• # question_answer In Fig. 3, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If $OP=2r$, show that $\angle OTS=\angle OST=30{}^\circ$.

 We have, $OP=2r$ Let                                $\angle TOP=\theta$ In $\Delta \,OTP,$                      $\cos \theta =\frac{OT}{OP}=\frac{r}{2r}=\frac{1}{2}$ $\because$                               $\theta =60{}^\circ$ Hence,                          $\angle TOS=2\theta =2\times 60{}^\circ =120{}^\circ$ In $\Delta \,TOS$ $\angle \,TOS+\angle OTS+\angle OST=180{}^\circ$ $120{}^\circ +2\angle OTS=180{}^\circ (\because \,\angle OTS=\angle OST)$ $2\angle OTS=180{}^\circ -120{}^\circ$ $\angle OTS=30{}^\circ$ Hence,              $\angle OTS=\angle OST=30{}^\circ$              Hence Proved.