Let, ∠ZYQ = ∠QYP = a°
Since, XYP is a straight line,
⇒ ∠XYZ + ∠ZYP = 180° (Linear Pair)
⇒∠ZYP = 180°- ∠XYZ
⇒∠ZYP = 180° - 64°
⇒∠ZYP = 116°
As it's given, YQ is the bisector of ∠ZYP,
⇒∠ZYQ = ∠QYP = ½ ZYP
= ½ × 116°
= 58°
⇒∠XYQ = ∠XYZ + ∠ZYQ
= 64° + 58°
= 112°
[Escape (∠XYQ value) if ya want because in Question it's not mentioned to find the value of ∠XYQ but we need need to find the value of ∠ZYQ, in order to find the reflex of it]
Now, We've to also find the reflex of
∠ZYQ,
⇒ Reflex of ∠ZYQ = 360° - ∠ZYQ
= 360° - 58°
= 302°
