BSEB Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.3 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

BSEB Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.3 Book Answers

BSEB Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 books are published by the Bihar Board Publishers. These Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 9th Maths Chapter 6 Lines and Angles Ex 6.3 book solutions pdf online from this page.

Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 Textbooks Solutions PDF

Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.3 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 9th Maths Chapter 6 Lines and Angles Ex 6.3 Textbooks. These Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.

Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 Books Solutions

Board	BSEB
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths Chapter 6 Lines and Angles Ex 6.3
Chapters	All
Provider	Hsslive

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BSEB Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 Textbooks Solutions with Answer PDF Download

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BSEB Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.3

Question 1.
In figure, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:
We have,
∠QPR + ∠SPR = 180°
⇒ ∠QPR + 135° = 180°
∠QPR = 180° – 135° = 45°
Now, ∠TQP = ∠QPR + ∠PRQ [By exterior angle theorem]
⇒ 110° = 45° + ∠PRQ
⇒ ∠PRQ = 110° – 45° = 65°
Hence, ∠PRQ = 65°

Question 2.
In figure, ∠X = 62%, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆ XYZ, find ∠OZY and ∠YOZ.

Solution:
Consider ∆ XYZ,
∠YXZ + ∠XYZ + ∠XZY = 180° [Angle-sum property]
⇒ 62° + 54° + ∠X∠y
[∵∠YXZ = 62°, ∠XYZ = 54°]
⇒ ∠ XZY = 180°- 62° – 54° = 64°
Since YO and ZO are Therefore ∠XYZ and ∠XZY.
Therefore

Question 3.
In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:
Since AB || DE and transversal AE intersect them at A and E respectively.
∴ ∠DEA = ∠BAE [Alternative angles]
⇒ ∠DEC = 35° [∵ ∠DEA = ∠DEC and ∠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°.

Question 4.
In figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution:
In ∆ PRT, we have
∠PRT + ∠RTP + ∠TPR = 180° [Angle-sum Property]
⇒ 40° + ∠RTP + 95° = 180°
⇒ ∠RTP = 180° – 40° – 95° = 45°
⇒ ∠STQ = ∠RTP [Vertically opp. 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°.

Question 5.
In figure, if PQ ⊥ PS, P PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution:
Using exterior angle property in ∆ SRQ, we have
∠QRT = ∠RQS + ∠QSR
⇒ 65° = 28° + ∠QSR
[∵ ∠QRT = 65°, ∠RQS = 28°]
⇒ QSR = 65° – 28°
= 37°
Since PQ || SR and the transversal PS intersects them at P and S respectively.
∴ ∠PSR + ∠SPQ = 180° [Sum of consecutive interior angles is 180°]
⇒ (∠PSQ + ∠QSR) + 90° = 180°
⇒ y + 37° + 90° = 180°
⇒ y = 180° – 90° – 37°
= 53°
In the right ∆ SPQ, we have
∠PQS + ∠PSQ = 90°
⇒ x + 53° = 90°
⇒ x = 90° – 53°
= 37°
Hence, x = 37°
and y = 53°

Question 6.
In 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 = 12 ∠QPR.

Solution:
In ∆ PQR, we have
ext. ∠PRS = ∠P + ∠Q
⇒ 12 ext. ∠PRS = 12∠P + 12∠Q
⇒ ∠TRS = 12∠P + ∠TQR … (1)
[∵ 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 … (2)
From (1) and (2), we get
12∠P + ∠TQR + ∠T
⇒ 12∠P = ∠T
⇒ ∠QTR = 12∠QPR

BSEB Textbook Solutions PDF for Class 9th

Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 Textbooks for Exam Preparations

Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 Textbook Solutions can be of great help in your Bihar Board Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 exam preparation. The BSEB STD 9th Maths Chapter 6 Lines and Angles Ex 6.3 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 6 Lines and Angles Ex 6.3 Books State Board syllabus with maximum efficiency.

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