Question

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 $\angle BQR={70}^{\circ }$ , then $\angle$AQB is equal to
(A) 20°
(B) 40°
(C) 35°
(D) 45°

Answer 1

Given , AB $\parallel$ PR

$\therefore \angle ABQ=\angle BQR={70}^{\circ }\left[Alternateangles\right]Also,QDisperpendiculartoABandQDbisectsAB.In\Delta QDAand\Delta QDB,\angle QDA=\angle QDB\left[Each{90}^{\circ }\right]AD=BDQD=QD\left[Commonsides\right]\therefore \Delta ADQ\cong \Delta BDQ\left[BySASsimilarlycriterion\right]Then\angle QAD=\angle QBD\left[CPTC\right]\left(i\right)Also\angle ABQ=\angle BQR\left[Alternateinteriorangle\right]\therefore \angle ABQ={70}^{\circ }[\angle BQR={70}^{\circ }]Hence\angle QAB={70}^{\circ }Nowin\Delta ABQ,\angle A+\angle B+\angle Q={180}^{\circ }\Rightarrow \angle Q={180}^{\circ }-({70}^{\circ }+{70}^{\circ })={40}^{\circ }$