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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  \(\angle COE\).

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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:

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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:

4

 

\(x+y=w+z\)                                  …………….(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:

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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:

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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 \(AB\parallel CD\).

 

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Sol:

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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 \(x\).

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Sol:

 10

 

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 \(\angle AGE, \angle GEF \; and \; \angle FGE.\)

11

 

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 \(R\).

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Sol:

From the hint we draw a line \(RU\) parallel to \(ST\) through point \(R\).

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\(\angle RST+\angle SRU=180^{\circ}\)

(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 \(x and y.\)

14

 

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.)

 15

 Sol:

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 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 \(\angle PRQ\).

17

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 \(\angle OZY \; \; and \angle YOZ\).

 18

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 \(\angle DCE\).

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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 \(\angle PRT=40^{\circ} \angle RPT=95^{\circ}, \angle SQT=75^{\circ} , \;\; find \angle SQT\).

 20

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 \(x\;\; and\; \; y\).

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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 \(\angle QTR=\frac{1}{2}\angle QPR\).

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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

\(2\angle QTR=\angle QPR\) \(\Rightarrow \angle QTR=\frac{1}{2}\angle QPR\)

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