In the given figure, $\mathrm{POQ}$ is a line. Ray $\mathrm{OR}$ is perpendicular to line $\mathrm{PQ}$. OS is another ray lying between rays $\mathrm{OP}$ and $\mathrm{OR}$. Prove that

$\angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})$

Solution:

It is given that $O R \perp P Q$

$\therefore \angle P=00^{0}$

$\Rightarrow \angle \mathrm{POS}+\angle \mathrm{SOR}=90^{\circ}$

$\angle R O S=90^{\circ}-\angle P O S \ldots$(1)

$\angle Q O R=90^{\circ}($ As OR $\perp P Q)$

$\angle Q O S-\angle R O S=90^{\circ}$

$\angle \mathrm{ROS}=\angle \mathrm{QOS}-90^{\circ} \ldots$(2)

On adding equations (1) and (2), we obtain

$2 \angle \mathrm{ROS}=\angle \mathrm{QOS}-\angle \mathrm{POS}$

$\angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})$
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