NCERT Solutions Class 9 Maths exercise 6.3 of chapter 6-Lines and Angles

In this post, you can find a complete explanation of NCERT Solutions Class 9 Maths exercise 6.3 of chapter 6-Lines and Angles which belongs to Chapter 6 Lines and angles. Before you go through maths exercise 6.3 class 9, you should read and study all theorems of chapter 6 Lines and Angles.

 

Class 9 Maths Chapter 6 Exercise 6.3

Here you will find complete solutions of class 9 maths chapter 6 exercise 6.3 along with the video. You can watch videos related to ex 6.3 class 9 which will help you to understand the topic and questions of class 9 ex 6.3 lines and angles exercise 6.3.Questions of class 9 maths chapter 6 exercise 6.3 maths are asked in the examination based on theorems of chapter 6 Lines and angles.

Class 9 Maths Chapter 6 Exercise 6.3 Question1

 Q1. In Fig, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR=135° and ∠PQT=110°, find ∠PRQ.

Class 9 maths ex 6.3 solutions Question1

We have :

∠QPR+ ∠SPR=180°   (Linear pair)

∴ ∠QPR+135°=180° 

Or ∠QPR=180°-135°=45°

Now ∠TQP=∠QPR+ ∠PRQ (by exterior angle theorem)

⇒110°=45°+∠PRQ 

⇒∠PRQ=110°-45°=65° 

Hence ∠PRQ=65° 

 Class 9 maths ex 6.3 solutions Question2

Q2. In the figure,  ∠X=62°,∠XYZ=54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find∠OZY and ∠YOZ. 

Class 9 maths ex 6.3 solutions Question2

Consider ∆XYZ

∠YXZ+∠XYZ+∠XZY=180° (angle-sum property)

⟹62°+54°∠XZY=180°  (∵∠YXZ=62°,∠XYZ=54°)

⟹∠XZY=180°-62°-54°=64° 

Since YO and ZO are bisects of∠XYZand ∠XZY, therefore

`\angle OYZ=1/2\times\angle XYZ` 

`\angle OYZ=1/2\times54°=27°` 

`\angle OZY=1/2\times\angle XZY` 

`\angle OYZ=1/2\times64°=32°` 

Now in ∆OYZ,we have:

∠YOZ+∠OYZ+∠OZY=180° (Angle sum property)

∠YOZ+27°+32°=180° 

⟹∠YOZ=180°-27°-32°=121° 

Hence, ∠OZY=32° and ∠YOZ=121°

Class 9 Maths Chapter 6 Exercise 6.3 Question 3

Q3. In the figure, if AB||DE, ∠BAC=35°, and ∠CDE=53°, find∠DCE.

Class 9 maths ex 6.3 solutions Question 3

Since AB||DE and transversal AE intersects them at points A and E respectively, therefore

∠DEA=∠BAE      (Alternate angles) 

∠DEC=35°            (∵∠DEA=∠DEC=∠BAE=35°)

In ∆DEC, we have:

∠DCE+∠DEC+∠CDE=180° (angle sum property)

⟹∠DCE+35°+53°=180° 

⟹∠DCE=180°-35°-53°=92° 

Hence, ∠DCE=92° 

Class 9 Maths Chapter 6 Exercise 6.3 Question 4

Q4. In the figure , if lines PQ and RS intersect at point T, such that ∠PRT=40°,∠RPT=95° and ∠TSQ=75°,find ∠SQT.

Class 9 maths ex 6.3 solutions Question 4

In ∆PRT,we have

∠PRT+∠RPT+∠TPR=180° (Angle sum property)

⟹40°+∠RTP +95°=180° 

⟹∠RTP =180°-40°-95°=45° 

Now ∠STQ=∠RTP   (vertically opposite angles)

∴∠STQ=45°  (∵∠RTP=45° proved)

 In ∆TQS,we have:

∠SQT+∠STQ+∠TSQ=180° (Angle sum property)

⟹∠SQT+ 45°+75°=180° (∵∠STQ=45° proved)

⟹∠SQT=180°-45°-75°=60° 

Hence, ∠SQT=60° 

Class 9 Maths Chapter 6 Exercise 6.3 Question 5

Q5. In the figure , if PQ⊥PS,PQ||SR,∠SQR=28° and ∠QRT=65°,then find the  value f x and y.

Class 9 maths ex 6.3 solutions Question 5

Using exterior angle property in ∆SRQ, we have:

∠QRT=∠RQS+∠QSR 

⇒65°=28°+∠QSR  (∵∠QRT=65°,∠RQS=28°)

⇒∠QSR=65°-28°=37°  

Now , PQ ||SR and the transversal PS intersects them at P and S respectively.

∴∠PSR+∠SPQ=180° (sum of the interior angles on the same side of the transversal is 180°)

⟹(∠PSQ+∠QSR)+90°=180° 

⟹y+37°+90°=180° 

⟹y=180°-37°-90°=53° 

In the right triangle SPQ , we have

∠PQS+∠PSQ=90° 

x+53°=90° 

⟹x=90°-53°=37° 

Hence , x=37° and y=53°

Class 9 Maths Chapter 6 Exercise 6.3 Question 6

Q6. In the figure , the side QR of ∆PQR is produced to a point S. if the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR=1/2∠QPR.

Class 9 maths ex 6.3 solutions Question 6

In ∆PQR , we have:

ext.∠PRS=∠P+∠Q 

`\Rightarrow1/2\angle PRS=1/2\angle P+1/2\angle Q` 

`\Rightarrow\ \ \ \angle TRS=1/2\angle P+\angle TQR ………………….(i)`

(∵QT and RT are bisectors of ∠Q and∠PRS,

respectively.) 

∴ ∠Q=2∠TQR and ext.∠PRS=2∠TRS 

In ∆QRT , we have:

ext.∠TRS= ∠TQR+∠T …………(ii)

From (i) and (ii), we get:

`1/2\angle P+\angle TQR=\angle TQR+\angle T `

`\Rightarrow1/2\angle P=\angle T` 

`\Rightarrow\angle QTR=1/2\angle QPR`  

Hence proved

We hope this explanation of   exercise 6.3 class 9 maths will be very useful for you. When you solve the questions of class 9 chapter 6 maths exercise 6.3, you can secure good rank in the examination. We recommend that you should watch the video to better understand the concepts of solution of exercise 6.3 class 9 maths.