Question

Secants containing chords RS and PQ of a circle intersects each other at point A in the exterior of a circle as shown in the given figure.
if m(arc APC) = 26$°$ and m(arc QDS) = 48$°$
then find (i) $\angle$AQR
(ii) $\angle$SPQ
(iii) $\angle$RAQ

Answer 1

Construction: Join OR, OP, OS and OQ.

We know that the angle subtended by the arc at the centre is twice the angle subtended at any part of the circle.

(i) $\angle$AQR = $\frac{1}{2}\mathrm{m}(\mathrm{arc}\mathrm{PCR})=\frac{1}{2}\times 26°=13°$

(ii) $\angle$SPQ = $\frac{1}{2}\mathrm{m}(\mathrm{arc}QDS)=\frac{1}{2}\times 48°=24°$

(iii)
$$\begin{gathered} \mathrm{In}△\mathrm{ARQ},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{QRS}=\angle \mathrm{RAQ}+\angle \mathrm{AQR}(\mathrm{Exterior}\mathrm{angle}\mathrm{property}) \\ \Rightarrow \angle \mathrm{SPQ}=\angle \mathrm{RAQ}+\angle \mathrm{AQR}(\because \angle \mathrm{QRS}=\angle \mathrm{SPQ},\mathrm{angle}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{segment}) \\ \Rightarrow 24°=\angle \mathrm{RAQ}+13° \\ \Rightarrow \angle \mathrm{RAQ}=24°-13°=11° \end{gathered}$$