NCERT Solutions for Class 9 Maths Chapter 6: Lines and Angles - Exercise 6.1

EXERCISE NO: 6.1

1. In the given figure, lines AB and CD intersect at O. If $\angle \text{AOC+}\angle \text{BOE}={{70}^{\circ }}$ and $\angle \text{BOD}=\text{4}{{\text{0}}^{\circ }}$ find\[\angle \text{BOE and reflex }\angle \text{COE}\].

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Ans:

Given: The line $\text{AB}$ and $\text{CD}$ intersect at $\text{O}$.

\[\angle \text{AOC+}\angle \text{BOE}={{70}^{\circ }}\]

$\angle \text{BOD}=\text{4}{{\text{0}}^{\circ }}$.

And \[\text{AB}\] is a straight line; rays on the line are \[\text{OC}\] and \[\text{OE}\]. 

If, $\angle \text{AOC + }\angle \text{COE + }\angle \text{BOE = 18}{{\text{0}}^{\text{o}}}$ 

$\Rightarrow \left( \angle \text{AOC + }\angle \text{BOE} \right)\text{+}\angle \text{COE = 18}{{\text{0}}^{\text{o}}}$ 

$\Rightarrow \text{7}{{\text{0}}^{\text{o}}}\text{+}\angle \text{COE = 18}{{\text{0}}^{\text{o}}}$ 

\[\Rightarrow \angle \text{COE = 18}{{\text{0}}^{\text{o}}}-7{{\text{0}}^{\text{o}}}=\text{11}{{\text{0}}^{\text{o}}}\] 

\[\text{Reflex }\angle \text{COE = 36}{{\text{0}}^{\text{o}}}\text{-11}{{\text{0}}^{\text{o}}}\text{=25}{{\text{0}}^{\text{o}}}\] 

\[\therefore \text{Reflex }\angle \text{COE =25}{{\text{0}}^{\text{o}}}\]

Then \[CD\] is a straight line; rays on the line are \[OE\] and \[OB\].

And $\angle \text{COE + }\angle \text{BOE + }\angle \text{BOD = 18}{{\text{0}}^{\text{o}}}$ 

$\Rightarrow \text{11}{{\text{0}}^{\text{o}}}\text{+}\angle \text{BOE+ 4}{{\text{0}}^{\text{o}}}\text{= 18}{{\text{0}}^{\text{o}}}$ 

\[\Rightarrow \angle \text{BOE = 18}{{\text{0}}^{\text{o}}}\text{-15}{{\text{0}}^{\text{o}}}\text{=3}{{\text{0}}^{\text{o}}}\] 

Hence, \[\angle \text{BOE = 3}{{\text{0}}^{\text{o}}}\] and \[\text{Reflex }\angle \text{COE = 25}{{\text{0}}^{\text{o}}}\].

2. In the given figure, lines XY and MN intersect at O. If $\angle \text{POY=9}{{\text{0}}^{\text{o}}}$ and $\text{a:b}\,\text{=}\,\text{2:3}$ , find c.

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Ans:

Given: The line $\text{XY}$ and $\text{MN}$ intersect at $\text{O}$.

Then, $\angle \text{POY=9}{{\text{0}}^{\text{o}}}$ and $\text{a:b}\,\text{=}\,\text{2:3}$.

Let us assume the common ratio between a and b be x.

$\therefore \text{a = 2x,}$ and $\text{b = 3x}$ 

And XY is a straight line, rays on the line are \[OM\] and \[OP\].

$\because \angle \text{XOM + }\angle \text{MOP + }\angle \text{POY = 18}{{\text{0}}^{\text{o}}}$ 

$\text{b + a + }\angle \text{POY = 18}{{\text{0}}^{\text{o}}}$ 

\[\text{3x + 2x + 9}{{\text{0}}^{\text{o}}}\text{ = 18}{{\text{0}}^{\text{o}}}\] 

\[\text{5x = 9}{{\text{0}}^{\text{o}}}\] 

\[\therefore \text{x = 1}{{\text{8}}^{\text{o}}}\] 

Then,

$\text{a = 2x = 2 }\!\!\times\!\!\text{ 18 = 3}{{\text{6}}^{\text{o}}}$ 

$\text{b = 3x = 3 }\!\!\times\!\!\text{ 18 = 5}{{\text{4}}^{\text{o}}}$ 

Now, \[MN\] is a straight line. Ray on the line is \[OX\].

The $\text{Linear Pair}$ is,

$\text{b + c = 18}{{\text{0}}^{\text{o}}}$

${{54}^{\text{o}}}\text{+ c = 18}{{\text{0}}^{\text{o}}}$

\[\text{c = 18}{{\text{0}}^{\text{o}}}-{{54}^{\text{o}}}\]

\[={{126}^{\text{o}}}\]

\[\therefore \text{c}={{126}^{\text{o}}}\]

3. In the given figure, $\angle \text{PQR = }\angle \text{PRQ}$, then prove that $\angle \text{PQS = }\angle \text{PRT}$.

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Ans: 

In the given figure, ST is a straight line and ray QP stands on it.

The $\text{Linear Pair}$ is,      

$\therefore \angle \text{PQS + }\angle \text{PQR = 18}{{\text{0}}^{\text{o}}}$ 

$\angle \text{PQR = 18}{{\text{0}}^{\text{o}}}\text{-}\,\angle \text{PQS}$        …… (1)

$\therefore \angle \text{PRT + }\angle \text{PRQ = 18}{{\text{0}}^{\text{o}}}$

\[\angle \text{PRQ = 18}{{\text{0}}^{\text{o}}}\text{-}\,\,\angle \text{PRT }\]      …… (2)

It is denoted as $\angle \text{PQR = }\angle \text{PRQ}$ . 

Now, equating equations (1) and (2). We get, 

$\text{18}{{\text{0}}^{\text{o}}}\text{-}\,\,\angle \text{PQS = 18}{{\text{0}}^{\text{o}}}\text{- }\angle \text{PRT}$

$\therefore \angle \text{PQS = }\angle \text{PRT}$

Hence proved.

4. In the given figure, if $\text{x + y = w + z}$ then prove that AOB is a line.

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Ans:

Given: $\text{x + y = w + z}$.

A complete angle becomes,

\[\text{x + y + z + w =36}{{\text{0}}^{\text{o}}}\]  

Then,

$\text{x + y + x + y  = 36}{{\text{0}}^{\circ }}$ 

$\text{2}\left( \text{x + y} \right)\text{ = 36}{{\text{0}}^{\text{o}}}$ 

$\therefore \text{x + y}=18{{\text{0}}^{\text{o}}}$ 

Since x and y form a linear pair, therefore, AOB is a line. 

Hence proved.

5. In the given figure, \[~\mathbf{POQ}\] is a line. Ray \[\mathbf{OR}\] is perpendicular to line \[\mathbf{PQ}\].\[\mathbf{OS}\] is another ray lying between rays \[\mathbf{OP}\] and \[\mathbf{OR}\]. Prove that$\angle \text{ROS = }\dfrac{\text{1}}{\text{2}}\left( \angle \text{QOS - }\angle \text{POS} \right)$

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Ans:

Given: \[\text{OR}\bot \text{PQ}\] 

$\angle \text{POR = 9}{{\text{0}}^{\text{o}}}$ 

$\angle \text{POS +}\angle \text{SOR= 9}{{\text{0}}^{\text{o}}}$ 

$\angle \text{ROS = 9}{{\text{0}}^{\text{o}}}\text{+}\angle \text{POS}$    …… (1)

$\because \text{OR}\bot \text{PQ}$. Then,

$\angle \text{QOR = 9}{{\text{0}}^{\text{o}}}$ 

$\angle \text{QOS - }\angle \text{ROS = 9}{{\text{0}}^{\text{o}}}$ 

\[\angle \text{ROS = }\angle \text{QOS - 9}{{\text{0}}^{\text{o}}}\]   …… (2)

Add equations (1) and (2), we obtain

\[2\angle \text{ROS = }\angle \text{QOS - }\angle \text{POS}\]

\[\therefore \angle \text{ROS = }\dfrac{1}{2}\left( \angle \text{QOS - }\angle \text{POS} \right)\]

Hence proved.

6. It is given that $\angle \text{XYZ = 6}{{\text{4}}^{\text{o}}}$ and \[\mathbf{XY}\] is produced to point \[\mathbf{P}\]. Draw a figure from the given information. If ray \[\mathbf{YQ}\] bisects $\angle \text{ZYP}$ , find $\angle \text{XYQ}$ and reflex $\angle \text{QYP}$.

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Ans:

Given: the line \[YQ\] bisects $\angle \text{PYZ}$.

Hence, $\angle \text{QYP = }\angle \text{ZYQ}$

It can be observed that \[PX\] is a line. Rays on the line are \[YQ\] and \[YZ\].

Then, $\angle \text{XYZ + }\angle \text{ZYQ + }\angle \text{QYP = 18}{{\text{0}}^{\text{o}}}$

${{64}^{\circ }}+2\angle \text{QYP}={{180}^{\circ }}$ 

\[2\angle \text{QYP}={{180}^{\circ }}\text{- }{{64}^{\circ }}={{116}^{\circ }}\]

\[\angle \text{QYP}={{58}^{\circ }}\] 

$\text{Also, }\angle \text{ZYQ = }\angle \text{QYP = 5}{{\text{8}}^{\text{o}}}\text{ }$ 

$\text{Reflex }\angle \text{QYP = 36}{{\text{0}}^{\text{o}}}\text{- 5}{{\text{8}}^{\text{o}}}={{302}^{\text{o}}}$ 

$\angle \text{XYQ = }\angle \text{XYZ + }\angle \text{ZYQ }$

\[={{64}^{\circ }}+{{58}^{\circ }}={{122}^{\circ }}\] 

$\therefore \angle \text{XYQ}={{122}^{\circ }}$ and $\text{Reflex }\angle \text{QYP}={{302}^{\circ }}$.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles (Ex 6.1) Exercise 6.1

Opting for the NCERT solutions for Ex 6.1 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.1 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Chapter wise NCERT Solutions for Class 9 Maths

• Chapter 1 - Number Systems

• Chapter 2 - Polynomials

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introduction to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelograms and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's Formula

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Class 9 Maths Chapter 6 Includes:

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 6 Exercise 6.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 6 Exercise 6.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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