NCERT Solutions for Class 9 Math Chapter 6.1 Lines and Angles

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EXERCISE 6.2 EXERCISE 6.3

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1.    In Fig. 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

solution :

   ∠AOC + ∠COE + ∠BOE = 180°[ Linear pair of angels ]

   But,

   ∠AOC + ∠BOE = 70° [ Given ]

   ∠BOD = 40° [ Given ]

   Therefore,

   ∠AOC + ∠BOE + ∠COE = 180°

   ⇒ 70° + ∠COE = 180°[ ∴ ∠AOC + ∠BOC = 70° (Given) ]

   ⇒ ∠COE = 180° − 70°

   ⇒ ∠COE = 110°

   Now,

   ∠AOC = ∠BOC [ Vertically opposite angels ]

   ⇒ ∠AOC = 40°[ ∴ ∠BOC = 40° (Given) ]

   Similarly,

   ∠AOC + ∠BOE + ∠COE = 180°

   ⇒ 40° + ∠BOE + 110° = 180°

   ⇒ 150° + ∠BOE = 180°

   ⇒ ∠BOE = 180° − 150°

   ⇒ ∠BOE = 30°

   So,

   ∠BOE = 30° and ∠COE = 110°

2.    In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

solution :

   b + a + ∠POY = 180°[ Linear pair of angels ]

   But,

   a : b = 2 : 3[ Given ]

   ∠POY = 90° [ Given ]

   Therefore,

   b + a + ∠POY = 180°

   ⇒ b + a + 90° = 180°

   ⇒ b + a = 180° − 90°

   ⇒ b + a = 90°

   Then,

   a : b = 2 : 3[ Given ]

   `\a = \frac{2} {5} × 90° \ `

   ⇒ `\a = 2 × 18° \ `

   ⇒ `\a = 36° \ `

   Similarly,

   a : b = 2 : 3[ Given ]

   `\b = \frac{3} {5} × 90° \ `

   ⇒ `\b = 3 × 18° \ `

   ⇒ `\b = 54° \ `

   Now,

   ∠MOY = `\c \ `[ Vertically opposite angels ]

   Therefore,

   ∠MOY = a + ∠POY

   ⇒ ∠MOY = 36° + 90°

   ⇒ ∠MOY = 126°

   ∴ c = 126°

   So,

   c = 126°

3.    In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

solution :

   ∠PQR = ∠PRQ [ Given ]

   ∠PQR + ∠PQS = 180°--------(i)[ Linear pair of angles ]

   Similarly,

   ∠PRQ + ∠PRT = 180°--------(ii)[ Linear pair of angles ]

   From (i) and (ii),

   We have,

   ∠PQR + ∠PQS = ∠PRQ + ∠PRT

   But,

   ∠PQR + ∠PRQ [ Given ]

   ∴ ∠PQS = ∠PRT

4.    In Fig. 6.16, if `\x+y = w+z\ `, then prove that AOB is a line.

solution :

   We know,

   The angles around a point are 360°.

   `\x+y+z+w=360°\ `

   But,

   `\x+y = w+z \ `[ Given ]

   So,

   `\(x+y)+(x+y)=360°\ `

   ⇒ `\2(x+y)=360°\ `

   ⇒ `\(x+y)=\frac{360°} {2}\ `

   ⇒ `\(x+y)=180°\ `

   ⇒ `\x+y=180°\ `

   Now,

   I say AOB is a line.

5.    In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
`\∠ROS=\frac{1} {2}(∠QOS-∠POS)\ `.

solution :

   ∠POS + ∠ROS + ∠ROQ = 180°[ Linear pairs of angels ]

   But,

   ∠ROQ = 90°[ Given ]

   Then,

   ∠POS + ∠ROS + ∠ROQ = 180°[ Linear pairs of angels ]

   ∠POS + ∠ROS + 90° = 180°

   ⇒ ∠POS + ∠ROS = 180° − 90°

   ⇒ ∠POS + ∠ROS = 90°--------(i)

   Now,

   ∠QOS = ∠ROQ + ∠ROS

   ⇒ ∠QOS = 90° + ∠ROS

   ⇒ ∠QOS − ∠ROS = 90°--------(ii)

   From (i) and (ii),

   We have,

   ∠POS + ∠ROS = ∠QOS − ∠ROS

   ⇒ ∠POS + ∠ROS + ∠ROS = ∠QOS

   ⇒ ∠POS + 2 ∠ROS = ∠QOS

   ⇒ 2 ∠ROS = ∠QOS − ∠POS

   ⇒`\ ∠ROS = \frac {∠QOS − ∠POS} {2} \ `

   So,

   `\∠ROS = \frac{1} {2} (∠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.

solution :

   ∠XYZ + ∠ZYQ + ∠QYP = 180°[ Linear pairs of angels ]

   ⇒ 64° + ∠ZYQ + ∠QYP = 180°[ ∴ ∠XYZ = 64° (Given) ]

   ⇒ ∠ZYQ + ∠QYP = 180° − 64°

   ⇒ ∠ZYQ + ∠QYP = 116°

   ⇒ ∠ZYQ + ∠ZYQ = 116°[ YQ bisects ∠ZYP So, ∠QYP = ∠ZYQ ]

   ⇒ 2 ∠ZYQ = 116°

   ⇒ ∠ZYQ = `\frac{116°} {2} \ `

   ⇒ ∠ZYQ = 58°

   So,

   ∴ ∠QYP = 58°

   ∴ Reflex ∠QYP = 360° − 58° = 302°

   Now,

   ∠XYQ = ∠XYZ + ∠ZYQ

   ⇒ ∠XYQ = 64° + ∠QYP[ ∠XYZ = 64° (Given) ∠QYP = ∠ZYQ ]

   ⇒ ∠XYQ = 64° + 58° = 122° [ ∠QYP = 58° ]

   Thus,

   ∠XYQ = 122° and reflex ∠QYP = 302°

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EXERCISE 6.2 EXERCISE 6.3