NCERT Solutions for Exercise 6.3 Class 9 Maths Chapter 6

Vishal kumarUpdated on 06 Oct 2023, 09:22 AM IST

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3- Download Free PDF

NCERT Solutions Class 9 Maths Chapter 6 Exercise 6.3 Lines and Angles- We'll explore the creation of lines and angles in 9th class maths exercise 6.3 answers of NCERT Solutions. This chapter explains how lines and angles interact and teaches us key rules and theorems. We may improve our problem-solving skills and see how lines and angles are utilised in maths and in the real world by undertaking the tasks in this chapter.

The class 9 maths chapter 6 exercise 6.3 have been carefully crafted by subject experts using simple language and clear explanations. There are a total of six questions that cover important concepts related to lines and angles. Additionally, students have the option to download the PDF version of these exercise 6.3 class 9 mathssolution for offline use, all without any cost. These resources are designed to make learning easier and more accessible for students, Along with NCERT book Class 9 Maths, chapter 6 exercise 6.3 the following exercises are also present.

NCERT Solutions Class 9 Maths Chapter 6 Exercise 6.3 Lines and Angles

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Access Lines and Angles Class 9 Maths Chapter 6 Exercise: 6.3

Q1 In Fig. 6.39, sides QP and RQ of $\Delta$ PQR are produced to points S and T respectively. If $\angle$ SPR = 135° and $\angle$ PQT = 110°, find $\angle$ PRQ.

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Answer:

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Given,
PQR is a triangle, $\angle$ SPR = $135^0$ , $\angle$ PQT = $110^0$

Now, $\angle$ TQP + $\angle$ PQR = $180^0$ (Linear pair)
So, $\angle$ PQR = $180^0-110^0 = 70^0$

Since the side of QP of the triangle, PQR is produced to S
So, $\angle$ PQR + $\angle$ PRQ = $135^0$ (Exterior angle property of triangle)
$\therefore \angle PRQ = 135^0 - 70^0 = 65^0$

Q2 In Fig. 6.40, $\angle$ X = 62°, $\angle$ XYZ = 54°. If YO and ZO are the bisectors of $\angle$ XYZ and $\angle$ XZY respectively of $\Delta$ XYZ, find $\angle$ OZY and $\angle$ YOZ.

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Answer:

We have
$\angle$ X = $62^0$ , $\angle$ XYZ = $54^0$
YO and ZO bisects the $\angle$ XYZ and $\angle$ XZY
Now, In $\Delta$ XYZ, by using angle sum property
$\angle$ XYZ + $\angle$ YZX + $\angle$ ZXY = $180^0$

So, $\angle$ YZX = $180^0-54^0-62^0$
$\angle$ YZX = $64^0$

and, $\angle$ OYZ = $54^0/2 = 27^0$ also, $\angle$ OZY = $32^0$

Now, in $\Delta$ OYZ
$\angle$ Y + $\angle$ O + $\angle$ Z = $180^0$ [ $\angle$ Y = $32^0$ and $\angle$ Z = $64^0$ ]
So, $\angle$ YOZ = $121^0$

Q3 In Fig. 6.41, if AB $||$ DE, $\angle$ BAC = 35° and $\angle$ CDE = 53°, find $\angle$ DCE.

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Answer:

We have,
AB || DE, $\angle$ BAC = 35° and $\angle$ CDE = 53°

AE is a transversal so, $\angle$ BAC = $\angle$ AED = $35^0$

Now, In $\Delta$ CDE,
$\angle$ CDE + $\angle$ DEC + $\angle$ ECD = $180^0$ (By angle sum property)
Therefore, $\angle$ ECD = $180^0-53^0-35^0$
$\angle ECD = 92^0$

Q4 In Fig. 6.42, if lines PQ and RS intersect at point T, such that $\angle$ PRT = 40°, $\angle$ RPT = 95° and $\angle$ TSQ = 75°, find $\angle$ SQT.

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Answer:

We have,
lines PQ and RS intersect at point T, such that $\angle$ PRT = 40°, $\angle$ RPT = 95° and $\angle$ TSQ = 75°

In $\Delta$ PRT, by using angle sum property
$\angle$ PRT + $\angle$ PTR + $\angle$ TPR = $180^0$
So, $\angle$ PTR = $180^0 -95^0-40^0$
$\Rightarrow \angle PTR = 45^0$

Since lines, PQ and RS intersect at point T
therefore, $\angle$ PTR = $\angle$ QTS (Vertically opposite angles)
$\angle$ QTS = $45^0$

Now, in $\Delta$ QTS,
By using angle sum property
$\angle$ TSQ + $\angle$ STQ + $\angle$ SQT = $180^0$
So, $\angle$ SQT = $180^0-45^0-75^0$
$\therefore \angle SQT = 60^0$

Q5 In Fig. 6.43, if PQ $\bot$ PS, PQ $||$ SR, $\angle$ SQR = 28° and $\angle$ QRT = 65°, then find the values of x and y .

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Answer:

We have,
PQ $\bot$ PS, PQ || SR, $\angle$ SQR = 28° and $\angle$ QRT = 65°
Now, In $\Delta$ QRS, the side SR produced to T and PQ || RS
therefore, $\angle$ QRT = $28^0+x$ = $65^0$
So, $x = 37^0$

Also, $\angle$ QRT = $\angle$ RSQ + $\angle$ SQR (By exterior angle property of a triangle)
Therefore, $\angle$ RSQ = $\angle$ QRT - $\angle$ SQR
$\angle RSQ = 65^0-28^0 = 37^0$

Now, in $\Delta$ PQS,
$\angle$ P + $\angle$ PQS + $\angle$ PSQ = $180^0$
$y=180^0-90^0-37^0$
$y=52^0$

Q6 In Fig. 6.44, the side QR of $\Delta$ PQR is produced to a point S. If the bisectors of $\angle$ PQR and $\angle$ PRS meet at point T, then prove that $\angle$ QTR = $\frac{1}{2}$ $\angle$ QPR.

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Answer:

We have,
$\Delta$ PQR is produced to a point S and bisectors of $\angle$ PQR and $\angle$ PRS meet at point T,
By exterior angle sum property,
$\angle$ PRS = $\angle$ P + $\angle$ PQR
Now, $\frac{1}{2}\angle PRS =\frac{1}{2}\angle P + \frac{1}{2} \angle PQR$
$\Rightarrow \angle TRS = \frac{1}{2}\angle P + \angle TQR$ ................(i)

Since QT and QR are the bisectors of $\angle$ PQR and $\angle$ PRS respectively.

Now, in $\Delta$ QRT,
$\angle TRS = \angle T + \angle TQR$ ..............(ii)

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

$\frac{1}{2}\angle P= \angle T$
$\Rightarrow \frac{1}{2}\angle QPR= \angle QTR$
Hence proved.

The angle sum property in triangles is proved using the theorems and axioms of parallel lines in NCERT Solutions for Class 9 Maths chapter 6 exercise 6.3 Lines and Angles. The triangle angle sum property states that the sum of a triangle's angles is 180 degrees.

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If a triangle side is formed, the exterior angle formed is equal to the sum of the two interior opposite angles. i.e. An exterior angle of a triangle = sum of opposite internal angles.

We can also prove this theorem very easily just by equating two equations.

We know that the sum of all interior angles of triangles is 180. So, from the above figure.

(1)

And we also know that the angle of a straight line is 180.

(2)

We will now equate equations 1 and 2 because the right-hand side of both equations is 180.

Hence,

We subtract angle 3 from both sides. So,

Hence, the exterior angle is equal to the sum of two opposite interior angles.

If a triangle side is formed, the exterior angle formed is equal to the sum of the two interior opposite angles.

Benefits of NCERT Solutions for Class 9 Maths Exercise 6.3

From NCERT syllabus Class 9 Maths chapter 6 exercise 6.3 we learn new formulas to calculate the mean and class mark.
Exercise 6.3 Class 9 Maths, is based on the angle sum property of a triangle
Understanding the concepts from Class 9 Maths chapter 6 exercise 6.3 will allow us to understand the concepts related to higher standard questions related to angle sum property of a triangle

Key Features of Class 9 Maths Chapter 6 Exercise 6.3

Comprehensive Coverage: The exercise 6.3 class 9 maths covers various concepts related to lines and angles, ensuring a thorough understanding of this fundamental geometry topic.
Expert-Crafted Solutions: The class 9 maths ex 6.3 solutions provided are written by subject experts, offering clear and concise explanations to help students grasp the concepts easily.
Easy-to-Understand Language: The 9th class maths exercise 6.3 answers are presented in simple and easy-to-understand language, making it accessible to students of all levels.
Variety of Questions: The class 9 ex 6.3 consists of a total of six questions, offering a variety of problem types to test students' understanding and problem-solving skills.
Conceptual Clarity: By solving these ex 6.3 class 9 questions, students can gain a deeper understanding of the relationships between lines and angles, helping them build a strong foundation in geometry.
Offline Accessibility: Students have the option to download the PDF version of the solutions, allowing them to access and use the resources offline without any additional cost.

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