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

Class 9 Mathematics by Full Circle

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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 ΔPQR are produced to points S and T respectively. If ∠ SPR = 135º and ∠ PQT = 110º, find ∠ PRQ.
 Solution: 
∵ TQR is a straight line,

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

∴ ∠ TQP + ∠ PQR = 180º                             [Linear pair]
⇒ 110º + ∠ PQR = 180º
⇒ ∠ PQR = 180º - 110º = 70º
Since, the side QP of ΔPQR is produced to S.
∴ Exterior angle so formed is equal to the sum of interior opposite angles.
∴ ∠ PQR + ∠ PRQ = 135º
⇒ 70º + ∠ PRQ = 135º                    [∵ ∠ PQR = 70º]
⇒ ∠ PRQ = 135 - 70º
⇒ ∠ PRQ = 65º


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

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

But ∠ XYZ = 54º and ∠ ZXY = 62°                          [Given]
∴ 54º + ∠ YZX + 62º = 180º
⇒ ∠ YZX = 180° - 54° - 62° = 64°
∵ YO and ZO are the bisectors ∠ XYZ and ∠ XZY respectively,             [Given]
∠ OYZ =(1/2)∠ XYZ =(1/2)(54º) = 27º
and  ∠ OZY =(1/2)∠ YZX = (1/2) (64º) = 32º
Now, in ΔOYZ, we have:
∠ YOZ + ∠ OYZ + ∠ OZY = 180º                 [By the angle sum property]
⇒ ∠ YOZ + 27º + 32º = 180º
⇒ ∠ YOZ = 180º - 27º - 32º = 121º
Thus, ∠ OZY = 32º and ∠ YOZ = 121º


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

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

Solution: ∵ AB || DE and AE is a transversal.                 [Given ]
∴ ∠ BAC = ∠ AED              [Interior alternate angles]
But ∠ BAC = 35º                [Given]
∴ ∠ AED = 35º
Now, in ΔCDE, we have
∠ CDE + ∠ DEC + ∠ DCE = 180º                       [Using the angle sum property]
∴ 53º + 35º + ∠ DCE = 180º                      [∵ ∠ DEC = ∠ AED = 35º and ∠ CDE = 53       (Given)]
⇒ ∠ DCE = 180º - 53º - 35º = 92º Thus, DCE = 92º


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

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

⇒ 95º + 40º + ∠ PTR = 180º
⇒ ∠ PTR = 180º - 95º - 40º = 45º
But PQ and RS intersect at T,
∴ ∠ PTR = ∠ QTS                       [Vertically opposite angles]
∴ ∠ QTS = 45º Now, in ΔTQS, we have
∠ TSQ + ∠ STQ + ∠ SQT = 180º              [By angle sum property]
∴ 75º + 45º + ∠ SQT = 180º                      [∵ ∠ TSQ = 75º and ∠ STQ = 45º]
⇒ ∠ SQT = 180º - 75º - 45º = 60º
Thus, ∠ SQT = 60º


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

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

∴ Exterior ∠ QRT = ∠ RQS + ∠ RSQ
But ∠ RQS = 28º and ∠ QRT = 65º
∴ From ∠ RQS + ∠ RSQ = ∠ QRT, we have
28º + ∠ RSQ = 65º
⇒ ∠ RSQ = 65º ∠ 28º = 37º
Since, PQ || SR and QS is a transversal.                    [Given ]
∴ ∠ PQS = ∠ RSQ                            [interior alternate angles]
⇒ x = 37º Again, PQ ⊥ PS [Given]
∴ ∠ P = 90º Now, in ΔPQS, we have
∠ P + ∠ PQS + ∠ PSQ = 180º                 [By angle sum property]
⇒ 90º + x + y = 180º
⇒ 90º + 37° + y = 180º                              [∵ x = 37º]
⇒ y = 180º ∠ 90º ∠ 37º = 53º
Thus, x = 37º and y = 53º


Question 6. In the adjoining 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 = (1/2)  ∠ QPR.

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

Solution: In ΔPQR, the side QR is produced to S.
∴ Exterior ∠ PRS = Sum of the interior opposite angles
⇒ ∠ PRS = ∠ P + ∠ Q
Since QT and RT are bisectors of ∠ Q and ∠ PRS respectively,
∴ (1/2)∠ PRS =(1/2)∠ P +(1/2)∠ Q

⇒ ∠ TRS =  (1/2) ∠ P + ∠ TQR                 …(1)
Now, In ΔQRT, we have
Exterior ∠ TRS = ∠ TQR + ∠ T                    ...(2)
From (1) and (2), we have
∠ TQR + (1/2)∠ P= ∠ TQR + ∠ T
⇒  (1/2)∠ P= ∠ T
i.e. (1/2) ∠ QPR = ∠ QTR
 or
 ∠ QTR = 
(1/2) ∠ QPR

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