Question

In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR, such that ∠BQR = 70°. Then, ∠AQB = ?


(a) 20°
(b) 35°
(c) 40°
(d) 45°

Answer 1

(c) 40°

$$\begin{gathered} \mathrm{S}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{}AB\parallel PR,BQ\mathrm{is}\mathrm{}\mathrm{transversal}. \\ \angle BQR=\angle ABQ={70}^0\left[\text{A}\mathrm{lternate}\mathrm{}\mathrm{angles}\right] \\ OQ\perp PQR(\text{T}\mathrm{angents}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{from}\mathrm{}\mathrm{an}\mathrm{}\mathrm{external}\mathrm{}\mathrm{point}\mathrm{}\mathrm{are}\mathrm{}\mathrm{perpendicular} \\ \mathrm{to}\mathrm{}\mathrm{the}\mathrm{}\mathrm{radius}\mathrm{}\mathrm{at}\mathrm{}\mathrm{the}\mathrm{}\mathrm{point}\mathrm{}\mathrm{of}\mathrm{}\mathrm{contact}) \\ andAB\parallel PQR \\ \therefore QL\perp AB;\mathrm{}\mathrm{so},OL\perp AB \\ \therefore OL\mathrm{bisects}\mathrm{}\mathrm{chord}AB\left[\text{P}\mathrm{erpendicular}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{from}\mathrm{}\mathrm{the}\mathrm{}\mathrm{centre}\mathrm{}\mathrm{bisects}\mathrm{}\mathrm{the}\mathrm{}\mathrm{chord}\right] \\ \mathrm{From}\mathrm{}∆QLAand∆QLB: \\ \angle QLA=\angle QLB={90}^0 \\ LA=LB\left(OL\mathrm{bisects}\mathrm{}\mathrm{chord}AB\right) \\ QL\mathrm{}\mathrm{is}\mathrm{}\mathrm{the}\mathrm{}\mathrm{common}\mathrm{}\mathrm{side}. \\ \therefore ∆QLA\cong ∆QLB\left[\mathrm{By}\mathrm{SAS}\mathrm{congruency}\right] \\ \therefore \angle QAL=\angle QBL \\ \Rightarrow \angle QAB=\angle QBA \\ \therefore ∆AQB\mathrm{}\mathrm{is}\mathrm{}\mathrm{isosceles}. \\ \therefore \angle LQA=\angle LQR \\ \angle LQP=\angle LQR={90}^0 \\ \angle LQB=\left({90}^0-{70}^0\right)={20}^0 \\ \therefore \angle LQA=\angle LQB={20}^0 \\ \Rightarrow \angle AQB=\angle LQA+\angle LQB \\ ={40}^0 \end{gathered}$$