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In the given figure, `angleXYZ=56^(@)` and XY is produced to a point P. If ray YQ bisects `angleZYP`, find `angleXYQ` and reflex `angleQYP`.

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Since XY is produced to point P, it follows that XP is a straight line and ray YZ stands on it.
`:. angleXYZ+angleZYP=180^(@)`
`implies 56^(@)+angleZYP=180^(@)`
`implies angleZYP=(180^(@)-56^(@))=124^(@)`
`implies angleZTQ=1/2 angleZTP=(1/2xx124^(@))=62^(@)`.
`:. angleXYQ=angleXYZ+angleZYQ=(56^(@)+62^(@))=118^(@)`.
Also, `angleQYP=1/2 angleZYP=(1/2xx124^(@))=62^(@)`
`:.` reflex `angleQYP=360^(@)-angleQYP=(360-62)^(@)=298^(@)`.

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