From an external point P , …

Given,

In {tex}\triangle{/tex}OTS,
{tex}​​​OT =OS  {/tex}
{tex}\Rightarrow \quad \angle O T S = \angle O S T{/tex} ...(i)
In right {tex}\triangle OTP,{/tex}
 {tex}\frac { \mathrm { OT } } { \mathrm { OP } } = \sin \angle \mathrm { TPO }{/tex}
{tex}\Rightarrow \frac { r } { 2 r } = \sin \angle \mathrm { TPO }{/tex}
{tex}\sin \angle \mathrm { TPO } = \frac { 1 } { 2 } \Rightarrow \angle \mathrm { TPO } = 30 ^ { \circ }{/tex}
Similarly {tex}\angle \mathrm { OPS } = 30 ^ { \circ }{/tex}
{tex}\Rightarrow \angle T P S = 30 ^ { \circ } + 30 ^ { \circ } = 60 ^ { \circ }{/tex}
Also {tex}\angle \mathrm { TPS } + \angle \mathrm { SOT } = 180 ^ { \circ }{/tex}
{tex}\Rightarrow \quad \angle \mathrm { SOT } = 120 ^ { \circ }{/tex}
In {tex}\triangle{/tex}SOT,
{tex}\angle \mathrm { SOT } + \angle \mathrm { OTS } + \angle \mathrm { OST } = 180 ^ { \circ }{/tex}
{tex}\Rightarrow 120 ^ { \circ } + 2 \angle \mathrm { OTS } = 180 ^ { \circ }{/tex}
{tex}\Rightarrow \angle \mathrm { OTS } = 30 ^ { \circ }{/tex} ...(ii)
From (i) and (ii) 
{tex}\angle \mathrm { OTS } = \angle \mathrm { OST } = 30 ^ { \circ }{/tex}