Class IX Science NCERT Solutions For Lines and Angles 6.1

Exercise 6.1 (Page 96)
1.  In the following figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.


Ans. Since AB is a straight line,
          ∴ ∠AOC + ∠COE + ∠EOB = 180°
          or (∠AOC + ∠BOE) + ∠COE = 180°
          or 70° + ∠COE = 180°
[∴ ∠AOC + ∠BOC = 70° (Given)]
          or ∠COE = 180° – 70° = 110°
          ∴ Reflex ∠COE = 360° – 110° = 250°
          ∴ AB and CD intersect at O.
          ∴ ∠COA = ∠BOD
[Vertically opposite angles]
          But ∠BOD = 40°[Given]
          ∴ ∠COA = 40°
          Also ∠AOC + ∠BOE = 70°
          ∴ 40° + ∠BOE = 70°
          or ∠BOE = 70° – 40° = 30°
          Thus, ∠BOE = 30° and reflex ∠COE = 250.
2.  In the following figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Ans. XOY is a straight line.
          ∴ ∠b + ∠a + ∠POY = 180°
          But ∠OPY = 90°                                        [Given]
          ∴ ∠b + ∠a = 180° – 90° = 90°.
          Also a : b = 2 : 3
          
          Since XY and MN intersect at O,
          ∴ ∠c = [∠a + ∠POY]
[Vertically opposite angles]
          or ∠c = 36° + 90° = 126°
          Thus, the required measure of ‘c’
3.  In the following figure, ∠PAR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Ans. ∴ ST is a straight line,
          ∴ ∠PQS + ∠PAR = 180°                                        ...(1)
          Similarly, ∠PRT + ∠PRQ = 180°                                        ...(2)
          From (1) and (2), we have
                ∠PQS + ∠PQR = ∠PRT + ∠PRQ
          But ∠PQR = ∠PRQ                                        [Given]
          ∴ ∠PQS = ∠PRT
4.  In the following figure, if x + y = w + z, then prove that AOB is a line.
Ans. ∴ Sum of all the angles at a point = 360°
          ∴ x + y + z + w = 360°
          or (x + y) + (z + w) = 360°
          But (x + y) = (z + w)                                        [Given]
          ∴ (x + y) + (x + y) = 360°
          or 2(x + y) = 360°
          
          ∴ AOB is a straight line.
5.  In the adjoining figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that 
Ans. ∴ POQ is a straight line.                                                                                                                                                      [Given]
          ∴ ∠POS + ∠ROS + ∠ROQ = 180°
          But OR ⊥ PQ
          ∴ ∠ROQ = 90°
          ∴ ∠POS + ∠ROS + 90° = 180°
          ⇒ ∠POS + ∠ROS = 90° ...(1)
          Now, we have ∠ROS + ∠ROQ = ∠QOS
          ⇒ ∠ROS + 90° = ∠QOS ...(2)
          From (1) and (2), we have
          ∠ROS + [∠POS + ∠ROS] = ∠QOS
          ⇒ 2∠ROS + ∠POS = ∠QOS
          ⇒ 2∠ROS = [∠QOS – ∠POS]
          
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.
Ans. ∴ XYP is a straight line.
          ∠XYZ + ∠ZYQ + ∠QYP = 180°
          ⇒ 64° + ∠ZYQ + ∠QYP = 180°
[∴ YQ, bisects ∠ZYP, so ∠QYP = ∠ZYQ]
          ⇒ 64° + 2∠QYP = 180°
          ⇒ 2∠QYP = 180° – 64° = 116°
          
          ∴ Reflex ∠QYP = 360° – 58° = 302°
          Since ∠XYQ = ∠XYZ + ∠ZYQ
          ⇒ ∠XYQ = 64° + ∠QYP
[∴ ∠XYZ = 64° (Given) and ∠ZYQ = ∠ QYP]
[∴ ∠QYP = 58°]
          ⇒ ∠XYQ = 64° + 58°
                = 122°
          Thus, ∠XYQ = 122° and reflex ∠QYP = 302°

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