In the figure,ray OS stands on a line POQ. Ra | vedclass

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(B) Ray $OS$ stands on the line $POQ$.Therefore,$\angle POS + \angle SOQ = 180^o$ (Linear pair axiom).Given that $\angle POS = x$.Therefore,$x + \angl

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(B) Ray $OS$ stands on the line $POQ$.Therefore,$\angle POS + \angle SOQ = 180^o$ (Linear pair axiom).Given that $\angle POS = x$.Therefore,$x + \angle SOQ = 180^o$,which implies $\angle SOQ = 180^o - x$.Now,ray $OR$ bisects $\angle POS$,therefore $\angle ROS = \frac{1}{2} 	imes \angle POS = \frac{x}{2}$.Similarly,ray $OT$ bisects $\angle SOQ$,therefore $\angle SOT = \frac{1}{2} 	imes \angle SOQ = \frac{1}{2} 	imes (180^o - x) = 90^o - \frac{x}{2}$.Now,$\angle ROT = \angle ROS + \angle SOT$.Substituting the values,$\angle ROT = \frac{x}{2} + 90^o - \frac{x}{2} = 90^o$.

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 $\angle ROT$. (in $^o$)

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