NCERT Solutions For Class 9 Maths Chapter 6 Lines and Angles (2025-26)

1. In the given figure, lines AB and CD intersect at O. if \[\angle \text{AOC+}\angle \text{BOE=7}{{\text{0}}^{\text{o}}}\] and $\angle \text{BOD=4}{{\text{0}}^{\text{o}}}$find $\angle \text{BOE}$ and reflex $\angle \text{COE}$

Ans:

AB is a straight line, OC and OE are rays from O.

We know that a straight line covers ${{180}^{\circ }}$ 

\[\text{    }\Rightarrow \angle \text{AOC+}\angle \text{COE+}\angle \text{BOE=18}{{\text{0}}^{\circ }}\] 

By clubbing $\angle \text{AOC and }\angle \text{BOE}$ together we can rewrite the above equation as

\[\Rightarrow \left( \angle \text{AOC+}\angle \text{BOE} \right)+\angle \text{COE=18}{{\text{0}}^{\circ }}\] 

Putting \[\text{         }\angle \text{AOC+}\angle \text{BOE=7}{{\text{0}}^{\circ }}\]

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

$\text{                    }\Rightarrow \angle \text{COE=18}{{\text{0}}^{\circ }}\text{-}{{70}^{\circ }}$

$\text{                    }\Rightarrow \angle \text{COE=11}{{\text{0}}^{\circ }}$

Hence  reflex $\text{    }\angle \text{COE=36}{{\text{0}}^{\circ }}\text{-11}{{\text{0}}^{\circ }}$

$\text{                 reflex}\angle \text{COE=25}{{\text{0}}^{\circ }}\text{       }$

Similarly CD is a straight line, OB and OE are rays from O.

We know that a straight line covers ${{180}^{\circ }}$ 

\[\text{    }\Rightarrow \angle B\text{OD+}\angle \text{COE+}\angle \text{BOE=18}{{\text{0}}^{\circ }}\]

\[\text{    }\Rightarrow \text{      }{{40}^{\circ }}\text{+}{{110}^{\circ }}\text{+}\angle \text{BOE=18}{{\text{0}}^{\circ }}\]

\[\text{    }\Rightarrow \text{                       }\angle \text{BOE=18}{{\text{0}}^{\circ }}\text{-}\left( \text{4}{{\text{0}}^{\circ }}\text{+11}{{\text{0}}^{\circ }} \right)\]

\[\text{    }\Rightarrow \text{                       }\angle \text{BOE=18}{{\text{0}}^{\circ }}\text{-}{{150}^{\circ }}\]

\[\text{    }\Rightarrow \text{                       }\angle \text{BOE=}{{30}^{\circ }}\]

Hence \[\angle \text{BOE=}{{30}^{\circ }}\] and reflex $\angle \text{COE=25}{{\text{0}}^{\circ }}$

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

Ans:

Let the common ratio between a and b be x.

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

XY is a straight line, OM and OP are rays from O.

We know that a straight line covers ${{180}^{\circ }}$ 

\[\angle \text{XOM +}\angle \text{MOP +}\angle \text{POY = 18}{{\text{0}}^{\text{o}}}\] 

Putting values for \[\angle \text{XOM=b and}\angle \text{MOP=a}\]

\[\Rightarrow \text{              b + a +}\angle \text{POY = 18}{{\text{0}}^{\text{o}}}\] 

\[\Rightarrow \text{          3x + 2x +}\angle \text{POY = 18}{{\text{0}}^{\text{o}}}\]

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

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

\[\therefore \text{                                  a = 2x}\]

\[\Rightarrow \text{                                a = 2}\times \text{1}{{\text{8}}^{\circ }}\]

\[\text{                                       = 3}{{\text{6}}^{\circ }}\]

\[\therefore \text{                                  b = 3x}\]

\[\Rightarrow \text{                                b = 3}\times \text{1}{{\text{8}}^{\circ }}\]

\[\text{                                       = 5}{{\text{4}}^{\circ }}\]

Similarly MN is a straight line, OX is a ray from O 

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

\[\text{5}{{\text{4}}^{\text{o}}}\text{ + c = 18}{{\text{0}}^{\text{o}}}\] 

\[\text{          }c=\text{ }{{180}^{\circ }}\text{ }-\text{ }{{54}^{\circ }}\] 

\[\text{          }c=\text{ 12}{{\text{6}}^{\circ }}\] 

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

Ans:

ST is a straight line, QP is a line segment from Q in ST to any point P

by Linear Pair property 

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

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

Similarly

$\angle PRT+\angle PRQ={{180}^{\circ }}$ 

$\Rightarrow \text{          }\angle \text{PRQ=18}{{\text{0}}^{\text{o}}}\text{-}\angle \text{PRT}$ ......($2$ )

Now in the question it is given that $\angle \text{PQR=}\angle \text{PRQ}$

Therefore on equating equation ($1$ ) and ($2$ ) we get

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

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

Hence proved

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

Ans:

It can be observed that,

Since there are ${{360}^{\circ }}$ around a point therefore we can write 

$x+y+z+w={{360}^{\circ }}$ 

It is given that,

$\text{x+y=w+z}$

Therefore writing$\text{x+y}$in place of $\text{w+z}$so that we can eliminate w and z, we get

$x+y+x+y={{360}^{\circ }}$ 

$2\left( x+y \right)={{360}^{\circ }}$ 

$x+y={{180}^{\circ }}$

Since x and y form a linear pair, hence we can say that AOB is a line.

Hence proved

5. In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

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

Ans: 

Since $\text{OR}\bot \text{PQ}$  therefore 

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

$\angle POS+ROS={{90}^{\circ }}$ 

$\Rightarrow \text{         }\angle \text{ROS=9}{{\text{0}}^{\text{o}}}\text{-}\angle \text{POS}$ ......   ($1$ )

Similarly $\angle \text{QOR=9}{{\text{0}}^{\text{o}}}$ (Since $\text{OR }\!\!\hat{\ }\!\!\text{ PQ}$)

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

$\Rightarrow \text{         }\angle \text{ROS=}\angle \text{QOS-9}{{\text{0}}^{\text{o}}}$......($2$)

We can clearly see that on adding equation ($1$ ) and ($2$) ${{90}^{\circ }}$ get canceled out

$\Rightarrow \text{2}\angle \text{ROS=}\angle \text{QOS-}\angle \text{POS}$ 

Which can easily be written as

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

Hence proved

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

Ans:

It is given that line YQ bisects $\angle \text{ZYP}$.

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

It can easily be understood that PX is a line, YQ and YZ being rays standing on it.

\[\angle \text{XYZ+}\angle \text{ZYQ+}\angle \text{QYP=18}{{\text{0}}^{\text{o}}}\] 

From above relation $\angle \text{QYP=}\angle \text{ZYQ}$ we can write

$\text{6}{{\text{4}}^{\text{o}}}\text{+2}\angle \text{QYP=18}{{\text{0}}^{\text{o}}}$ 

$\Rightarrow \text{     2}\angle \text{QYP=18}{{\text{0}}^{\text{o}}}\text{-6}{{\text{4}}^{\text{o}}}$ 

$\Rightarrow \text{        }\angle \text{QYP=5}{{\text{8}}^{\circ }}$

Therefore $\angle \text{ZYQ=5}{{\text{8}}^{\circ }}$

Also Reflex $\angle \text{QYP=30}{{\text{2}}^{\circ }}$

Now we can write $\angle \text{XYQ}$ as below 

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

${{64}^{\circ }}+{{58}^{\circ }}={{122}^{\circ }}$ 

Therefore we found $\angle \text{XYQ=12}{{\text{2}}^{\circ }}$ and so the Reflex $\angle \text{QYP=30}{{\text{2}}^{\circ }}$

Exercise-6.2

1. In the given figure, if AB \[\parallel \] CD, CD \[\parallel \] EF and y: z = 3:7, find x.

Ans:

It is given that AB\[\parallel \] CD and CD \[\parallel \] EF

$\therefore $ AB\[\parallel \]CD\[\parallel \] EF (Lines parallel to other fixed line are parallel to each other)

It can easily be understood that

$\text{x=z}$  (since alternate interior angles are equal) ...... ($1$)

It is given that y: z = 3: 7

Without any loss of generality we can let $\text{y=3a}$and $\text{z=7a}$  

Also, $\text{x+y=18}{{\text{0}}^{\text{o}}}$ (Co-interior angles together sum up to ${{180}^{\circ }}$)

From equation ($1$) we can write z in place of x as shown

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

It can further written as shown

$7a+3a={{180}^{\circ }}$ 

$\Rightarrow \text{     10a=18}{{\text{0}}^{\text{o}}}$  

$\Rightarrow \text{         a=1}{{\text{8}}^{\text{o}}}$

$\therefore \text{          x=7 }\!\!\times\!\!\text{ 1}{{\text{8}}^{\text{o}}}$ 

$\therefore \text{          x=12}{{\text{6}}^{\text{o}}}$ 

2. In the given figure, If AB\[\parallel \]CD, EF$\bot $ CD and$\angle \text{GED=12}{{\text{6}}^{\text{o}}}$ , find$\angle \text{AGE}$ , $\angle \text{GEF}$and $\angle \text{FGE}$.

Ans:

We are given that,

AB $\parallel $ CD and EF$\bot $ CD and

$\text{             }\angle \text{GED=12}{{\text{6}}^{\text{o}}}$

Which can be written as 

$\Rightarrow \angle \text{GEF+}\angle \text{FED=12}{{\text{6}}^{\text{o}}}$ 

$\Rightarrow \text{     }\angle \text{GEF+9}{{\text{0}}^{\text{o}}}\text{=12}{{\text{6}}^{\text{o}}}$ 

Hence we can obtain $\text{GEF}$as shown below

$\Rightarrow \text{             }\angle \text{GEF=12}{{\text{6}}^{\text{o}}}\text{-9}{{\text{0}}^{\text{o}}}$

$\Rightarrow \text{             }\angle \text{GEF=3}{{\text{6}}^{\circ }}$

$\angle \text{AGE=}\angle \text{GED=12}{{\text{6}}^{\text{o}}}$ (∠AGE and ∠GED are alternate interior angles)

But 

$\angle \text{AGE+}\angle \text{FGE=18}{{\text{0}}^{\text{o}}}$ (because these form a linear pair)

$\Rightarrow \text{12}{{\text{6}}^{\text{o}}}\text{+}\angle \text{FGE=18}{{\text{0}}^{\text{o}}}$ 

$\Rightarrow \angle \text{FGE=18}{{\text{0}}^{\text{o}}}\text{-12}{{\text{6}}^{\text{o}}}$ 

$\Rightarrow \angle \text{FGE=5}{{\text{4}}^{\text{o}}}$

Hence, we found $\angle \text{AGE=12}{{\text{6}}^{\text{o}}}$, $\angle \text{GEF=3}{{\text{6}}^{\text{o}}}$ , $\angle \text{FGE=5}{{\text{4}}^{\text{o}}}$ 

3. In the given figure, if PQ $\parallel $ ST, $\angle \text{PQR=11}{{\text{0}}^{\text{o}}}$ and$\angle \text{RST=13}{{\text{0}}^{\text{o}}}$ , find $\angle \text{QRS}$.

(Hint: Draw a line parallel to ST through point R.)

Ans:

In this question we will have some construction of our own, we draw a line XY parallel to ST and so parallel to PQ passing through point R.

$\angle \text{PQR+}\angle \text{QRX=18}{{\text{0}}^{\text{o}}}$ (Co-interior angles on the same side of transversal QR together sum up to ${{180}^{\circ }}$)

$\therefore \text{11}{{\text{0}}^{\text{o}}}\text{+}\angle \text{QRX=18}{{\text{0}}^{\text{o}}}$ 

$\Rightarrow \text{         }\angle \text{QRX=7}{{\text{0}}^{\text{o}}}$ 

Also,

$\angle \text{RST+}\angle \text{SRY=18}{{\text{0}}^{\circ }}$  (sum of Co-interior angles on the same side of transversal SR equals $\text{18}{{\text{0}}^{\circ }}$)

$\Rightarrow \text{       }\angle \text{SRY=18}{{\text{0}}^{\circ }}\text{-13}{{\text{0}}^{\circ }}$ 

$\Rightarrow \text{       }\angle \text{SRY=}{{50}^{\circ }}$

Now from the construction XY is a straight line. RQ and RS are rays from it.

$\angle \text{QRX+}\angle \text{QRS+}\angle \text{SRY=18}{{\text{0}}^{\circ }}$ 

$\Rightarrow \text{  7}{{\text{0}}^{\circ }}+\angle QRS+{{50}^{\circ }}={{180}^{\circ }}$ 

Hence we found that 

$\angle \text{QRS=6}{{\text{0}}^{\circ }}$

4. In the given figure, if AB $\parallel $ CD, $\angle APQ={{50}^{\circ }}$ and$\angle PRD={{127}^{\circ }}$ , find x and y.

Ans:

$\angle APR=PRD$ (since alternate interior angles are equal)

$\therefore \text{5}{{\text{0}}^{\circ }}+y={{127}^{\circ }}$ (since $\angle APR=\angle \text{APQ+}\angle \text{PQR}$)

$\Rightarrow \text{y=7}{{\text{7}}^{\circ }}$ 

similarly,

$\angle \text{APQ=PQR}$ ∠APQ = ∠PQR (since alternate interior angles are equal)

$\therefore \text{x=5}{{\text{0}}^{\circ }}$ 

Therefore we found that $\text{x=5}{{\text{0}}^{\circ }}$ and $\text{y=7}{{\text{7}}^{\circ }}$

5. In the given figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS atC and again reflects back along CD. Prove that AB||CD.

Ans:

Let us construct BM$\bot $ PQ and CN$\bot $RS.

Since  PQ$\parallel $ RS, and so BM $\parallel $ CN

Therefore, CN and BM are two parallel lines and a transversal line BC cuts them at B and C respectively.

∠2 = ∠3......($1$) (since alternate interior angles are equal)

But, by laws of reflection in Physics

 ∠1 = ∠2 and ∠3 = ∠4 

Now from equation ($1$)

∠1 = ∠2 = ∠3 = ∠4

Therefore 

∠1 + ∠2 = ∠3 + ∠4

$\angle \text{ABC=}\sum \text{DCB}$ 

But, these are alternate interior angles.

$\therefore $ AB$\parallel $ CD

Hence  proved.

Overview of Deleted Syllabus for Class 9 Maths Chapter 6 Lines and Angles

Conclusion

NCERT Solutions for Chapter 6 Class 9 Math Lines and Angles, covers important concepts and theorems for mastering basic geometry. The chapter emphasizes understanding how lines and angles interact and proving lines parallel using corresponding, alternate, and co-interior angles. Mastering these concepts is crucial for future mathematics topics. The detailed solutions help students approach and solve geometric problems, enhancing their analytical and problem-solving skills. Mastering this chapter is essential not only for exams but also for a deeper understanding of geometry.

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