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
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Solution
Given , AB ∥ PR ∴∠ABQ=∠BQR=70∘[Alternateangles]Also,QDisperpendiculartoABandQDbisectsAB.InΔQDAandΔQDB,∠QDA=∠QDB[Each90∘]AD=BDQD=QD[Commonsides]∴ΔADQ≅ΔBDQ[BySASsimilarlycriterion]Then∠QAD=∠QBD[CPTC](i)Also∠ABQ=∠BQR[Alternateinteriorangle]∴∠ABQ=70∘[∠BQR=70∘]Hence∠QAB=70∘NowinΔABQ,∠A+∠B+∠Q=180∘⇒∠Q=180∘−(70∘+70∘)=40∘