Given: \(∠\)XYZ = 64° and Ray YQ bisects \(∠\)PYZ.
To Find: \(∠\)XYQ and Reflex \(∠\)QYP
It can be observed that PX is a line. Rays YQ and YZ stand on it.
Hence, \(∠\)XYZ +\(∠\)ZYP = 180°
By substituting, \(∠\)XYZ = 64°, we get
64° +\(∠\)ZYP = 180°
∴ \(∠\)ZYP = 116°
As YQ bisects \(∠\)ZYP,
\(∠\)ZYQ = \(∠\)QYP
Or \(∠\)ZYP = 2\(∠\)ZYQ
∴ \(∠\)ZYQ = \(∠\)QYP = 58°
Then \(∠\)XYQ = \(∠\)XYZ + \(∠\)ZYQ
\(∠\)XYQ = a + 64°
\(⇒\) \(∠\)XYQ = 58° + 64° = 122°.
\(∠\)XYQ = 122°
Now, reflex \(∠\)QYP = 180°+\(∠\)XYQ
Since \(∠\)XYQ = 122°
we have
∠QYP = 180°+122°
∴ \(∠\)QYP = 302°