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Answer :
Here, as PQ || SR, by alternate angles axiom, we get,
\(\angle{PQR}\) = \(\angle{QRT}\)
It is given that,
\(\angle{PQS}\) = x, \(\angle{SQR}\) = \(28^\circ\) and \(\angle{QRT}\) = \(65^\circ\)
\(\therefore \) \(\angle{PQS}\) + \(\angle{SQR}\) = \(\angle{QRT}\)
\(\Rightarrow \) x + \(28^\circ\) = \(65^\circ\)
\(\Rightarrow \) x = \(37^\circ\) ...(i)
Now, considering right angled triangle PQS,
\(\angle{SPQ}\) = \(90^\circ\)
\(\angle{SPQ}\) + x + y = \(180^\circ\) ....(Since Sum of all angles of a triangle is equal to 180)
\(\therefore \) \(90^\circ\) + \(37^\circ\) + y = \(180^\circ\)
\(\Rightarrow \) y = \(180^\circ\) - \(127^\circ\)
\(\Rightarrow \) y = \(53^\circ\)