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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Updated on: 10-Oct-2022

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