Question

In fig 9.18, 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 $\angle RQSindegrees$.Hint: Draw a line through Q and perpendicular to QP.

Answer 1

It is given that, $\angle RPQ={30}^o$ and $PR$ and $PQ$ are tangents drawn from $P$ to the same circle.

Hence $PR=PQ$ [Since tangents drawn from an external point to a circle are equal in length]

$\therefore$ $\angle PRQ=\angle PQR$ [Angles opposite to equal sides are equal in a triangle. ]

In $△PQR,$

$\angle RQP+\angle QRP+\angle RPQ={180}^o$ [Angle sum property of a triangle ]

$\Rightarrow$ $2\angle RQP+{30}^o={180}^o$

$\Rightarrow$ $2\angle RQP={150}^o$

$\Rightarrow$ $\angle RQP={75}^o$

so $\angle RQP=\angle QRP={75}^o$

$\Rightarrow$ $\angle RQP=\angle RSQ={75}^o$ [ By Alternate Segment Theorem]

Given, $RS\parallel PQ$

$\therefore$ $\angle RQP=\angle SRQ={75}^o$ [Alternate angles]

$\Rightarrow$ $\angle RSQ=\angle SRQ={75}^o$

$\therefore$ $QRS$ is also an isosceles triangle. [Since sides opposite to equal angles of a triangle are equal.]

$\Rightarrow$ $\angle RSQ+\angle SRQ+\angle RQS={180}^o$ [Angle sum property of a triangle]

$\Rightarrow$ ${75}^o+{75}^o+\angle RQS={180}^o$

$\Rightarrow$ ${150}^o+\angle RQS={180}^o$

$\therefore$ $\angle RQS={30}^o$