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# 5.In figure below, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $$\angle{ROS}$$ = $$\frac{1}{2}$$ ($$\angle{QOS}$$ - $$\angle{POS}$$).

Since it is given that, OR is perpendicular to PQ, we have,$$\angle{PQR}$$ = $$\angle{ROQ}$$ = $$90^\circ$$
Also, we can say that,
$$\angle{POS}$$ + $$\angle{ROS}$$ = $$90^\circ$$
$$\Rightarrow$$ $$\angle{ROS}$$ = $$90^\circ$$ - $$\angle{POS}$$ ...(i)

Now, by adding, $$\angle{ROS}$$ on both the side, we get,

2$$\angle{ROS}$$ = $$90^\circ$$ - $$\angle{POS}$$ + $$\angle{ROS}$$
$$\Rightarrow$$ 2$$\angle{ROS}$$ = ($$90^\circ$$ + $$\angle{ROS}$$) - $$\angle{POS}$$
$$\Rightarrow$$ 2$$\angle{ROS}$$ = $$\angle{QOS}$$ - $$\angle{POS}$$
$$\Rightarrow$$ $$\angle{ROS}$$ = $$\frac{1}{2}$$ ($$\angle{QOS}$$ - $$\angle{POS}$$).
Hence, proved.