Answered

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Pls tell the answer fast please​

Pls Tell The Answer Fast Please class=

Sagot :

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°

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