In Fig. 6.1, if AB || CD || EF, PQ || RS, RQD = 25° and CQP = 60°, then QRS is equal to


From the given figure, we have


AB || CD || EF


PQ || RS


RQD = 25°


CQP = 60°


Now, we have PQ || RS.


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


Since, PQ || RS


PQC = BRS


We have PQC = 60°


BRS = 60° - - - - (i)


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


Since, AB || CD


DQR = QRA


We have DQR = 25°


QRA = 25° - - - - (ii)


By linear pair axiom,


ARS + BRS = 180°


ARS = 180° - BRS


ARS = 180° - 60° (From (i), BRS = 60°)


ARS = 120° - - - - (iii)


Now, QRS = QRA + ARS


From equations (ii) and (iii), we have QRA = 25° and ARS = 120°


Thus, the above equation can be written as:


QRS = 25° + 120°


QRS = 145°


Thus, the option (C) is correct.


Option (A) is not correct because 85°, i.e., 60° + 25° is not the correct angle combination for QRS.


Option (B) is not correct because 135°, is not the exact solution of QRS.


Option (D) is not correct because 110°, is not the correct solution of QRS.

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