Chapter 6: Lines and Angles

EXERCISE 6.1

1. In Fig. 6.13 , lines AB and CD intersect at O . If ∠AOC+∠BOE=70° and ∠BOD=40° , find ∠BOE and reflex ∠COE .    Page : 96  photo

Solution:  Given, ∠AOC+∠BOE=70° and ∠BOD=40° .

  Since , AB is a straight line .

    ∠AOC+∠COE+∠BOE=180° 

⇒∠AOC+∠BOE+∠COE=180°  

 ⇒70°+∠COE=180°

 ⇒∠COE=180°-70°

 ⇒∠COE=110° 

and CD is a straight line .

 ∠COE+∠BOE+∠BOD=180°

 ⇒110°+∠BOE+40°=180°

 ⇒∠BOE+150°=180°

 ⇒∠BOE=180°-150°

⇒∠BOE=30° 

Reflex ∠COE=360°-110°=250° 

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

Solution : Given, ∠POY=90° and a :b=2 :3 .

 Let a=2x and b=3x .

Since,  XY is a straight line .

 2x+3x+∠POY=180°

 ⇒5x+90°=180°

 ⇒5x=180°-90°

 ⇒5x=90°

⇒x=90°5

 ⇒x=18°

   a=2x=2×18°=36°  and  b=3x=3×18°=54°

and MN is a straight line .

    b+c=180°

 ⇒54°+c=180°

 ⇒c=180°-54°

⇒c=126° 

Therefore, the value of c is 126° .

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

Solution:  Given , ∠PQR=∠PRQ , then we prove that ∠PQS=∠PRT .

Proof : Since, ray PQ  stands on line SR .

∠PQS+∠PQR=180° ………. (i)

And ray PR  stands on line QT .

∠PRQ+∠PRT=180° …………. (ii)  

From (i) and (ii) , we get  ∠PQS+∠PQR=∠PRQ+∠PRT 

But , ∠PQR=PRQ

∠PQS+∠PQR=∠PQR+∠PRT 

 ∠PQS=∠PRT  Proved .