Chapter 9.3, Problem 21WE

To determine

To Prove: that $OR$ is tangent to $\odot Q$ and $QR$ is tangent to $\odot O$ shown in figure.

Explanation of Solution

Given:

Figure is drawn as.

  

  $m\stackrel{⌢}{RVS}=60\text{ and }m\stackrel{⌢}{RUS}=120$.

Calculation:

  $m\stackrel{⌢}{RVS}=60$ means central angle in the circle is given as.

  $\angle ROS=60$and $\angle RQS=120$.

In triangle ROQ and SOQ.

  $\begin{array}{l}QS=QS\\ OR=OS\\ RQ=SQ\end{array}$

By SSS postulate, $\Delta ROQ\sim \Delta SOQ.$

Therefore, by CPCT.

  $\angle ROQ\cong \angle SOQ\text{and }\angle RQO\cong \angle SQO$

In triangle ROQ.

  $\begin{array}{l}\angle ROQ+\angle R+\angle RQO=180\\ 30+\angle R+60=180\therefore \angle ROQ=\frac{1}{2}\angle ROS=30\text{ and }\angle RQO=\frac{1}{2}\angle RQS=60\\ \angle R={90}^{\circ }\end{array}$

In triangle SOQ.

  $\begin{array}{l}\angle SOQ+\angle S+\angle SQO=180\\ 30+\angle S+60=180\therefore \angle SOQ=\frac{1}{2}\angle ROS=30\text{ and }\angle SQO=\frac{1}{2}\angle RQS=60\\ \angle S={90}^{\circ }\end{array}$

Since, $\angle R={90}^{\circ }$ and $\angle S={90}^{\circ }$ therefore, $OR$ is tangent to $\odot Q$ and $QR$ is tangent to $\odot O$ .

It has been proved.