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 \(\triangle{XYZ}\), \(\angle{X}\) + \(\angle{Y}\) + \(\angle{Z}\) = \(180^\circ\)

because, Sum of all angles of triangle is equal to 180
\(\therefore \) \(62^\circ\) + \(\angle{Y}\) + \(\angle{Z}\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{Y}\) + \(\angle{Z}\) = \(118^\circ\)

Now, so as to find bisected angles, multiply both sides by \(\frac{1}{2} \) We get,
\(\frac{1}{2} \) [\(\angle{Y}\) + \(\angle{Z}\)] = \(\frac{1}{2} \) × \(118^\circ\) = \(59^\circ\)

Thus we get,
\(\angle{OYZ}\) + \(\angle{OZY}\) = \(59^\circ\)
....(as YO and ZO are the bisectors of \(\angle{XYZ}\) and \(\angle{XZY}\))

Also, it is given that, \(\angle{XYZ}\) = \(54^\circ\) and we have
\(\Rightarrow \) \(\angle{OYZ}\) = \(\frac{1}{2} \) × \(\angle{XYZ}\)
\(\Rightarrow \) \(\angle{OZY}\) + \(\frac{1}{2} \) × \(54^\circ\) = \(59^\circ\)
\(\Rightarrow \) \(\angle{OZY}\) = \(59^\circ\) - \(27^\circ\) = \(32^\circ\)

Also, in \(\triangle{YOZ}\),

\(\angle{OYZ}\) + \(\angle{YOZ}\) + \(\angle{OZY}\) = \(180^\circ\)

because, Sum of all angles of triangle is equal to 180
\(\therefore \) \(27^\circ\) + \(\angle{YOZ}\) + \(32^\circ\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{YOZ}\) + \(59^\circ\) = \(180^\circ\)
\(\Rightarrow \) \(\angle{YOZ}\) = \(180^\circ\) - \(59^\circ\)
\(\Rightarrow \) \(\angle{YOZ}\) = \(121^\circ\)