 # Lines and Angles

## Class 9 NCERT Maths

### NCERT

1   In the given figure, lines $AB$ and $CD$ intersect at $O$. If $\angle AOC +\angle BOE = 70^o$ and $\angle BOD =40^o$ find $\angle BOE$ and reflex $\angle COE.$ $AB$ is a straight line, rays $OC$ and $OE$ stand on it.$\\$ $\therefore \angle AOC+\angle COE+\angle BOE=180^o\\ \Rightarrow (\angle AOC+\angle BOE)+\angle COE=180^o\\ \Rightarrow 70^o+\angle COE=180^o\\ \Rightarrow \angle COE=180^o-70^o=110^o$$\\ Reflex \angle COE=360^o-110^o=250^o$$\\$ $CD$ is a straight line, rays $OE$ and $OB$ stand on it.$\\$ $\therefore \angle COE+\angle BOE+\angle BOD=180^o\\ \Rightarrow 110^o+\angle BOE+40^o=180^o\\ \Rightarrow \angle BOE=180^o-150^o=30^o$$\\ Hence, \angle BOE = 30^o and Reflex \angle COE = 250^o 2 In the given figure, lines XY and MN intersect at O. If \angle POY = 90^o and a: b = 2:3, find c. ##### Solution : Let the common ratio between a and b be x.$$\\$ $\therefore a=2x$ and $b=3x$$\\ XY is a straight line, rays OM and OP stand on it.\\ \therefore \angle XOM+\angle MOP+\angle POY=180^o$$\\$ $b+a+\angle POY =180^o$$\\ 3x+2x+90^o=180^o$$\\$ $5x=90^o$$\\ x=18^o$$\\$ $a=2x=2*18=36^o$$\\ b=3x=3*18=54^o$$\\$ $MN$ is a straight line. Ray $OX$ stands on it.$\\$ $\therefore b+c=180^o$(Linear Pair)$\\$ $54^o+c=180^o$$\\ c=180^o-54^o=126^o$$\\$ $\therefore c=126^o$

3   In the given figure, $\angle PQR = \angle PRQ,$ then prove that $\angle PQS = \angle PRT.$ ##### Solution :

In the given figure, $ST$ is a straight line and ray $QP$ stands on it.$\\$ $\therefore \angle PQS + \angle PQR = 180^o$(Linear Pair)$\\$ $\angle PQR = 180^o - \angle PQS ... (1)$$\\ \angle PRT + \angle PRQ = 180^o (Linear Pair)\\ \angle PRQ = 180^o - \angle PRT ... (2)$$\\$ It is given that $\angle PQR = \angle PRQ.$$\\ Equating Equations (1) and (2), we obtain\\ 180^o -\angle PQS = 180 - \angle PRT$$\\$ $\angle PQS = \angle PRT$(Proved)

4   In the given figure, if $x + y = w + z$ then prove that $AOB$ is a line. ##### Solution :

It can be observed that,$\\$ $x + y + z + w = 360^o$ (Complete angle)$\\$ It is given that,$\\$ $x + y = z + w$$\\ \therefore x + y + x + y = 360^o$$\\$ $2(x + y) = 360^o$$\\ x + y = 180^o$$\\$ Since $x$ and $y$ form a linear pair, therefore, $AOB$ is a line. (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$\\$ $\angle ROS=\dfrac{1}{2}(\angle QOS-\angle POS)$ It is given that $OR \perp PQ$$\\ \angle POR = 90^o\\ \angle POS + \angle SOR = 90^o\\ \angle ROS = 90^o - \angle POS ... (1)\\ \angle QOR = 90^o (As \ OR \perp PQ)\\ \angle QOS - \angle ROS = 90^o\\ \angle ROS = \angle QOS - 90^o ... (2)$$\\$ On adding Equations (1) and (2), we obtain$\\$ $2 \angle ROS = \angle QOS - \angle POS$$\\ \angle ROS=\dfrac{1}{2}(\angle QOS-\angle POS) 6 It is given that \angle XYZ = 64 ^o and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects \angle ZYP , find \angle XYQ and reflex \angle QYP. ##### Solution : It is given that line YQ bisects \angle PYZ. \\ Hence, \angle QYP = \angle ZYQ \\ It can be observed that PX is a line. Rays YQ and YZ stand on it. \\ \angle XYZ + \angle ZYQ + \angle QYP = 180^o \\ 64^o + 2\angle QYP = 180^o \\ 2 \angle QYP = 180^o - 64^o = 116^o \\ \angle QYP = 58^o \\ Also, \angle ZYQ = \angle QYP = 58^o \\ Reflex \angle QYP = 360^o - 58^o = 302^o \\ \angle XYQ = \angle XYZ + \angle ZYQ$$\\$ $= 64^o + 58^o = 122^o$ $\\$ Hence, $\angle XYQ = 122^o,$ Reflex $\angle QYP = 302^o$

7   In the given figure, find the values of $x$ and $y$ and then show that $AB || CD.$

##### Solution :

It can be observed that,$\\$ $50^o + x = 180^o$ (Linear pair) $\\$ $x = 130^o$ ... (1)$\\$ Also,$y = 130^o$ (Vertically opposite angles)$\\$ As $x$ and $y$ are alternate interior angles for lines $AB$ and $CD$ and also measures of these angles are equal to each other, therefore, line $AB || CD.$

8   In the given figure, if $AB || CD, CD || EF$ and $y: z = 3:7$, find $x$.

##### Solution :

It is given that $AB || CD$ and $CD || EF$ $\\$ $\therefore AB || CD || EF$ (Lines parallel to the same line are parallel to each other)$\\$ It can be observed that$\\$ $x = z$ (Alternate interior angles) ... (1)$\\$ It is given that $y: z = 3: 7$ $\\$ Let the common ratio between $y$ and $z$ be a.$\\$ $\therefore y = 3a$ and $z = 7a$ $\\$ Also,$x + y = 180^o$ (Co-interior angles on the same side of the transversal)$\\$ $z + y = 180^o$ [Using Equation (1)]$\\$ $7a + 3a = 180^o$ $\\$ $10a = 180^o$ $\\$ $a = 18^o$ $\\$ $\therefore x = 7a = 7 × 18^o = 126^o$ $\\$

9   In the given figure, If $AB || CD, EF \perp CD$ and $\angle GED = 126^o,$ find $\angle AGE, \angle GEF$ and $\angle FGE.$

##### Solution :

It is given that,$\\$ $AB || CD$ and $EF \perp CD$ $\\$ $\angle GED = 126^o$ $\\$ $\angle GEF + \angle FED = 126^o$ $\\$ $\angle GEF + 90^o = 126^o$ $\\$ $\angle GEF = 36^o$ $\\$ As $\angle AGE$ and $\angle GED$ are alternate interior angles.$\\$ $\angle AGE = \angle GED = 126^o$ $\\$ However, $\angle AGE + \angle FGE = 18^o$ (Linear pair)$\\$ $126^o + \angle FGE = 180^o$ $\\$ $\angle FGE = 180^o - 126^o = 54^o$ $\\$ Hence, $\angle AGE = 126^o, \angle GEF = 36^o, \angle FGE = 54^o$ $\\$

10   In the given figure, if $PQ || ST, \angle PQR = 110^o$ and $\angle RST = 130^o$, find $\angle QRS.$$\\ [Hint: Draw a line parallel to ST through point R.] ##### Solution : Let us draw a line XY parallel to ST and passing through point R. \angle PQR + \angle QRX = 180^o (Co-interior angles on the same side of transversal QR)\\ 110^o + \angle QRX = 180^o \\ \angle QRX = 70^o \\ Also,\\ \angle RST + \angle SRY = 180^o (Co-interior angles on the same side of transversal SR)\\ 130^o + \angle SRY = 180^o$$\\$ $\angle SRY = 50^o$ $\\$ $XY$ is a straight line. $RQ$ and $RS$ stand on it.$\\$ $\angle QRX + \angle QRS + \angle SRY = 180^o$ $\\$ $70^o + \angle QRS + 50^o = 180^o$ $\\$ $\angle QRS = 180^o - 120^o = 60^o$ $\\$

11   In the given figure, if $AB || CD, \angle APQ = 50^o$ and $\angle PRD = 127^o,$ find $x$ and $y.$

##### Solution :

$\angle APR = \angle PRD$ (Alternate interior angles) $50^o + y = 127^o$ $\\$ $y = 127^o -50^o$ $\\$ $y = 77^o$ $\\$ Also,$\\$ $\angle APQ = \angle PQR$ (Alternate interior angles) $50^o = x$ $\\$ Therefore, $x = 50^o$ and $y = 77^o$

12   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 $RS$ at $C$ and again reflects back along $CD$. Prove that $AB || CD.$

Let us draw $BM \perp PQ$ and $CN \perp RS.$ $\\$ As $PQ || RS,$$\\ Therefore, BM || CN \\ Thus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.\\ \angle 2 = \angle 3 (Alternate interior angles)\\ However, \angle 1 = \angle 2 and \angle 3 = \angle 4 (By laws of reflection)\\ \angle 1 = \angle 2 = \angle 3 = \angle 4 \\ Also,\\ \angle 1 + \angle 2 = \angle 3 + \angle 4 \angle ABC = \angle DCB \\ However, these are alternate interior angles. \\ \therefore AB || CD 13 In the given figure, sides QP and RQ of \delta PQR are produced to points S and T respectively. If \angle SPR = 135^o and \angle PQT = 110^o, find \angle PRQ. ##### Solution : It is given that,\\ \angle SPR = 135^o and \angle PQT = 110^o \\ \angle SPR + \angle QPR = 180^o (Linear pair angles) 135^o + \angle QPR = 180^o \\ \angle QPR = 45^o \\ Also, \angle PQT + \angle PQR = 180^o (Linear pair angles) 110^o + \angle PQR = 180^o \\ \angle PQR = 70^o \\ As the sum of all interior angles of a triangle is 180o, therefore, for \Delta PQR, \\ \angle QPR + \angle PQR + \angle PRQ = 180^o \\ 45^o + 70^o + \angle PRQ = 180^o \\ \angle PRQ = 180^o - 115^o \\ \angle PRQ = 65^o \\ 14 In the given figure, \angle X = 62^o, \angle XYZ = 54^o. If YO and ZO are the bisectors of \angle XYZ and \angle XZY respectively of \Delta XYZ, find \angle OZY and \angle YOZ. ##### Solution : As the sum of all interior angles of a triangle is 180^o, therefore, for \Delta XYZ, \angle X + \angle XYZ + \angle XZY = 180^o \\ 62^o + 54^o + \angle XZY = 180^o \\ \angle XZY = 180^o - 116^o \\ \angle XZY = 64^o \\ \angle OZY = \dfrac{64}{2} = 32^o (OZ is the angle bisector of \angle XZY) Similarly, \angle OYZ = \dfrac{54}{2} = 27^o \\ Using angle sum property for \Delta OYZ , we obtain \angle OYZ + \angle YOZ + \angle OZY = 180^o \\ 27^o + \angle YOZ + 32^o = 180^o \\ \angle YOZ = 180^o - 59^o \\ \angle YOZ = 121^o \\ Hence, \angle OZY = 32^o and \angle YOZ = 121^o \\ 15 Question 3: In the given figure, if AB || DE, \angle BAC = 35^o and \angle CDE = 53^o, find \angle DCE. ##### Solution : AB || DE and AE is a transversal.\\ \angle BAC = \angle CED (Alternate interior angles)\\ \angle CED = 35^o \\ In \Delta CDE, \\ \angle CDE + \angle CED + \angle DCE = 180^o (Angle sum property of a triangle) \\ 53^o + 35^o + \angle DCE = 180^o \\ \angle DCE = 180^o - 88^o \\ \angle DCE = 92^o \\ 16 In the given figure, if lines PQ and RS intersect at point T, such that \angle PRT = 40^o, \\ \angle RPT = 95^o and \angle TSQ = 75^o, find \angle SQT. ##### Solution : Using angle sum property for \Delta PRT, we obtain\\ \angle PRT + \angle RPT + \angle PTR = 180^o 40^o + 95^o + \angle PTR = 180^o \\ \angle PTR = 180o - 135^o \\ \angle PTR = 45^o \\ \angle STQ = \angle PTR = 45^o (Vertically opposite angles) \angle STQ = 45^o$$\\$ By using angle sum property for $\Delta STQ,$ we obtain$\\$ $\angle STQ + \angle SQT + \angle QST = 180^o 45^o + \angle SQT + 75^o = 180^o$ $\\$ $\angle SQT = 180^o - 120^o$ $\\$ $\angle SQT = 60^o$ $\\$

17   In the given figure, if $PQ \perp PS, PQ || SR, \angle SQR = 28^o$ and $\angle QRT = 65^o,$ then find the values of $x$ and $y.$

##### Solution :

It is given that $PQ || SR$ and $QR$ is a transversal line. $\angle PQR = \angle QRT$ (Alternate interior angles)$\\$ $x + 28^o = 65^o$ $\\$ $x = 65^o - 28^o$ $\\$ $x = 37^o$ $\\$ By using the angle sum property for $\Delta SPQ,$ we obtain $\angle SPQ + x + y = 180^o$ $\\$ $90^o + 37^o + y = 180^o$ $\\$ $y = 180^o - 127^o$ $\\$ $y = 53^o$ $\\$ $x = 37^o$ and $y = 53^o$

18   In the given figure, 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 $\angle QTR= \dfrac{1}{2} \angle QPR$

##### Solution :

In $\Delta QTR, \angle TRS$ is an exterior angle. $\therefore \angle QTR + \angle TQR = \angle TRS$ $\\$ $\angle QTR = \angle TRS - \angle TQR ... (1)$ $\\$ For $\Delta PQR, \angle PRS$ is an external angle.$\\$ $\therefore \angle QPR + \angle PQR = \angle PRS$ $\\$ $\angle QPR + 2\angle TQR = 2\angle TRS$ (As $QT$ and $RT$ are angle bisectors) $\\$ $\angle QPR = 2(\angle TRS - \angle TQR)$ $\\$ $\angle QPR = 2\angle QTR$[By using Equation (1)]$\\$ $\angle QTR = \dfrac{1}{2} \angle QPR$