Q. 44.1( 72 Votes )

In fig 3.27, the

Answer :

(1)In ΔAPR,


AP = RP {Radius of the circle with centre P}


PAR = PRA … (1)


In ΔRQB,


RQ = QB {Radius of the circle with centre Q}


QRB = QBR …. (2)


PRA = QRB {Vertically Opposite Angle} ….(3)


PAR = QBR {From (1), (2) and (3)}


Alternate interior angles are equal.


AP || BQ


Hence, proved.


(2) In ∆ APR and ∆ RQB,


PAR = QBR and PRA = QRB {From (1) and (2)}


∆ APR ~ ∆ RQB {AA}


Hence, proved.


(4) Given: PAR = 35°


QBR = 35° = QRB {Proved previously}


In ∆ RQB,


RQB + QRB + QBR = 180° {Angle sum property of the triangle}


⇒∠ RQB + 35° + 35° = 180°


⇒∠ RQB = 180°- 70° = 110°


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