NCERT Class 9 Maths Solutions Chapter - 6 Lines And Angles, Ex - 6.3

Ex - 6.3

Question 1.  In the given figure, sides QP and RQ of PQR are produced to points S and T respectively. If SPR = 135� and PQT = 110�, find PRQ.



Solution

Given that
    SPR = 135� and PQT = 110�
    Now, SPR + QPR = 180�             (linear pair angles)
     135� + QPR = 180� 
     QPR = 45�                         
    Also, PQT + PQR = 180�             (linear pair angles)
     110� + PQR = 180�
    PQR = 70�     
    As we know that sum of all interior angles of a triangle is 180�, so, for PQR  
QPR + PQR + PRQ = 180� 
 45� + 70� + PRQ = 180� 
PRQ = 180� - 115�
 PRQ = 65�

Question 2.  In the given figure, X = 62�, XYZ = 54�. If YO and ZO are the bisectors of XYZ and XZY respectively of XYZ, find OZY and YOZ



Solution

As we know that sum of all interior angles of a triangle is 180�, so for XYZ
X + XYZ + XZY = 180�  
62� + 54� + XZY = 180�
XZY = 180� - 116�
XZY = 64�

OZY =   = 32�         (OZ is angle bisector of XZY)
Similarly, OYZ =  = 27�
Using angle sum property for OYZ, we have
OYZ + YOZ + OZY = 180º
27� + YOZ + 32� = 180�
YOZ = 180� - 59�
YOZ = 121�

Question 3.  In the given figure, if AB || DE, BAC = 35� and CDE = 53�, find DCE.

Solution

AB || DE and AE is a transversal                    
BAC =CED                 (alternate interior angle)
CED = 35�
In CDE,
CDE + CED + DCE = 180�         (angle sum properly of a triangle)
53� + 35� + DCE = 180�
DCE = 180� - 88�
DCE = 92�

Question 4.  In the given figure, if lines PQ and RS intersect at point T, such that PRT = 40�, RPT = 95� and

TSQ = 75�, find SQT.

Solution

Using angle sum property for PRT, we have
PRT + RPT + PTR = 180� 
40� + 95� + PTR = 180�
PTR = 180� - 135�
PTR = 45�
STQ = PTR = 45�             (vertically opposite angles)
STQ = 45�
By using angle sum property for STQ, we have
STQ + SQT + QST = 180� 
45� + SQT + 75� = 180�
SQT = 180� - 120�
SQT = 60� 

Question 5.  In the given figure, if PQ  PS, PQ || SR, SQR = 28� and QRT = 65�, then find the values of x and y.

Solution

Given that PQ || SR and QR is a transversal line
PQR = QRT             (alternate interior angles)
x + 28� = 65�
x = 65� - 28�
x = 37�
By using angle sum property for SPQ, we have
SPQ + x + y = 180� 
90� + 37� + y = 180�
y = 180� - 127�
y = 53�
 x = 37� and y = 53�.

Question 6.  In the given figure, the side QR of PQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that

QTR= QPR.

Solution

In QTR, TRS is an exterior angle.
 QTR + TQR = TRS
QTR = TRS - TQR        (1)
For PQR, PRS is external angle

QPR + PQR = PRS

QPR + 2TQR = 2TRS    (As QT and RT are angle bisectors)
QPR = 2(TRS - TQR)
QPR = 2QTR            [By using equation (1)]
QTR =  QPR

Ex-6.2  

Ex-7.1