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

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 $\angle QYP$.
Sol.

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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