NCERT Solutions Class 9 Maths Chapter 6 Lines and Angles

Ever wondered why the world seems so structured and orderly? It's all thanks to the invisible lines and angles that shape our surroundings. In Class 9 Maths Chapter 6: Lines and Angles, we delve into the fascinating world of geometry, where understanding the relationships between lines and angles forms the backbone of mathematical reasoning. 

With the help of ALLEN’s NCERT solutions for Class 9 Maths Chapter 6, you can dive deeper into these concepts with ease. The step-by-step solutions are designed to clarify even the hard problems, making your learning journey smooth and efficient to get good marks in your school examinations. 

1.0Download Class 9 Maths Chapter 6 NCERT Solutions PDF Online

NCERT Solutions Maths for class 9 chapter 6 is a detailed guide to studying the properties of lines and angles. For preparing the questions that appear in these NCERT exercises, click on the links below which are created by experienced teachers of ALLEN’.

2.0NCERT Questions with Solutions for Class 9 Maths Chapter 6 - Detailed Solutions

Exercise : 6.1

• In figure, lines $AB$ and $CD$ intersect at 0 . If $\angle \mathrm{AOC}+\angle \mathrm{BOE}=70^{\circ }$ and $\angle \mathrm{BOD}=40^{\circ }$, find $\angle \mathrm{BOE}$ and reflex $\angle \mathrm{COE}$.

• Sol. $\angle AOC=\angle BOD$ [Vertically opposite angles] $\Rightarrow \angle \mathrm{AOC}=40^{\circ }$ $[\because \angle \mathrm{BOD}=40^{\circ }$ is given $]$ Now, $\angle \mathrm{AOC}+\angle \mathrm{BOE}=70^{\circ }$ [Given] $\Rightarrow 40^{\circ }+\angle \mathrm{BOE}=70^{\circ }$ $\Rightarrow \angle \mathrm{BOE}=30^{\circ }$ $\angle \mathrm{AOE}+\angle \mathrm{BOE}=180^{\circ }$ [Linear pair of angles] $\Rightarrow \angle \mathrm{AOE}+30^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{AOE}=150^{\circ }$ $\Rightarrow \angle \mathrm{AOC}+\angle \mathrm{COE}=150^{\circ }$ $\Rightarrow 40^{\circ }+\angle \mathrm{COE}=150^{\circ }$ $\Rightarrow \angle \mathrm{COE}=110^{\circ }$ Reflex $\angle \mathrm{COE}=360^{\circ }-110^{\circ }=250^{\circ }$
• In figure, lines $XY$ and $MN$ intersect at 0 . If $\angle \mathrm{POY}=90^{\circ }$ and $\mathrm{a}:\mathrm{b}=2:3$, find c .

• Sol. Ray OP stands on line XY $\angle \mathrm{POX}+\angle \mathrm{POY}=180^{\circ }$ $\angle \mathrm{POX}+90^{\circ }=180^{\circ }$ $\angle \mathrm{POX}=90^{\circ }$ $\angle \mathrm{POM}+\angle \mathrm{XOM}=90^{\circ }$ $\mathrm{a}+\mathrm{b}=90^{\circ }$ $\mathrm{a}:\mathrm{b}=2:3$ $\frac{\mathrm{a}}{2}=\frac{\mathrm{b}}{3}=\mathrm{k}$ $\mathrm{a}=2\mathrm{k},\mathrm{b}=3\mathrm{k}$ $3\mathrm{k}+2\mathrm{k}=90^{\circ }$ $\mathrm{k}=18^{\circ }$ $\Rightarrow \mathrm{a}=36^{\circ },\mathrm{b}=54^{\circ }$ $\therefore$ Ray OX stands on line MN $\angle \mathrm{XOM}+\angle \mathrm{XON}=180^{\circ }$ $\mathrm{b}+\mathrm{c}=180^{\circ }$ $54^{\circ }+\mathrm{c}=180^{\circ }\Rightarrow \mathrm{c}=126^{\circ }$
• In figure, $\angle \mathrm{PQR}=\angle \mathrm{PRQ}$, then prove that $\angle \mathrm{PQS}=\angle \mathrm{PRT}$.

• Sol. $\angle \mathrm{PQR}=\angle \mathrm{PRQ}=\mathrm{x}$ (say) $\dots$ (1) Now, $\angle \mathrm{PQS}+\angle \mathrm{PQR}=180^{\circ }[$ Linear pair of angles] And $\angle \mathrm{PRT}+\angle \mathrm{PRQ}=180^{\circ }$ [Linear pair of angles] $\Rightarrow \angle \mathrm{PQS}+\angle \mathrm{PQR}=\angle \mathrm{PR}+\angle \mathrm{PRQ}$ $\left[\because \text{ each }=180^{\circ }\right]$ $\Rightarrow \angle \mathrm{PQS}+\mathrm{x}=\angle \mathrm{PRT}+\mathrm{x}[\mathrm{By}(1)]$ $\Rightarrow \angle \mathrm{PQS}=\angle \mathrm{PRT}$
• In figure, if $x+y=w+z$, then prove that AOB is a line.

• Sol. $\mathrm{x}+\mathrm{y}=\mathrm{w}+\mathrm{z}$ $x+y+w+z=360^{\circ }$ [Complete angle] $\Rightarrow 2(\mathrm{x}+\mathrm{y})=360^{\circ }$, $x+y=180^{\circ }$ [From (1)] $\Rightarrow AOB$ is a line.
• In figure, POQ is a line. Ray OR is perpendicular to line PQ . $OS$ is another ray lying between rays OP and OR. Prove that $\angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})$.

• Sol. $\angle \mathrm{POR}=\angle \mathrm{QOR}=90^{\circ }$ $[\because \mathrm{OR}\perp \mathrm{PQ}$ at 0$]$ Now, $\angle \mathrm{QOS}=\angle \mathrm{QOR}+\angle \mathrm{ROS}$ $\Rightarrow \angle \mathrm{QOS}=90^{\circ }+\angle \mathrm{ROS}$ $\angle \mathrm{POS}+\angle \mathrm{ROS}=\angle \mathrm{POR}$ $\Rightarrow \angle \mathrm{POS}=\angle \mathrm{POR}-\angle \mathrm{ROS}$ $\Rightarrow \angle \mathrm{POS}=90^{\circ }-\angle \mathrm{ROS}$ Subtracting (3) from (2), $\angle \mathrm{QOS}-\angle \mathrm{POS}=\left\{90^{\circ }+\angle \mathrm{ROS}\right\}-\{90^{\circ }-$ $\angle$ ROS $\}$ $=2\times \angle$ ROS $\Rightarrow 2\times \angle \mathrm{ROS}=\{\angle \mathrm{QOS}-\angle \mathrm{POS}\}$ i.e., $\angle \mathrm{ROS}=\frac{1}{2}\{\angle \mathrm{QOS}-\angle \mathrm{POS}\}$
• It is given that $\angle \mathrm{XYZ}=64^{\circ }$ and XY is produced to point P. Draw a figure from the given information. if ray YQ bisects $\angle \mathrm{ZYP}$, find $\angle \mathrm{XYQ}$ and reflex $\angle \mathrm{QYP}$. Sol. $\angle \mathrm{XYZ}+\angle \mathrm{ZYP}=180^{\circ }$ [Linear pair] $\Rightarrow 64^{\circ }+\angle \mathrm{ZYP}=180^{\circ }$ $\Rightarrow \angle \mathrm{ZYP}=116^{\circ }$
  
  Ray YQ bisects angle $\angle \mathrm{ZYP}$ $\Rightarrow \angle \mathrm{PYQ}=\angle \mathrm{ZYQ}=\frac{116^{\circ }}{2}=58^{\circ }$ Reflex $\angle \mathrm{QYP}=360^{\circ }-58^{\circ }=302^{\circ }$ $\angle \mathrm{XYQ}=\angle \mathrm{XYZ}+\angle \mathrm{ZYQ}$ $=64^{\circ }+58^{\circ }=122^{\circ }$

Exercise : 6.2

• In figure, find the values of $x$ and $y$ and then show that $AB\Vert CD$.
  
  Sol. $x+50=180^{\circ }$ (Linear pair of angles) $\Rightarrow \mathrm{x}=130^{\circ }$ $y=130^{\circ }$ (Vertically opposite angles) Now, $\mathrm{x}=\mathrm{y}$ and the two angles form a pair of alternate angles made by a transversal intersecting the lines AB and CD Therefore, $\mathrm{AB}|\mid CD$.
• In figure, if $\mathrm{AB}\Vert \mathrm{CD},\mathrm{CD}\Vert EF$ and $\mathrm{y}:\mathrm{z}=3$ : 7 , find x .
  
  Sol. $AB\Vert CD$ and $CD\Vert EF$ $\Rightarrow \mathrm{AB}\Vert \mathrm{EF}$ $\Rightarrow \mathrm{x}=\mathrm{z}$ (Alternate angles) Now, $\mathrm{x}+\mathrm{y}=180^{\circ }$ (Pair of interior angles on the same side of the transversal) $\Rightarrow \mathrm{z}+\mathrm{y}=180^{\circ }$ i.e, $\mathrm{y}+\mathrm{z}=180^{\circ }$ Also, we are given that, $\mathrm{y}:\mathrm{z}=3:7$ Then $\mathrm{y}=\frac{3}{10}\times 180^{\circ }=54^{\circ }$ and $\mathrm{z}=\frac{7}{10}\times 180^{\circ }=126^{\circ }$ We have $x=z=126^{\circ }$ Therefore, $x=126^{\circ }$
• In figure, if $\mathrm{AB}\Vert \mathrm{CD},\mathrm{EF}\perp \mathrm{CD}$ and $\angle \mathrm{GED}=$ $126^{\circ }$, find $\angle \mathrm{AGE},\angle \mathrm{GEF}$ and $\angle \mathrm{FGE}$.
  
  Sol. AB||CD [given] $\angle \mathrm{AGE}=\angle \mathrm{GED}=126^{\circ }$ [Alternate angles] $\Rightarrow \angle \mathrm{GEF}+90^{\circ }=126^{\circ }$ $\angle \mathrm{GEF}=36^{\circ }$ $\angle \mathrm{GEC}+\angle \mathrm{GEF}+\angle \mathrm{FED}=180^{\circ }$ [Straight line] $\angle \mathrm{GEC}+126^{\circ }=180^{\circ }$ $\angle \mathrm{GEC}=180^{\circ }-126^{\circ }=54^{\circ }$ $\angle \mathrm{FGE}=\angle \mathrm{GEC}=54^{\circ }$ [Alternate angles]
• In figure, if $\mathrm{PQ}\Vert \mathrm{ST},\angle \mathrm{PQR}=110^{\circ }$ and $\angle \mathrm{RST}=130^{\circ }$, find $\angle \mathrm{QRS}$.
  
  Sol. Through R, we draw XRY | PQ .
  
  $\Rightarrow \mathrm{XRY}\Vert \mathrm{ST}$ $(\because \mathrm{PR}\Vert \mathrm{ST})$ $\angle \mathrm{QRX}+110^{\circ }=180^{\circ }$ and $\angle \mathrm{YRS}+130^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{QRX}=70^{\circ }$ And $\angle \mathrm{YRS}=50^{\circ }$ Now, $\angle \mathrm{QRX}+\angle \mathrm{QRS}+\angle \mathrm{YRS}=180^{\circ }$ $\Rightarrow 70^{\circ }+\angle \mathrm{QRS}+50^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{QRS}=60^{\circ }$
• In figure, if $\mathrm{AB}\Vert \mathrm{CD},\angle \mathrm{APQ}=50^{\circ }$ and $\angle PRD=127^{\circ }$, find $x$ and $y$.
  
  Sol. $\mathrm{AB}\Vert +\mathrm{CD}$ [given] $\mathrm{x}=\angle \mathrm{APQ}=50^{\circ }$ [Alternate angles] $\angle \mathrm{APQ}+\mathrm{y}=\angle \mathrm{PRD}=127^{\circ }$ [Alternate angles] $50^{\circ }+y=127^{\circ }$ $y=127^{\circ }-50^{\circ }=77^{\circ }$
• In 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 at C and again reflects back along CD. Prove that AB || CD.
  
  Sol. We draw $BE\perp RS$, then $BE$ is also $\perp PQ$ $(\because \mathrm{PQ}\Vert \mathrm{RS})$ We draw $\mathrm{CF}\perp \mathrm{PQ}$. Here, also $\mathrm{CF}\perp \mathrm{RS}$
  
  Here, if we consider $PQ$ as transversal intersecting lines BE and CF , then each pair of corresponding angles is equal. (each equal to $90^{\circ }$ ) Thus, we have $\mathrm{BE}\Vert \mathrm{CF}$. Now, $\angle \mathrm{ABE}=\angle \mathrm{CBE}$ (Angle of incidence $=$ Angle of reflection) $\Rightarrow \angle \mathrm{ABE}=\angle \mathrm{CBE}=\frac{1}{2}\times \angle \mathrm{ABC}$ Similarly, $\angle \mathrm{BCF}=\angle \mathrm{FCD}=\frac{1}{2}\times \angle \mathrm{DCB}$ Now, BE || CF $\Rightarrow \angle \mathrm{CBE}=\angle \mathrm{BCF}$ (alternate angles) $\Rightarrow \frac{1}{2}\times \angle \mathrm{ABC}=\frac{1}{2}\times \angle \mathrm{DCB}\{$ by (1) and (2) $\}$ $\Rightarrow \angle \mathrm{ABC}=\angle \mathrm{DCB}$ $\Rightarrow \mathrm{AB}\Vert \mathrm{CD}$

Exercise: 6.3

• In figure, sides $QP$ and $RQ$ of $△PQR$ are produced to points $S$ and $T$ respectively. If $\angle \mathrm{SPR}=135^{\circ }$ and $\angle \mathrm{PQT}=110^{\circ }$, find $\angle PRQ$.
  
  Sol. $\angle \mathrm{QPR}+135^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{QPR}=45^{\circ }$ Now, $\angle \mathrm{QRP}+\angle \mathrm{QPR}=\angle \mathrm{PQT}$ $\Rightarrow \angle \mathrm{QRP}+45^{\circ }=110^{\circ }$ $\Rightarrow \angle \mathrm{QRP}=65^{\circ }$
• In figure, $\angle \mathrm{X}=62^{\circ },\angle \mathrm{XYZ}=54^{\circ }$. If YO and ZO are the bisectors of $\angle \mathrm{XYZ}$ and $\angle \mathrm{XZY}$ respectively of $△XYZ$, find $\angle OZY$ and $\angle \mathrm{YOZ}$.
  
  Sol. In $△\mathrm{XYZ}$ $\angle \mathrm{XYZ}+\angle \mathrm{YZX}+\angle \mathrm{ZXY}=180^{\circ }$ $54^{\circ }+\angle \mathrm{YZX}+62^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{YZX}=64^{\circ }$ $\because \mathrm{YO}$ is bisector $\Rightarrow \angle \mathrm{XYO}=\angle \mathrm{OYZ}=\frac{54^{\circ }}{2}=27^{\circ }$ ZO is bisector $\Rightarrow \angle \mathrm{XZO}=\angle \mathrm{OZY}=\frac{64^{\circ }}{2}=32^{\circ }$ In $△\mathrm{OYZ}$ $\angle \mathrm{OYZ}+\angle \mathrm{OZY}+\angle \mathrm{YOZ}=180^{\circ }$ $27^{\circ }+32^{\circ }+\angle \mathrm{YOZ}=180^{\circ }$ $\Rightarrow \angle \mathrm{YOZ}=121^{\circ }$
• In figure, if $\mathrm{AB}\Vert \mathrm{DE},\angle \mathrm{BAC}=35^{\circ }$ and $\angle \mathrm{CDE}=53^{\circ }$, find $\angle \mathrm{DCE}$.
  
  Sol. $\mathrm{AB}|\mid \mathrm{DE}$ (given) $\angle \mathrm{DEC}=\angle \mathrm{BAC}=35^{\circ }$ (Alternate angles) $\angle \mathrm{CDE}=53^{\circ }$ In $△$ CDE $\angle \mathrm{CDE}+\angle \mathrm{DEC}+\angle \mathrm{DCE}=180^{\circ }$ $53^{\circ }+35^{\circ }+\angle \mathrm{DCE}=180^{\circ }$ $\angle \mathrm{DCE}=180^{\circ }-88^{\circ }=92^{\circ }$
• In figure, if lines $PQ$ and RS intersect at point $T$, such that $\angle \mathrm{PRT}=40^{\circ },\angle \mathrm{RPT}=95^{\circ }$ and $\angle \mathrm{TSQ}=75^{\circ }$, find $\angle \mathrm{SQT}$.
  
  Sol. In $△\mathrm{PRT}$ $\angle \mathrm{PTR}+\angle \mathrm{PRT}+\angle \mathrm{RPT}=180^{\circ }$ $\angle \mathrm{PTR}+40^{\circ }+95^{\circ }=180^{\circ }$ $\angle \mathrm{PTR}=45^{\circ }$ $\angle \mathrm{QTS}=\angle \mathrm{PTR}=45^{\circ }$ [Vertically opposite angles] In $△$ TSQ $\angle \mathrm{QTS}+\angle \mathrm{TSQ}+\angle \mathrm{SQT}=180^{\circ }$ $45^{\circ }+75^{\circ }+\angle \mathrm{SQT}=180^{\circ }$ $\angle \mathrm{SQT}=60^{\circ }$
• In figure, if $\mathrm{PQ}\perp \mathrm{PS},\mathrm{PQ}\Vert \mathrm{SR},\angle \mathrm{SQR}=28^{\circ }$ and $\angle \mathrm{QRT}=65^{\circ }$, then find the values of x and $y$.
  
  Sol. $\angle \mathrm{QSR}=\mathrm{x}$ (alternate angles) Now, $\angle \mathrm{QSR}+28^{\circ }=65^{\circ }$ $\Rightarrow \mathrm{x}+28^{\circ }=65^{\circ }$ $\Rightarrow \mathrm{x}=37^{\circ }$ In $△$ PQS, $90^{\circ }+x+y=180^{\circ }$ $\Rightarrow 90^{\circ }+37^{\circ }+y=180^{\circ }$ $\Rightarrow \mathrm{y}=53^{\circ }$
• In figure, the side QR of $△\mathrm{PQR}$ is produced to a point $S$. If the bisectors of $\angle \mathrm{PQR}$ and $\angle \mathrm{PRS}$ meet at point T , then prove that $\angle \mathrm{QTR}=\frac{1}{2}\angle \mathrm{QPR}$.
  
  Sol. $\angle \mathrm{PRS}=\angle \mathrm{PQR}+\angle \mathrm{QPR}$ Now, $\angle \mathrm{QTR}+\angle \mathrm{TQR}=\angle \mathrm{TRS}$ $\Rightarrow \angle \mathrm{QTR}+\frac{1}{2}\times \angle \mathrm{PQR}=\frac{1}{2}\angle \mathrm{PRS}$ $\Rightarrow 2\angle \mathrm{QTR}+\angle \mathrm{PQR}=\angle \mathrm{PRS}$ From (1) and (2), $2\angle \mathrm{QRT}+\angle \mathrm{PQR}=\angle \mathrm{PQR}+\angle \mathrm{QPR}$ $\Rightarrow 2\angle \mathrm{QTR}=\angle \mathrm{QPR}$ $\Rightarrow \angle \mathrm{QTR}=\frac{1}{2}\angle \mathrm{QPR}$

3.0Quick Insights of Chapter 6 NCERT Maths - Lines and Angles

• Chapter 6 of Class 9th Maths briefly introduces the concepts of a line segment and ray, collinear points, and the formation of angles.
• It relates the angles formed when a transversal cuts between parallel lines to the connections between those angles and to the angle relationships that you should use as evidence that two lines are parallel. The connections are corresponding, alternate interior, and alternate exterior angles.
• Also emphasizes the study of lines and angles, properties of angles, proofs, and applications of geometrical theorems such as the Linear Pair Postulate and the Vertical Angle Theorem.
• Gives an overview of Vertically opposite angles are formed and equal measurements which occur when two lines converge.