In triangle PST, PS = PT \[ \therefore \angle PST = \angle PTS = \cfrac{180^\circ - 80^\circ}{2} = 50^\circ \] Again, since QS \(\parallel\) PT \[ \therefore \angle PSQ = 180^\circ - 80^\circ = 100^\circ \] Therefore, \[ \angle QST = \angle PSQ - \angle PST = 100^\circ - 50^\circ = 50^\circ \]