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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,

\(\Rightarrow \) \(\angle{TSQ}\) + \(\angle{SQT}\) = \(\angle{PTS}\)

\(\Rightarrow \) \(75^\circ\) + \(\angle{SQT}\) = \(135^\circ\)

\(\Rightarrow \) \(\angle{SQT}\) = \(135^\circ\) - \(75^\circ\)

\(\Rightarrow \) \(\angle{SQT}\) = \(60^\circ\)

- 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 \(\angle{PRQ}\).
- 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 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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