Class IX Science NCERT Solutions For Lines and Angles 6.3

Exercise 6.3 (Page 107)
1.  In the adjoining figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Ans. ∵ TQR is a straight line,
          ∴ ∠TQP + ∠PQR = 180°
[Linear pair]
          ⇒ 110° + ∠PQR = 180°
          ⇒ ∠PQR = 180° – 110° = 70°
          Since, the side QP of ΔPQR is produced to S.
          ∴ Exterior angles so formed is equal to the sum of interior opposite angles.
          ∴ ∠PQR + ∠PRQ = 135°
          ⇒ 70° + ∠PRQ = 135°
[∴ ∠PQR = 70°]
          ⇒ ∠PRQ = 135° – 70°
          ⇒ ∠PRQ = 65°
2.  In the adjoining figure, ∠X = 65°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.
Ans. In ∠XYZ,
          ∠XYZ + ∠YZX + ∠ZXY = 180°
          [∵ Sum of angles of a triangle is 180°]
          But ∠XYZ = 54° and ∠ZXY = 62°
[Given]
          ∴ 54° + ∠YZX + 62° = 180°
          ⇒ ∠YZX = 180° – 54° – 62° = 64°
          ∴ YO and ZO are the bisectors ∠XYZ and ∠XZY respectively,
[Given]
          
          Now, in ΔOYZ, we have:
          ∠YOZ + ∠OYZ + ∠OZY = 180°
[By the angle sum property]
          ⇒∠YOZ + 27° + 32° = 180°
          ⇒ ∠YOZ = 180° – 27° – 32° = 121°
          Thus, ∠OZY = 32° and ∠YOZ = 121°
3.  In the following figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Ans. ∴ AB || DE and AE is a transversal.
[Given]
          ∴ ∠BAC = ∠AED [Interior alternate angles]
          But ∠BAC = 35°
          ∴ ∠AED = 35°
          Now, in ΔCDE, we have
          ∠CDE + ∠DEC + ∠DCE = 180°
[Using the angle sum property]
          ∴ 53° + 35° + ∠DCE = 180°
[∴ ∠DEC = ∠AED = 35° and ∠CDE = 53 (Given)]
          ⇒ ∠DCE = 180° – 53° – 35° = 92°
          Thus, DCE = 92°
4.  In the figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Ans. In ΔPRT
          ∠P + ∠R + ∠PTR = 180°
[By the angle sum property]
          ⇒ 95° + 40° + ∠PTR = 180°
          ⇒ ∠PTR = 180° – 95° – 40° = 45°
          But PQ and RS intersect at T,
          ∴ ∠PTR = ∠QTS
[Vertically opposite angles]
          ∴ ∠QTS = 45°
          Now, in ΔTQS, we have
          ∠TSQ + ∠STQ + ∠SQT = 180°
[By angle sum property]
          ∴ 75° + 45° + ∠SQT = 180°
[∴ ∠TSQ = 75° and ∠STQ = 45°]
          ⇒ ∠SQT = 180° – 75° – 45° = 60°
          Thus, ∠SQT = 60°
5.  In the adjoining figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Ans. In ΔQRS, the side SR is produced to T.
          ∴ Exterior ∠QRT = ∠RQS + ∠RSQ
          But ∠RQS = 28° and ∠QRT = 65°
          ∴ From ∠RQS + ∠RSQ = ∠QRT, we have
          28° + ∠RSQ = 65°
          ⇒ ∠RSQ = 65° – 28° = 37°
          Since, PQ || SR and QS is a transversal.
[Given]
          ∴ ∠PQS = ∠RSQ
[Interior alternate angles]
          ⇒ x = 37°
          Again, PQ ⊥ PS
[Given]
          ∴ ∠P = 90°
          Now, in ΔPQS, we have
          ∠P + ∠PQS + ∠PSQ = 180°
[By angle sum property]
          ⇒ 90° + x + y = 180°
          ⇒ 90° + 37° + y = 180°
[∴ x = 37°]
          ⇒ y = 180° – 90° – 37° = 53°
          Thus, x = 37° and y = 53°
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 
Ans. 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,
          
          Now, In ΔQRT, we have
                  Exterior ∠TRS = ∠TQR + ∠T
...(2)
          From (1) and (2), we have
          

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