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Answer :

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.

\(\because \) \(\angle{RPQ}\) + \(\angle{PRQ}\) = \(\angle{PQT}\)

\(\Rightarrow \)\(45^\circ\) + \(\angle{PRQ}\) = \(110^\circ\)

\(\Rightarrow \)\(\angle{PRQ}\) = \(110^\circ\) - \(45^\circ\)

\(\Rightarrow \) \(\angle{PRQ}\) = \(65^\circ\).

- In figure, \(\angle{X}\) = \(60^\circ\) , \(\angle{XYZ}\) = \(54^\circ\) , if YO and ZO are the bisectors of \(\angle{XYZ}\) and \(\angle{XZY}\) respectively of \(\triangle{XYZ}\) , find \(\angle{OZY}\) and \(\angle{YOZ}\).
- In figure, if AB || DE , \(\angle{BAC}\) = \(35^\circ\) and \(\angle{CDE}\) = \(53^\circ\) , find \(\angle{DCE}\).
- In figure, if lines PQ and RS interest at point T, such that \(\angle{PRT}\) = \(40^\circ\) ,\(\angle{RPT}\) = \(95^\circ\) and \(\angle{TSQ}\) = \(75^\circ\) , find \(\angle{SQT}\).
- In figure, if PQ is perpendicular to PS , PQ || SR , \(\angle{SQR}\) = \(28^\circ\) and \(\angle{QRT}\) = \(65^\circ\) , then find the values of x and y.
- In figure, the side QR of \(\angle{PQR}\) is produced to a point S. If the bisectors of \(\angle{PQR}\) and \(\angle{PRS}\) meet at point T, then prove that \(\angle{QTR}\) = (1/2) \(\angle{QPR}\)

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