(a)OR(b)

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that $\displaystyle \angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}$.

$\displaystyle \mathrm{TP}=\mathrm{TQ}$ $\displaystyle \Rightarrow \angle \mathrm{TPQ}=\angle \mathrm{TQP}$ Let $\displaystyle \angle \mathrm{PTQ}$ be $\displaystyle \theta$ $\displaystyle \Rightarrow \angle \mathrm{TPQ}=\angle \mathrm{TQP}=\frac{180^{\circ}-\theta}{2}=90^{\circ}-\frac{\theta}{2}$ Now $\displaystyle \angle \mathrm{OPT}=90^{\circ}$ $\displaystyle \Rightarrow \angle \mathrm{OPQ}=90^{\circ}-\left(90^{\circ}-\frac{\theta}{2}\right)=\frac{\theta}{2}$ $\displaystyle \angle \mathrm{PTQ}=2 \angle \mathrm{OPQ}$ $\displaystyle \mathrm{DR}=\mathrm{DS}=3 \mathrm{~cm}$ \[\begin{array}{l} \therefore \quad A R=A D-D R=17-3=14 \mathrm{~cm} \Rightarrow A Q=A R=14 \mathrm{~cm} \end{array} \] \[\therefore \quad Q B=A B-A Q=20-14=6 \mathrm{~cm} \] Since $\displaystyle \mathrm{QB}=\mathrm{OP}=\mathrm{r} \quad \therefore \quad$ radius $\displaystyle =6 \mathrm{~cm}$