It is given that \(\angle{XYZ}\) = \(64^\circ\) and XY is produced to point P. Draw a figure from the given information.
If ray YQ bisects \(\angle{ZYP}\) , find \(\angle{XYQ}\) and reflex\(\angle{QYP}\).

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\(\angle{QYP}\) = \(302^\circ\)