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 $\angle COE.$

Solution :

$AB$ 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$\\$ $\angle ROS=\dfrac{1}{2}(\angle QOS-\angle POS)$

Solution :

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 $AB || CD.$

Solution :

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 $[By using Equation (1)]$\\$ $\angle QTR = \dfrac{1}{2} \angle QPR$