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Lines and Angles : Exercise 6.1 (Mathematics NCERT Class 9th)

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Q.1    In 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.


Since OA and OB are opposite rays. Therefore AB is a line.
Since ray OC stands on AB. Therefore,
\angle AOC + \angle COB = 180^o [Linear Pairs]
 \Rightarrow \angle AOC + \angle COE + \angle BOE = 180^o [Since \angle COB = \angle COE + \angle BOE]
 \Rightarrow \left( {\angle AOC + \angle BOE} \right) + \angle COE = 180^o
 \Rightarrow 70^o + \angle COE = 180^o [Since \angle AOC + \angle BOE = 70^o (Given)]
 \Rightarrow \angle COE = 180^o - 70^o = 110^o
Therefore Reflex \angle COE = 360^o - 110^o = 250^o
Since OC and OD are opposite rays. Therefore CD is a line.
Since ray OE stands on CD. Therefore -
\angle COE + \angle EOD = 180^o [Linear Pairs]
 \Rightarrow \angle COE + \angle BOE + \angle BOD = 180^o[Since \angle EOD = \angle BOE + \angle BOD]
 \Rightarrow 110^o + \angle BOE + 40^o = 180^o
[Since \angle COE = 110^o\,\left( {proved\,\,above} \right),\angle BOD = 40^o (Given)]
 \Rightarrow \angle BOE = 180^o - 110^o - 40^o = 30^o
Hence, \angle BOE = 30^o
and reflex\,\angle COE = 250^o

Q.2       In figure , lines XY and MN intersect at O. If \,\angle POY = 90^o and a : b = 2 : 3, find c.


Since a : b = 2 : 3 and a + b = \angle POX = \angle POY = 90º and sum of ratios = 2 + 3 = 5
Therefore a = {2 \over 5} \times 90^o= 2 \times 18^o = 36^o
and b = {3 \over 5} \times 90^o= 3 \times 18^o = 54^o
Since OM and ON are opposite rays. Therefore MN is a line.
Since ray OX stands on MN. Therefore,
\angle MOX + \angle XON = 180{\,^o}\, [Linear Pairs]
 \Rightarrow \,\,\,b + c = {180^o}
 \Rightarrow 54^\circ + c = {180^o}
 \Rightarrow c = {180^o} - {54^o} = {126^o}
Hence , c = {126^o}

Q.3       In figure \angle PQR = \angle PRQ, then prove that \angle PQS = \angle PRT.


Since QS and QR are opposite rays. Therefore, SR is a line.
Since QP stands on the line SR.
Therefore \angle PQS + \angle PQR = 180^o [Linear Pair] ...(1)
Again RQ and RT are opposite rays. Therefore, QT is a line.
Since PR stands on the line QT.
Therefore \angle PRQ + \angle PRT = 180^o [Linear Pair] ... (2)
From (1) and (2), we have
\angle PQS + \angle PQR = \angle PRQ + \angle PRT [Since Each side = 180º] ... (3)
Also \angle PQR = \angle PRQ       (given)      ................(4)
Subtracting (4) from (3), we have
\angle PQS = \angle PRT

Q.4       In figure, if x + y = w + z, then prove that AOB is a line.


We know that the sum of all angles around a point is equal to  360º
Therefore \left( {\angle BOC + \angle COA} \right) + \left( {\angle BOD + \angle AOD} \right) = 360^o
 \Rightarrow \left( {x + y} \right) + \left( {w + z} \right) = 360^o
But x + y = w + z [Given]
Therefore x + y = w + z = {{{{360}^o}} \over 2} = {180^o}
Thus , \angle BOC\,\,and\,\,\angle COA,\;\;\angle BOD\,\,and\,\,\angle AOD form linear pairs. Consequently OA and OB are two opposite rays.
Therefore AOB is a straight line.

Q.5       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 ROS\, = {1 \over 2}\left( {\angle QOS - \angle POS} \right)


Since OR is perpendicular to the line PQ.
Therefore \angle POR = \angle ROQ [Since Each = 90º]
 \Rightarrow \angle POS + \angle ROS = \angle QOS - \angle ROS
 \Rightarrow 2\angle ROS = \angle QOS - \angle POS
 \Rightarrow \angle ROS = {1 \over 2}\left( {\angle QOS - \angle POS} \right)

Q.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.

Since XY is produced to point P. Therefore XP is a straight line.
Since YZ stands on XP.
Therefore \angle XYZ + \angle ZYP = 180^o [Linear Pair]
 \Rightarrow 64^o + \angle ZYP = 180^o [Since \angle XYZ = 64^o]
 \Rightarrow \angle ZYP = 180^o - 64^o = 116^o
Since ray YQ bisects \angle ZYP

Therefore , \angle QYP = \angle ZYQ = {{{{116}^o}} \over 2} = 58^\circ
Now , \angle XYQ = \angle XYZ + \angle ZYQ
 \Rightarrow \angle XYQ = 64^o + 58^o = 122^o
and reflex \angle QYP = 360^o - \angle QYP = 360^o- 58^o= 302^o


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