In Fig. 6.3, if OP||RS, OPQ = 110° and QRS = 130°, then PQR is equal to


In the figure, it is given to us that –


OP || RS


OPQ = 110°


QRS = 130°


We have to find PQR.


Let us extend OP, so as to intersect QR at point D. The figure is as -


Now, we have OP || RS and, DR is a transversal.


We know, if a transversal intersects two parallel lines, then each pair of alternate angles are equal.


DRS = RDP


It is given to us that QRS = 130°


DRS = 130°


RDP = 130° - - - - (i)


Now, we have QR as a line segment. By linear pair axiom,


PDQ + PDR = 180°


QDP + RDP = 180°


QDP = 180° - RDP


QDP = 180° - 130°


QDP = 50°


PDQ = 50° - - - - (ii)


In ΔPQD,


OPQ is an exterior angle.


Also, we know that exterior angle is equal to the sum of the two interior opposite angles.


OPQ = PQD + PDQ - - - - (iii)


We have OPQ = 110° and from equation (ii) we have PDQ = 50°.


So, equation (iii) can be written as –


110° = PQD + 50°


PQD = 60°


So, PQR = 60°


Thus, option (C) is correct.


Option (A) is not correct. If PQR is equal to 40°, then it won’t satisfy the linear axiom. So, PQR is not 40°.


Option (B) is not correct. We have got PDQ = 50°. If PQR = 50°, then the exterior angle, i.e., OPQ = 110° is not equal to the sum of the opposite interior angles, PDQ and PQR which are now 50° each. So, PQR is not equal to 50°.


Option (D) is not correct. We have, OPQ = 110°. By linear pair axiom, QPD = 70°. So, PQR is not equal to 700

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