It is given that $ \angle \mathrm{XYZ}=64^{\circ} $ and $ \mathrm{XY} $ is produced to point P. Draw a figure from the given information. If ray $ \mathrm{YQ} $ bisects $ \angle \mathrm{ZYP} $, find $ \angle \mathrm{XYQ} $ and reflex $ \angle Q Y P $.
Given:
It is given that $\angle XYZ=64^o$, $XY$ is produced to point $P$ and ray $YQ$ bisects $\angle ZYP$.
To do:
We have to draw a figure from the given information and find $\angle XYQ$ and reflex $\angle QYP$.
Solution:
$XYP$ is a line.
Therefore,
$\angle XYZ+\angle ZYP=180^o$
$64^o+\angle ZYP=180^o$ (since $\angle XYZ=64^o$)
This implies,
$\angle ZYP=180^o-64^o$
$\angle ZYP=116^o$
Since,
$YQ$ bisects $\angle ZYP$
We get,
$\angle ZYQ=\angle QYP$
and also,
$\angle ZYP=2\angle QYP$
This implies,
$116^o=2\angle QYP$
$\frac{116^o}{2}=\angle QYP$
$58^o=\angle QYP$
That is,
$\angle QYP=58^o$
Therefore,
$\angle ZYQ=\angle QYP=58^o$
Similarly, we get,
$\angle XYQ=\angle XYZ+\angle ZYQ$
This implies,
$\angle XYQ=64^o+58^o$
$\angle XYQ=122^o$
Now let us find,
Reflex $\angle QYP$
$\angle QYP=180^o+\angle XYQ$ (since $\angle QYP$ is reflex of $\angle XYQ$)
We have $\angle XYQ$ by substituting we get,
$\angle QYP=180^o+122^o$
This implies,
$\angle QYP=302^o$.
Hence, $\angle QYP=302^o$.
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