Exercise 6.1 Class 9 maths 6. Lines and Angles - ncert solutions - Toppers Study

Exercise 6.1

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Q1.  In Fig. 6.13, lines AB and CD intersect at O. If  ∠AOC + ∠ BOE = 70° and ∠BOD = 40° find ∠BOE and reflex ∠COE.

Solution:

 ∠BOD = 40°

∠AOC  = ∠BOD (Vertically opposite Angle)

∠AOC = 40°

∠AOC  + ∠ BOE = 70° (Given)

∠BOE = 70°

∠BOE = 70° - 40°

∠BOE = 30°

AOB is straight line

∠AOC +  ∠COE +∠BOE = 180° (linear pair)

⇒ 70° + ∠COE = 180°

⇒ ∠COE = 180° - 70°

⇒ ∠COE = 110°

Reflex ∠COE = 360 - 110°

                      = 250°

Q2. In Fig. 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Solution:

∠POY=90° (given) 

Let  ∠a and ∠b = 2x and 3x

XOY is a straight line

∠a + ∠b + ∠POY = 180°

⇒2x + 3x + 90°= 180°

⇒5x  = 180° ­­- 90°

⇒5x = 90°

⇒x = 90°/5

⇒x = 18°

Now ∠a = 2 x 18°

             = 36°

         ∠b =3 x 18°

               = 54°

MON is a straight line   

∠b + ∠c = 180°(linear pair)

∠54° + ∠c = 180°

⇒∠c = 180°- 54°

          =126°

Q3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT 

Solution : 

Given : ∠PQR = ∠PRQ

To prove : ∠PQS = ∠PRT

Proof : 

∠PQS + ∠PQR = 180°  .................. (1)  Linear pair

∠PRT + ∠PRQ = 180°  .................. (2)  Linear pair

From equation (1) and (2) 

       ∠PQS + ∠PQR = ∠PRT + ∠PRQ 

Or,  ∠PQS + ∠PQR = ∠PRT + ∠PQR    (∠PQR = ∠PRQ given) 

Or, ∠PQS = ∠PRT Proved 

Q4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line. 

Solution: 

Given : x + y = w + z 

To prove : AOB is a line.

Proof :  

We know that;

x + y + w + z = 360^०

(Angle Sustained on centre) 

x + y + x + y =  360^० (x + y = w + z given) 

2x + 2y = 360^० 

2 (x + y) = 360^० 

x + y = 180^० (linear pair) 

Therefore, AOB is a line   

Hence, Proved

Q5. 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

Solution:

Given:

POQ is a straight line. OR ⊥ PQ and OS is

another ray lying between rays OP and OR.

To prove: 

Proof: OR ⊥ PQ  (given)

∴ ∠QOR = 90^०  …………… (1)

 POQ is straight line

∴ ∠POR + ∠QOR = 180^० (linear pair)

⇒ ∠POR + 90^० = 180^०

⇒ ∠POR = 180^०– 90^०

⇒ ∠POR = 90^०…………… (2)

Now,  ∠ROS + ∠QOR = ∠QOS

Or,       ∠ROS = ∠QOS – ∠QOR  ……………. (3)

Again, ∠ROS + ∠POS = ∠POR

Or,       ∠ROS = ∠POR – ∠POS  ……………. (4)

Adding equation (1) and (2)

∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠POR – ∠POS 

2 ∠ROS = ∠QOS – 90^०+ 90^०– ∠POS

 2 ∠ROS = (∠QOS – ∠POS)

Hence Proved  

Q6.  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:  

Given: ∠ XYZ = 64°and XY is produced to point P. YQ bisects ∠ ZYP.

To Find: ∠XYQ and reflex ∠QYP. 

YQ bisects ∠ZYP 

∴ ∠ZYQ = ∠QYP   ................. (1) 

 ∵ XY is produced to point P.

∴ PX is a straight line. 

Now, ∠ XYZ + ∠ZYQ + ∠QYP = 180° (linear pair) 

Or,     64° + ∠ZYQ + ∠QYP = 180°

⇒      ∠ZYQ + ∠QYP = 180° - 64°

⇒      ∠ZYQ + ∠ZYQ = 116°                 [Using equation (1) ]

⇒      2∠ZYQ = 116°

⇒      ∠ZYQ = 116°/2

⇒      ∠ZYQ = 58°

∠ZYQ = ∠QYP = 58°

∠XYQ = ∠XYZ + ∠ZYQ 

           = 64° + 58°

           = 122°

∵ ∠QYP = 58°

∴ Reflex ∠QYP = 360° - 58°

                        = 302°

∠XYQ  = 122°, Reflex ∠QYP = 302°