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Answer :
Since, we know, exterior angle is equal to sum of interior opposite angles
Thus, we get,
\(\Rightarrow \) \(\angle{PTS}\) = \(\angle{RPT}\) + \(\angle{PRT}\)
\(\Rightarrow \) \(\angle{PTS}\) = \(95^\circ\) + \(40^\circ\)
\(\Rightarrow \) \(\angle{PTS}\) = \(135^\circ\) .....(given, \(\angle{PRT}\) = \(40^\circ\) and \(\angle{RPT}\) = \(95^\circ\))
Similarly,
\(\Rightarrow \) \(\angle{TSQ}\) + \(\angle{SQT}\) = \(\angle{PTS}\)
\(\Rightarrow \) \(75^\circ\) + \(\angle{SQT}\) = \(135^\circ\)
\(\Rightarrow \) \(\angle{SQT}\) = \(135^\circ\) - \(75^\circ\)
\(\Rightarrow \) \(\angle{SQT}\) = \(60^\circ\)