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Question 10
In figure. If PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and BQR=70 , then AQB is equal to

(A) 20°
(B) 40°
(C) 35°
(D) 45°

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Solution

Given , AB PR

ABQ=BQR=70 [Alternate angles]Also, QD is perpendicular to AB and QD bisects AB.In ΔQDA and ΔQDB,QDA=QDB [Each 90]AD=BDQD=QD [Common sides]ΔADQΔBDQ [By SAS similarly criterion]Then QAD=QBD [CPTC](i)Also ABQ=BQR [Alternate interior angle]ABQ=70 [BQR=70]Hence QAB=70Now in ΔABQ,A+B+Q=180Q=180(70+70)=40



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