Question

In the figure, tangents PQ and PR are drawn to a circle such that $\angle RPQ={30}^0$. A chord RS is drawn parallel to the tangent PQ. Find the $\angle$RQS.

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${75}^0$

Answer 1

Answer: (A)

${30}^0$

$PR=PQ$

$\Rightarrow \angle PRQ=\angle PQR=2\angle PQR.$

So, in $\Delta PQR$ we have,

$2\angle PQR+\angle QPR={180}^o$ ....(angle sum property of triangles).

$\therefore 2\angle PQR+{30}^o={180}^o$

$\Rightarrow \angle PQR={75}^o=\angle PRQ$ .........(i)

Again, $\angle PQR=\angle QSR={75}^o$ ...(by alternate segment theorem) ...(from i)....(ii)

Also, $SR\parallel PQ.\Rightarrow \angle PQR={75}^o=\angle QRS$ ......(iii)

$\therefore$ In $\Delta SRQ$, we have

$\angle RQS+\angle QSR+\angle SRQ={180}^o$ .....(Angle Sum property of triangles.)

$\Rightarrow \angle RQS={180}^o-(\angle QSR+\angle SRQ)$

$\Rightarrow \angle RQS={180}^o-({75}^o+{75}^o)={30}^o$ .... (from (i) & (ii))