Question

In the given figure, tangents PQ and PR are drawn from an external point to a circle with centre O, such that $\angle \mathrm{RPQ}=30°$. A chord RS is drawn parallel to the tangent PQ. Find $\angle \mathrm{RQS}$. [CBSE 2015]

Answer 1

It is given that tangents PQ and PR are drawn from an external point P to a circle with centre O, such that $\angle \mathrm{RPQ}=30°$. Also, RS || PQ.

Join OR and OS.

PQ and PR are tangents drawn from an external point P to a circle.

∴ PQ = PR (Lengths of tangents drawn from an external point to a circle are equal)

In ∆PQR,

PQ = PR

∴ $\angle \mathrm{PRQ}=\angle \mathrm{PQR}$ (In a triangle, equal sides have equal angles opposite to them)

Now,

$\angle \mathrm{PRQ}+\angle \mathrm{PQR}+\angle \mathrm{RPQ}=180°$ (Angle sum property)
$$\begin{gathered} \Rightarrow 2\angle \mathrm{PRQ}+30°=180° \\ \Rightarrow 2\angle \mathrm{PRQ}=180°-30°=150° \\ \Rightarrow \angle \mathrm{PRQ}=75° \\ \therefore \angle \mathrm{PQR}=\angle \mathrm{PRQ}=75°.....\left(1\right) \end{gathered}$$

Since SR || PQ and RQ is the transversal,

∴ $\angle \mathrm{SRQ}=\angle \mathrm{RQP}=75°$ (Alternate angles)

Now, PR is the tangent and OR is the radius through the point of contact R.

∴ $\angle \mathrm{ORP}=90°$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

$$\begin{gathered} \Rightarrow \angle \mathrm{ORQ}=\angle \mathrm{ORP}-\angle \mathrm{PRQ}=90°-75°=15°\left[\mathrm{Using}\left(1\right)\right] \\ \therefore \angle \mathrm{ORS}=\angle \mathrm{SRQ}-\angle \mathrm{ORQ}=75°-15°=60° \end{gathered}$$

In ∆SOR,

OR = OS (Radii of the circle)

∴ $\angle \mathrm{OSR}=\angle \mathrm{ORS}=60°$ (In a triangle, equal sides have equal angles opposite to them)

Now,

$\angle \mathrm{ORS}+\angle \mathrm{OSR}+\angle \mathrm{ROS}=180°$ (Angle sum property)
$$\begin{gathered} \Rightarrow 60°+60°+\angle \mathrm{ROS}=180° \\ \Rightarrow \angle \mathrm{ROS}=180°-120°=60° \end{gathered}$$

We know that the angle subtended by an arc at the centre is twice the angle subtended by it any point on the remaining part of the circle.

$$\begin{gathered} \therefore \angle \mathrm{ROS}=2\angle \mathrm{RQS} \\ \Rightarrow \angle \mathrm{RQS}=\frac{1}{2}\angle \mathrm{ROS}=\frac{1}{2}\times 60°=30° \end{gathered}$$