# Chapter 6: Lines and Angles

Q1: Lines AB and CD intersect at 0. If $$\angle AOC +\angle BOE= 70^{\circ}$$ and $$\angle BOD=40^{\circ}$$, find  $$\angle BOE$$ and reflex  $$\angle COE$$.

SOL:

Lines  $$AB and CD intersect at O$$.

$$∴ \angle AOC=\angle BOD$$                           (Vertically Opposite Angles)

But $$\angle BOD=40^{\circ}$$                              ………(i)

$$∴ \angle AOC=40^{\circ}$$                    ……….(ii)

Now, $$\angle AOC+\angle BOE=70^{\circ}\\ \Rightarrow 40^{\circ}+\angle BOE=70^{\circ}$$

$$\Rightarrow \angle BOE=30^{\circ}$$

Reflex $$\angle COE= \angle COD+\angle BOD+\angle BOE$$

=$$\angle COD-40^{\circ}+30^{\circ}$$                ………..(iii)     (using angle (i) and (ii)

$$\angle COD=180^{\circ}$$                                           (as it is a straight line)

Thus ,$$\angle COE= 180^{\circ}+40^{\circ}+30^{\circ}=270^{\circ}$$

Q2: In the figure, line $$XY$$ and $$MN$$ intersect at $$O$$.

If $$\angle POY=90^{\circ}$$ and $$a:b=2:3$$, find c.

Sol:

Ray $$OP$$ stands on line $$XY$$

$$∴ \angle POX+\angle POY=180^{\circ}$$        (as is linear pair axiom)

Since $$\angle POY=90^{\circ}$$   (given)

$$\Rightarrow \angle POX=90^{\circ}$$ $$\angle POM+\angle XOM=90^{\circ}$$

$$\Rightarrow a+b=90^{\circ}$$                 …………………….(i)

$$a:b=2:3\Rightarrow \frac{a}{b}=\frac{2}{3}\\ or \; \; \frac{a}{2}=\frac{b}{3}=k$$ $$\Rightarrow a=2k, b=3k$$

Substituting the values of a and b in (i), we get

$$2k+3k=90^{\circ}$$ $$\Rightarrow 5k=90^{\circ}\\ \Rightarrow k=18^{\circ}$$

$$∴ a=2k=2(18^{\circ})=36^{\circ}\\ b=3k=3(18^{\circ})=54^{\circ}$$    …….(ii)

Ray $$OX$$ stands on line MN              (Linear pair axiom)

$$\Rightarrow b+c=180^{\circ}$$ $$\Rightarrow 54^{\circ}+c=180^{\circ}$$

$$\Rightarrow c=180^{\circ}- 54^{\circ}$$        (using equation (ii)

$$\Rightarrow c=126^{\circ}$$

Q3: Prove that $$\angle PQS=\angle PRT$$ in the given isosceles triangle.

Sol:

Ray $$QP$$ stands on line $$ST$$.

$$∴ \angle PQS +\angle PQR=180^{\circ}$$           ………….(i)

(as linear pair axiom ray $$QP$$ stands on line $$ST$$)

$$∴ \angle PQS+\angle PQR=180^{\circ}$$           …………..(ii)

From (i) and (ii), we get

$$\angle PQS+\angle PQR= \angle PRQ+\angle PRT$$ $$\Rightarrow \angle PQS=\angle PRT$$

(since $$\angle PQR=\angle PRQ$$)

Q4: In the given figure if $$x+y=w+z$$ then prove that $$AOB$$ is a line.

Sol:

$$x+y=w+z$$                                  …………….(i)

We know, the sum of all the angles round a point is equal to $$360^{\circ}$$

So, $$x+y+w+z=360^{\circ}$$

Putting the value of w+z in above equation we have,

$$x+y+x+y=360^{\circ}$$ $$\Rightarrow 2(x+y)=360^{\circ}$$ $$\Rightarrow (x+y)=180^{\circ}$$

$$∴ AOB$$ is a line.

Q5: In the figure $$POQ$$  is a line. Ray $$QR$$ is a perpendicular to line $$PQ$$. $$OS$$ is another ray lying between rays $$OP and OR$$. Prove that $$\angle ROS=\boldsymbol{\frac{1}{2}}(\angle QOS-\angle POS)$$.

Sol:

Given, ray $$QR$$ is perpendicular to line $$PQ$$.

$$∴ \angle QOR=\angle POR=90^{\circ}$$                  ……………(i)

$$\angle QOS=\angle QOR+\angle ROS$$                             …………….(ii)

$$\angle QOS=\angle POR-\angle ROS$$                               ……………(iii)

From equation (ii) and (iii), we have

$$\angle QOS-\angle POS=\angle QOR+\angle ROS-\angle POR+\angle ROS$$ $$= (\angle QOR-\angle POR)+2\angle ROS$$ $$=2\angle ROS$$

(As we know $$\angle QOR=\angle POR$$ from equation (i) )

$$\Rightarrow \angle ROS=\frac{1}{2}(\angle QOS-\angle POS)$$

Q6: It is given that $$\angle XYZ=64^{\circ}$$ and $$XY$$ is produced to point $$P$$. Draw a figure from the given information.If ray $$YQ$$ bisect $$\angle ZYP$$, find $$\angle XYQ$$ and reflex $$\angle QYP$$.

Sol:

Given, ray $$YZ$$ stands on line $$PX$$.

$$∴ \angle XYZ+\angle ZYP=180^{\circ}$$

(As forms a linear pair axiom)

$$\Rightarrow 64^{\circ}+\angle ZYP=180^{\circ}$$

$$\Rightarrow \angle ZYP=116^{\circ}$$                                ……………(i)

Given ray $$YQ$$ bisect $$\angle ZYP$$

$$∴\angle PYQ=\angle ZYQ=\frac{1}{2}\angle ZYP=\frac{1}{2}(116^{\circ})=58^{\circ}$$                                  ……………..(ii)

$$∴ reflex \angle QYP=360^{\circ}-58^{\circ}=302^{\circ}$$

Again ,$$\angle XYQ=\angle XYZ+\angle ZYQ$$

$$=64^{\circ}+58^{\circ}$$                  (From eq (ii))

$$=122^{\circ}$$

Q7: In the given figure, find the values of $$x and y$$ and then show that $$AB\parallel CD$$.

Sol:

Ray $$AE$$ stands on line $$GH$$

$$∴ \angle AEG+\angle AEH=180^{\circ}$$

(as forms Linear pair axioms)

$$\Rightarrow 50^{\circ}+x=180^{\circ}$$  (as $$\angle AEG$$ is given)

$$\Rightarrow x=180^{\circ}-50^{\circ}=130^{\circ}$$     …….(i)

$$y=130^{\circ}$$                           ……..(ii)    (vertically opposite angles)

From (i) and (ii) we have,

$$x=y$$

But these angles are alternate interior angles and as they are equal.

So we can say that $$AB\parallel CD$$

Q8: In the figure, if $$AB\parallel CD$$, $$CD\parallel EF$$ and $$y:z=3:7$$, find $$x$$.

Sol:

Given $$AB\parallel CD$$, $$CD\parallel EF$$

$$AB\parallel EF$$

( as lines parallel to the same line are parallel to each other)

$$x=z$$                    ………….(i)         (alternate interior angles)

$$x+y=180^{\circ}$$ …………(ii)        (consecutive interior angles on the same side of the transversal $$GH$$ to parallel lines $$AB$$ and $$CD$$.

From (i) and (ii), we have

$$z+y=180^{\circ}$$ $$y:z=3:7$$

Now solving the ratios, we get

Sum of ratios =3+7=10

$$\Rightarrow y=\frac{3}{10}\times 180^{\circ}=54^{\circ}$$ and

$$z=\frac{7}{10}\times 180^{\circ}=126^{\circ}$$

As we know $$x=z$$

$$∴ x=z=126^{\circ}$$

Q9: In the figure, if $$AB\parallel CD$$, $$EF\perp CD$$ and $$\angle GED=126^{\circ}$$, find $$\angle AGE, \angle GEF \; and \; \angle FGE.$$

Sol:

(i) $$\angle AGE=\angle GED=126^{\circ}$$  (as forms alternate interior angles)

(ii) $$\angle GED=\angle GEF+\angle FED=126^{\circ}$$

$$\Rightarrow \angle GEF+90^{\circ}=126^{\circ}$$

(as given that $$EF\perp CD$$)

$$\Rightarrow \angle GEF=126^{\circ}-90^{\circ}=36^{\circ}$$

(iii) $$\angle CEG+\angle GED=180^{\circ}$$

And its given that $$\angle GED=126^{\circ}$$

$$\Rightarrow \angle CEG+126^{\circ}=180^{\circ}$$ $$\Rightarrow \angle CEG=180^{\circ}-126^{\circ}$$ $$\Rightarrow \angle CEG=54^{\circ}$$

$$\angle FGE=\angle CEG=54^{\circ}$$        (alternate angles)

Q10: In the figure, if $$PQ\parallel ST$$, $$\angle PQR=110^{\circ}$$ and $$\angle RST=130^{\circ}$$, find $$\angle QRS.$$

[Hint:Draw a line parallel to $$ST$$ through point $$R$$.

Sol:

From the hint we draw a line $$RU$$ parallel to $$ST$$ through point $$R$$.

$$\angle RST+\angle SRU=180^{\circ}$$

(Sum of the consecutive interior angles on the same side of the transversal is $$180^{\circ}$$)

$$\Rightarrow 130^{\circ}+\angle SRU=180^{\circ}$$

$$\Rightarrow \angle SRU=180^{\circ}-130^{\circ}=50^{\circ}$$       ………(i)

$$\angle QRU=\angle PQR=110^{\circ}$$         (alternate interior angles)

$$\Rightarrow \angle QRS+\angle SRU=110^{\circ}$$

$$\Rightarrow \angle QRS+50^{\circ}=110^{\circ}$$     (using (i))

$$\Rightarrow \angle QRS=110^{\circ}-50^{\circ}=60^{\circ}$$

Q11: In the figure, if $$AB\parallel CD$$, $$\angle APQ=50^{\circ}$$ and $$\angle PRD=127^{\circ}$$, find $$x and y.$$

Sol:

From the figure we can see that

$$\angle APQ=x=50^{\circ}$$     (as forms alternate interior angles)

$$\angle PRD=x+y=127^{\circ}$$   (as we know that exterior angles of a triangle is equal to the sum of the two interior opposite  angles )

$$\Rightarrow 50^{\circ}+y=127^{\circ}$$ $$\Rightarrow y=127^{\circ}-50^{\circ}=77^{\circ}$$

Q12 : In the Fig, $$PQ$$ 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 $$AB\parallel CD$$.

(Hint: Draw perpendiculars at A and B to the two plane mirrors. Recall that the angle of the incidence is equal to angle of reflection.)

Sol:

Construction: Draw ray $$BL\perp PQ$$ and ray $$CM\perp RS$$.

$$BL\perp PQ$$, $$CM\perp RS$$ and $$PQ\parallel RS$$.

$$∴ BL\parallel CM$$

$$\angle LBC=\angle MCB$$     ……(i)     (Alternate interior angles)

$$\angle ABL=\angle LBC$$ ………(ii)       (angle of incidence= angle of reflection)

$$\angle MCB=\angle MCD$$  ……….(iii)       (angle of incidence = angle of reflection).

From (i), (ii) and (iii), we get $$\angle ABL=\angle MCD$$….(iv)

Adding (i) and (iv), we get

$$\angle LBC+\angle ABL=\angle MCB+ \angle MCD$$ $$\Rightarrow \angle ABC=\angle BCD$$

But these are alternate interior angles and they are equal.

So, $$AB\parallel CD$$

Q13: In the figure, sides $$QP\: \: and \: \: RQ$$ are produced to point $$S\: \: and \: \: T$$ respectively. If $$\angle SRP=135^{\circ}\; \; and \: \: \angle PQT=110^{\circ}$$, find $$\angle PRQ$$.

Sol:

TR is a line So, $$\angle PQT+\angle PQR=180^{\circ}$$

$$\Rightarrow 110^{\circ} +\angle PQR=180^{\circ}$$

$$\Rightarrow \angle PQR=180^{\circ}-110^{\circ} =70^{\circ}$$   ……(i)

$$QS$$ is a line.

$$∴ \angle SPR+\angle QRP=180^{\circ}$$ $$\Rightarrow 135^{\circ}+\angle QRP=180^{\circ}$$

$$\Rightarrow \angle QRP=180^{\circ}-135^{\circ}=45^{\circ}$$ …….(ii)

In $$\bigtriangleup PQR, \angle PQR+\angle QPR+\angle PRQ=180^{\circ}$$   (sum of all the interior angles of a traingle is $$180^{\circ}$$.

$$\Rightarrow 70^{\circ}+45^{\circ}\angle+ PRQ=180^{\circ}$$    (by equation (i) and (ii) )

$$\Rightarrow 115^{\circ}+ PRQ=180^{\circ}$$ $$\Rightarrow PRQ=180^{\circ}-115^{\circ}=65^{\circ}$$

Q14 :  In the figure, $$\angle X=62^{\circ}\;\; \angle XYZ=54^{\circ}$$. If $$YO$$ and $$ZO$$ of  $$\angle XYZ\;\; and \;\; \angle XZY$$ respectively of $$\bigtriangleup XYZ$$, find $$\angle OZY \; \; and \angle YOZ$$.

Sol:

In $$\bigtriangleup XYZ$$, $$\angle XYZ+ \angle YZX+ \angle ZXY=180^{\circ}$$.       (Sum of interior angles of a triangle is $$180^{\circ}$$)

$$\Rightarrow 116^{\circ}+\angle YZX=180^{\circ}$$

$$\Rightarrow \angle YZX=180^{\circ}-116^{\circ}=64^{\circ}$$  …..(i)

$$YO$$ is the bisector of $$\angle XYZ$$

$$\Rightarrow \angle XYO=\angle OYZ=\frac{1}{2}\angle XYZ=\frac{1}{2}(54^{\circ})=27^{\circ}$$        …….(ii)

$$\Rightarrow ZO \; is \; the\; bisector\; of \;\angle YZX$$

$$∴ \angle XZO =\angle OZY=\frac{1}{2}\angle YZX=\frac{1}{2}(64^{\circ})=32^{\circ}$$     ……(iii)   (from equation (i))

In triangle $$\bigtriangleup OYZ$$, $$\angle OYZ, \; \;\angle OZY \; \; and \angle YOZ=180^{\circ}$$     (sum of interior angle of a triangle is 180°)

$$\Rightarrow 27^{\circ}+32^{\circ}+\angle YOZ=180^{\circ}$$    ( using equation (i) and (ii))

$$\Rightarrow 59^{\circ}+\angle YOZ=180^{\circ}$$ $$\Rightarrow \angle YOZ=180^{\circ}-59^{\circ}=121^{\circ}$$

Q15 : In the figure,$$AB=\frac{1}{2}DE$$,$$\angle BAC=35^{\circ} and \angle CDE=53^{\circ}$$,find $$\angle DCE$$.

Sol:

$$\angle DEC=\angle BAC=35^{\circ}$$ ….(i)    (alternate interior angles)

$$\angle CDE=53^{\circ}$$       ……..(ii)       (given)

In $$\bigtriangleup CDE, \angle CDE+\angle DEC+\angle DCE=180°$$    (as sum of interior angles of a triangle is $$180^{\circ}$$.

$$\Rightarrow 53^{\circ}+35^{\circ}+\angle DCE=180^{\circ}$$ $$\Rightarrow 88^{\circ}+\angle DCE=180^{\circ}$$ $$\Rightarrow \angle DCE=180^{\circ}-88^{\circ}=92^{\circ}$$

Q16 : In the figure,if lines $$PQ\; \;and RS$$intersect at point $$T,$$such that $$\angle PRT=40^{\circ} \angle RPT=95^{\circ}, \angle SQT=75^{\circ} , \;\; find \angle SQT$$.

Sol:

In In $$\bigtriangleup PRT, \angle PTR+\angle PRT+\angle RPT=180°$$    (as sum of interior angles of a triangle is $$180°$$).

$$\Rightarrow \angle PTR+40^{\circ}+95^{\circ}=180^{\circ}$$ $$\Rightarrow \angle PTR+135^{\circ}=180^{\circ}$$ $$\Rightarrow \angle PTR=180^{\circ}-135^{\circ}$$ $$\Rightarrow \angle PTR=45^{\circ}$$

$$\Rightarrow \angle QTR=\angle PTR=45^{\circ}$$   (vertically opposite angles)

In $$\bigtriangleup TSQ, \angle QTS+\angle TSQ+\angle SQT=180^{\circ}$$    (as sum of interior angles of a triangle is $$180^{\circ}$$.

$$\Rightarrow 45^{\circ}+75^{\circ}+\angle SQT=180^{\circ}$$ $$\Rightarrow 120^{\circ}+\angle SQT=180^{\circ}$$ $$\Rightarrow \angle SQT=180^{\circ}-120^{\circ}$$ $$\Rightarrow \angle SQT=60^{\circ}$$

Q17 :  In the figure, if $$PQ\perp PS,\; \angle SQR=28^{\circ}\;\;and\; \; \angle QRT=65^{\circ}$$,then find the value of $$x\;\; and\; \; y$$.

Sol:

$$\angle QRT=\angle RQS+\angle QSR$$ (exterior angle is equal to sum of the two opposite interior angles)

$$\Rightarrow 65^{\circ}=28^{\circ}+\angle QSR$$ $$\Rightarrow \angle QSR=65^{\circ}-28^{\circ}=37^{\circ}$$

$$PQ\perp SP$$     (given)

$$\angle QPS+\angle PSR=180^{\circ}$$     (the sum of consecutive interior angles on the same side of the transversal in $$180^{\circ}$$ )

$$\Rightarrow 90^{\circ}+\angle PSR=180^{\circ}$$ $$\Rightarrow \angle PSR=180^{\circ}-90^{\circ}=90^{\circ}$$ $$\Rightarrow \angle PSQ+\angle QSR=90^{\circ}$$ $$\Rightarrow y+37^{\circ}=90^{\circ}$$ $$\Rightarrow Y=90^{\circ}-37^{\circ}=53^{\circ}$$

In $$\bigtriangleup PSQ, \angle PSQ+\angle QSP+\angle QPS=180^{\circ}$$    (as sum of interior angles of a triangle is $$180^{\circ}$$.

$$\Rightarrow x+y+90{\circ}=180^{\circ}$$ $$x+53^{\circ}+90^{\circ}=180^{\circ}$$ $$x+143^{\circ}=180^{\circ}$$ $$x=180^{\circ}-143^{\circ}=37^{\circ}$$

Q18 :  In the figure the side $$QR \;\; of \bigtriangleup PQR$$ is produced to a point $$S$$. If the bisectors of $$\angle PQR\;\; and \angle PRS$$ meet at point $$T$$, then prove that $$\angle QTR=\frac{1}{2}\angle QPR$$.

Sol:

$$\angle TRS$$ is an exterior angle of $$\bigtriangleup TQR$$

$$\angle TRS=\angle TQR+\angle QTR$$   …..(i)     (Since, the exterior angle is equal to the sum of the two interior opposite angles)

$$\angle PRS$$ is an exterior angle of $$\bigtriangleup PQR$$

$$\angle PRS=\angle PQR+\angle QPR$$   ……(ii)     (since, the exterior angle is equal to the sum of the two interior opposite angles)

$$\Rightarrow 2\angle TRS=2\angle TQR+\angle QPR$$   ($$QT$$ is the bisector of $$\angle PQR \;and\; RT$$ is the bisector of $$\angle PRS$$

$$\Rightarrow \angle TRS$$

$$\Rightarrow 2(\angle TRS-\angle TQR)=\angle QPR$$   ……(iii)

From (i), $$\angle TRS-\angle TQR=\angle QTR$$     ……(iv)

From (iii) and (iv) , we obtain

$$2\angle QTR=\angle QPR$$ $$\Rightarrow \angle QTR=\frac{1}{2}\angle QPR$$

## Practise This Question

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1. Shambhavi
March 17, 2018

Wow! This is really really helpful

May 25, 2018

In Question 5.2. below statement seems to be wrong. it should be angle 1+2 = 180, not angles 1+3 = 180
âˆ 1+âˆ 3=180âˆ˜