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Lines and Angles : Exercise 6.3 (Mathematics NCERT Class 9th)


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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.

18Sol.

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

19Sol.

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.

21Sol.

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.

20Sol.

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.

22Sol.

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.

23Sol.

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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