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Answer :
Given that, ray YQ bisects \(\angle{ZYP}\)
Thus, \(\angle{ZYQ}\) = \(\angle{QYP}\) = \(\frac{1}{2} \) (\(\angle{ZYP}\)) ...(i)
By Linear pair axiom,
\(\angle{XYZ}\) + \(\angle{ZYQ}\) + \(\angle{QYP}\) = \(180^\circ\)
But, it is given that,
\(\angle{XYZ}\) = \(64^\circ\) ...(ii)
Therefore, from (i) and (ii), we get,
\(\therefore \) \(64^\circ\) + \(\angle{ZYQ}\) + \(\angle{ZYQ}\) = \(180^\circ\)
\(\Rightarrow \) 2\(\angle{ZYQ}\) = \(180^\circ\) - \(64^\circ\)
\(\Rightarrow \) 2\(\angle{ZYQ}\) = \(116^\circ\)
\(\Rightarrow \) \(\angle{ZYQ}\) = \(58^\circ\)
Now,
\(\because \) \(\angle{XYQ}\) = \(\angle{XYZ}\) + \(\angle{ZYQ}\)
\(\Rightarrow \) \(\angle{XYQ}\) = \(64^\circ\) + \(58^\circ\) = \(122^\circ\)
Now,
\(\angle{QYP}\) + reflex\(\angle{QYP}\) = \(360^\circ\)
\(\Rightarrow \) \(58^\circ\) + reflex\(\angle{QYP}\) = \(360^\circ\)
\(\Rightarrow \) reflex\(\angle{QYP}\) = \(302^\circ\)