NCERT Solutions For Class 9 Maths Chapter 6 Lines And Angles Exercise 6.1 (2025-26)

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\[\angle \text{BOE and reflex }\angle \text{COE}\].

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.

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

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.

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

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 $\angle \text{QYP}$.

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 $\text{Reflex }\angle \text{QYP}={{302}^{\circ }}$.

Conclusion:

NCERT Solutions for Class 9 Ex 6.1 by Vedantu is a helpful resource that simplifies the understanding of basic geometry concepts. This Exercise focuses on fundamental definitions of Line, line-segment, angle, ray and different types of angles, pairs of angles. The important concepts include supplementary angles, complementary angles, vertically opposite angles and properties of angles. Vedantu’s detailed solutions and clear explanations ensure students can easily follow along and strengthen their foundational knowledge in geometry, which is crucial for tackling more complex topics in later exercises.

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