In the figure, ray $OS$ stands on a line $POQ$. Ray $OR$ and ray $OT$ are angle bisectors of $\angle POS$ and $\angle SOQ$ respectively. If $\angle POS = x$, find $\angle ROT$. "
Given:
Ray $OS$ stands on a line $POQ$.
Ray $OR$ and ray $OT$ are angle bisectors of $\angle POS$ and $\angle SOQ$ respectively.
$\angle POS = x$.
To do:
We have to find $\angle ROT$.
Solution:
$\angle POS = x$
Therefore,
$\angle POS + \angle QOS = 180^o$ (Linear pair)
$x + \angle QOS = 180^o$
$\angle QOS = 180^o - x$
$OR$ and $OT$ are the angle bisectors of $\angle POS$ and $\angle QOS$ respectively.
This implies,
$\angle \mathrm{ROS}=\frac{x}{2}$
$\angle \mathrm{TOS}=\frac{180^{\circ}-x}{2}$
$\angle \mathrm{ROT}=\angle \mathrm{ROS}+\angle \mathrm{TOS}$
$\angle \mathrm{ROT}=\frac{x}{2}+\frac{180^{\circ}-x}{2}$
$\angle \mathrm{ROT}=\frac{x+180^{\circ}-x}{2}$
$\angle \mathrm{ROT}=\frac{180^{\circ}}{2}$
$\angle \mathrm{ROT}=90^{\circ}$
Hence, $\angle \mathrm{ROT}=90^{\circ}$.
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