ + âˆ  YZX + âˆ  ZXY = 180Âº             [âˆµ Sum of angles of a triangle is 180Âº]

But âˆ  XYZ = 54Âº and âˆ  ZXY = 62Â°                          [Given]
âˆ´ 54Âº + âˆ  YZX + 62Âº = 180Âº
â‡’ âˆ  YZX = 180Â° - 54Â° - 62Â° = 64Â°
âˆµ YO and ZO are the bisectors âˆ  XYZ and âˆ  XZY respectively,             [Given]
âˆ  OYZ =(1/2)âˆ  XYZ =(1/2)(54Âº) = 27Âº
and  âˆ  OZY =(1/2)âˆ  YZX = (1/2) (64Âº) = 32Âº
Now, in Î”OYZ, we have:
âˆ  YOZ + âˆ  OYZ + âˆ  OZY = 180Âº                 [By the angle sum property]
â‡’ âˆ  YOZ + 27Âº + 32Âº = 180Âº
â‡’ âˆ  YOZ = 180Âº - 27Âº - 32Âº = 121Âº
Thus, âˆ  OZY = 32Âº and âˆ  YOZ = 121Âº

Question 3. In the following figure, if AB || DE, âˆ  BAC = 35Âº and âˆ  CDE = 53Âº, find âˆ  DCE.

Solution: âˆµ AB || DE and AE is a transversal.                 [Given ]
âˆ´ âˆ  BAC = âˆ  AED              [Interior alternate angles]
But âˆ  BAC = 35Âº                [Given]
âˆ´ âˆ  AED = 35Âº
Now, in Î”CDE, we have
âˆ  CDE + âˆ  DEC + âˆ  DCE = 180Âº                       [Using the angle sum property]
âˆ´ 53Âº + 35Âº + âˆ  DCE = 180Âº                      [âˆµ âˆ  DEC = âˆ  AED = 35Âº and âˆ  CDE = 53       (Given)]
â‡’ âˆ  DCE = 180Âº - 53Âº - 35Âº = 92Âº Thus, DCE = 92Âº

Question 4. In the figure, if lines PQ and RS intersect at point T, such that âˆ  PRT = 40Âº, âˆ  RPT = 95Âº and âˆ  TSQ = 75Âº, find âˆ  SQT.
Solution:
In Î”PRT âˆ  P + âˆ  R + âˆ  PTR = 180Âº [By the angle sum property]

â‡’ 95Âº + 40Âº + âˆ  PTR = 180Âº
â‡’ âˆ  PTR = 180Âº - 95Âº - 40Âº = 45Âº
But PQ and RS intersect at T,
âˆ´ âˆ  PTR = âˆ  QTS                       [Vertically opposite angles]
âˆ´ âˆ  QTS = 45Âº Now, in Î”TQS, we have
âˆ  TSQ + âˆ  STQ + âˆ  SQT = 180Âº              [By angle sum property]
âˆ´ 75Âº + 45Âº + âˆ  SQT = 180Âº                      [âˆµ âˆ  TSQ = 75Âº and âˆ  STQ = 45Âº]
â‡’ âˆ  SQT = 180Âº - 75Âº - 45Âº = 60Âº
Thus, âˆ  SQT = 60Âº

Question 5. In the adjoining figure, if PQ âŠ¥ PS, PQ ||SR, âˆ  SQR = 28Âº and âˆ  QRT = 65Âº, then find the values of x and y.
Solution:
In DQRS, the side SR is produced to T.

âˆ´ Exterior âˆ  QRT = âˆ  RQS + âˆ  RSQ
But âˆ  RQS = 28Âº and âˆ  QRT = 65Âº
âˆ´ From âˆ  RQS + âˆ  RSQ = âˆ  QRT, we have
28Âº + âˆ  RSQ = 65Âº
â‡’ âˆ  RSQ = 65Âº âˆ  28Âº = 37Âº
Since, PQ || SR and QS is a transversal.                    [Given ]
âˆ´ âˆ  PQS = âˆ  RSQ                            [interior alternate angles]
â‡’ x = 37Âº Again, PQ âŠ¥ PS [Given]
âˆ´ âˆ  P = 90Âº Now, in Î”PQS, we have
âˆ  P + âˆ  PQS + âˆ  PSQ = 180Âº                 [By angle sum property]
â‡’ 90Âº + x + y = 180Âº
â‡’ 90Âº + 37Â° + y = 180Âº                              [âˆµ x = 37Âº]
â‡’ y = 180Âº âˆ  90Âº âˆ  37Âº = 53Âº
Thus, x = 37Âº and y = 53Âº

Question 6. In the adjoining figure, the side QR of Î”PQR is produced to a point S. If the bisectors of âˆ  PQR and âˆ  PRS meet at point T, then prove that âˆ  QTR = (1/2)  âˆ  QPR.

Solution: In Î”PQR, the side QR is produced to S.
âˆ´ Exterior âˆ  PRS = Sum of the interior opposite angles
â‡’ âˆ  PRS = âˆ  P + âˆ  Q
Since QT and RT are bisectors of âˆ  Q and âˆ  PRS respectively,
âˆ´ (1/2)âˆ  PRS =(1/2)âˆ  P +(1/2)âˆ  Q

â‡’ âˆ  TRS =  (1/2) âˆ  P + âˆ  TQR                 â€¦(1)
Now, In Î”QRT, we have
Exterior âˆ  TRS = âˆ  TQR + âˆ  T                    ...(2)
From (1) and (2), we have
âˆ  TQR + (1/2)âˆ  P= âˆ  TQR + âˆ  T
â‡’  (1/2)âˆ  P= âˆ  T
i.e. (1/2) âˆ  QPR = âˆ  QTR
or
âˆ  QTR =
(1/2) âˆ  QPR

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