In figure, sides QP and RQ of \(\angle{PQR}\) are produced to points S and T, respectively. If \(\angle{SPR}\) = \(135^\circ\) and \(\angle{PQT}\) = \(110^\circ\) , find \(\angle{PRQ}\).

By linear pair axiom,
\(\because \) \(\angle{RPS}\) + \(\angle{RPQ}\) = \(180^\circ\)
\(\Rightarrow \) \(135^\circ\) + \(\angle{RPQ}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{RPQ}\) = \(180^\circ\) - \(135^\circ\)
\(\Rightarrow \), \(\angle{RPQ}\) = \(45^\circ\)

We also know that, Sum of interior opposite angles is equal to exterior angles.
\(\because \) \(\angle{RPQ}\) + \(\angle{PRQ}\) = \(\angle{PQT}\)
\(\Rightarrow \)\(45^\circ\) + \(\angle{PRQ}\) = \(110^\circ\)
\(\Rightarrow \)\(\angle{PRQ}\) = \(110^\circ\) - \(45^\circ\)
\(\Rightarrow \) \(\angle{PRQ}\) = \(65^\circ\).