PQRSTW is a regular hexagon and OS intersect RT at V. Calculate $\angle TVS$
60o
90o
120o
30o
$\begin{align} & \text{Since}PQRSTW~\text{isa regular},\text{ it means all the sides are equal } \\ & \text{and also all the angles are equal}. \\ & \overset{\wedge }{\mathop{R}}\,=\overset{\wedge }{\mathop{S}}\,=\overset{\wedge }{\mathop{T}}\,=\overset{\wedge }{\mathop{W}}\,=\overset{\wedge }{\mathop{P}}\,=\overset{\wedge }{\mathop{Q}}\,={{120}^{o}} \\ & PQ=QR=SR=ST=TW=WP\text{ } \\ & \text{ }\!\!\{\!\!\text{ Side of regular polygon }\!\!\}\!\!\text{ } \\ & \text{consider }\Delta QRS,\left| QR \right|=\left| SR \right|\text{ } \\ & \text{ }\!\!\{\!\!\text{ sides of an issoscele triangle }\!\!\}\!\!\text{ } \\ & Q\overset{\wedge }{\mathop{S}}\,R+S\overset{\wedge }{\mathop{R}}\,Q+S\overset{\wedge }{\mathop{Q}}\,R={{180}^{o}} \\ & x+{{120}^{o}}+x={{180}^{o}} \\ & x=30 \\ & T\overset{\wedge }{\mathop{S}}\,V=T\overset{\wedge }{\mathop{S}}\,R-Q\overset{\wedge }{\mathop{S}}\,R,\text{ }T\overset{\wedge }{\mathop{S}}\,V={{120}^{o}}-{{30}^{o}}={{90}^{o}} \\ & \text{Also, consider }\Delta TSR \\ & \left| TS \right|=\left| SR \right| \\ & \angle STR=\angle SRT\text{ }\!\!\{\!\!\text{ Base angle of issoscele triangle }\!\!\}\!\!\text{ } \\ & y+y+{{120}^{o}}={{180}^{o}} \\ & y={{30}^{o}} \\ & \text{Consider }\Delta TSV \\ & {{30}^{o}}+{{90}^{o}}+z=180 \\ & z={{60}^{o}} \\ & \angle TVS=z={{60}^{o}} \\\end{align}$