# Lines and Angles : Exercise 6.3 (Mathematics NCERT Class 9th) Q.1     In figure sides QP and RQ of $\Delta$ PQR are produced to points S and T respectively. If $\angle SPR =$ 135º and $\angle PQT =$ 110º, find $\angle PRQ.$

We have ,
$\angle QPR + \angle SPR =$ 180º [Linear pair]
$\Rightarrow$ $\angle QPR$ + 135º = 180º
$\Rightarrow$ $\angle QPR =$ 180º – 135º = 45º
Now , $\angle TQP = \angle QPR + \angle PRQ$ [By exterior angle theorem]
$\Rightarrow$ 110º = 45º + $\angle PRQ$
$\Rightarrow$ $\angle PRQ$ = 110º – 45º = 65º
Hence $\angle PRQ$ = 65º

Q.2      In figure, $\angle X =$ 62º, $\angle XYZ =$ 54º. If YO and ZO are the bisectors of $\angle XYZ\,\,and\,\,\angle XZY$ respectively of $\Delta$ XYZ, find $\angle OZY\,\,and\,\,\angle YOZ$

Consider $\Delta \,$ XYZ,
$\angle YXZ + \angle XYZ + \angle XZY$ = 180º [Angle - sum property]
$\Rightarrow$ 62º + 54º + $\angle XZY$ = 180º [Since $\angle YXZ =$ 62º , $\angle XYZ =$ 54º]
$\Rightarrow \angle XZY = 180^\circ - 62^\circ - 54^\circ = 64^\circ$
Since YO and ZO are bisectors of $\angle XYZ\,\,and\,\,\angle XZY$
Therefore
$\angle OYZ = {1 \over 2} \times \angle XYZ = {1 \over 2} \times 54 = 27^\circ$
and , $\angle OZY = {1 \over 2} \times \angle XZY = {1 \over 2} \times 64^\circ = 32^\circ$
In $\Delta$ OYZ , we have
$\angle YOZ + \angle OYZ + \angle OZY =$ 180º [Angle sum property]
$\Rightarrow$ $\angle YOZ +$ 27º + 32º = 180º
$\Rightarrow$ $\angle YOZ$ = 180º – 27º – 32º
= 180º – 59º = 121º
Hence, $\angle OZY =$ 32º
and $\angle YOZ =$ = 121º

Q.3      In figure if AB || DE, $\angle BAC$ = 35º and $\angle CDE =$ 53º, find $\angle DCE$.

Since AB || DE and transversal AE intersects them at A and E respectively.
Therefore $\angle DEA = \angle BAE$ [Alternate angles]
$\Rightarrow$ $\angle DEC =$ 35º [Since $\angle DEA = \angle DEC\,and\,\angle BAE =$ 35º]
In $\Delta$ DEC, we have
$\angle DCE + \angle DEC + \angle CDE =$ 180º [Angle sum property]
$\Rightarrow$ $\angle DCE +$ 35º + 53º = 180º
$\Rightarrow$ $\angle DCE =$ 180º – 35º – 53º
= 180º – 88º = 92º
Hence , $\angle DCE =$ 92º

Q.4      In figure, if lines PQ and RS intersect at point T, such that $\angle PRT =$ 40º, $\angle RPT =$ 95º and $\angle TSQ =$ 75º, find $\angle SQT$.

In $\Delta$ PRT, we have
$\angle PRT + \angle RTP + \angle TPR =$ 180º [Angle - sum property]
$\Rightarrow$ 40º + $\angle RTP +$ 95º = 180º
$\Rightarrow$ $\angle RTP =$ 180º – 40º – 95º
= 180º – 135º = 45º
$\angle STQ = \angle RTP$ [Vertically opp. angles]
$\Rightarrow$ $\angle STQ =$ 45º [Since RTP = 45º (proved)]
In $\Delta$ TQS we have
$\angle SQT + \angle STQ + \angle TSQ =$ 180º [Angle - sum property]
$\Rightarrow$ $\angle SQT +$ 45º + 75º = 180º [Since $\angle STQ = 45^ \circ$(proved) and $\angle TSQ = 75^ \circ$]
$\Rightarrow$ $\angle SQT =$ 180º – 45º – 75º
= 180º – 120º = 60º
Hence , $\angle SQT =$ 60º

Q.5      In figure if PQ $\bot$ PS, PQ|| SR, $\angle SQR$ = 28º and $\angle QRT$ = 65º, then find the values of x and y.

Using exterior angle property in $\Delta$ SRQ, we have
$\angle QRT = \angle RQS + \angle QSR$
$\Rightarrow$ 65º = 28º + $\angle QSR$ [Since $\angle QRT =$ 65º , $\angle RQS =$ 28º]
$\Rightarrow$ QSR = 65º – 28º = 37º
Since PQ|| SR and the transversal PS intersects them at P and S respectively.
Therefore $\angle PSR + \angle SPQ =$ 180º [Sum of consecutive interior angles is 180º]
$\Rightarrow$ $\left( {\angle PSQ + \angle QSR} \right) +$ 90º = 180º
$\Rightarrow$ y + 37º + 90º = 180º
$\Rightarrow$ y = 180º – 90º – 37º
= 180º – 127º = 53º
In the right $\Delta \,SPQ$, we have
$\angle PQS + \angle PSQ =$ 90º
$\Rightarrow$ x + 53º = 90º
$\Rightarrow$ x = 90º – 53º = 37º
Hence, x = 37º
and y = 53º

Q.6    In 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 $\angle QTR = {1 \over 2}\angle QPR$.

In $\Delta \,PQR\,$ we have ext. $\angle \,PRS = \angle P + \angle Q\,$
$\Rightarrow$ ${1 \over 2}ext.\angle PRS = {1 \over 2}\angle P + {1 \over 2}\angle Q$   ( By exterior angle theorem )
$\Rightarrow$ $\angle TRS = {1 \over 2}\angle P + \angle TQR$ ... (1)
[Since QT and RT are bisectors of $\angle Q\,and\,\angle PRS$ respectively therefore $\angle Q\, = 2\,\angle TQR\,\,and\,\,ext.\,\angle PRS = 2\angle TRS$]
In $\Delta$ QRT. we have ext. $\angle TRS = \angle TQR + \angle T$ ... (2)
From (1) and (2) , we get
${1 \over 2}\angle P + \angle TQR = \angle TQR + \angle T$
$\Rightarrow$ ${1 \over 2}\angle P = \angle T$
$\Rightarrow$ $\angle QTR = {1 \over 2}\angle QPR$ [ Since $\angle P = \angle QPR and \angle T= \angle QTR$]
Hence , $\angle QTR = {1 \over 2}\angle QPR$

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