Question

In the figure, 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$={30}^{\circ }$.

Answer 1

Given: $OT=OS=r$ and $OP=2r$

In $\Delta TOP$,

$\sin TPO=\frac{TO}{OP}$

= $\frac{r}{2r}=\frac{1}{2}$

Since, $\sin {30}^o=\frac{1}{2}$

Therefore, $\angle TPO={30}^o$

Similarly for $\angle OPS={30}^o$

Now,

$\angle TPS=\angle TPO+\angle OPS$
= ${30}^o+{30}^o={60}^o$

As we know that $\angle TPS+\angle TOS={180}^o$

So, $\angle TOS={180}^o-\angle TPS$

= ${180}^o-{60}^o={120}^o$

Now, in $\Delta TOS$, let $\angle OST=\angle OTS=x^o$

Also, $\angle TOS+x^o+x^o={180}^o$

${120}^o+2x^o={180}^o$

$2x^o={60}^o$

$x^o={30}^o$

Therefore, $\angle OST=\angle OTS={30}^o.$