Question

# $$POQ$$ is a line ray $$OR$$ is a perpendicular to line $$PQ$$. $$OS$$ is another ray lying between ray $$OP$$ and $$OR$$. Prove that  $$\angle ROS=\dfrac {1}{2}(\angle QOS -\angle POS)$$

Solution

## Given:- $$OR \perp PQ$$To prove:-$$\angle{ROS} = \cfrac{1}{2} \left( \angle{QOS} - \angle{POS} \right)$$Proof:-$$\because \; OR \perp PQ$$$$\therefore \; \angle{ROP} = 90°$$$$\angle{ROQ} = 90°$$$$\therefore$$ We can say that,$$\angle{ROP} = \angle{ROQ}$$$$\angle{POS} + \angle{ROS} = \angle{ROQ} \; \left[\because \angle{ROP} = \angle{POS} + \angle{ROS} \right]$$$$\Rightarrow \; \angle{POS} + \angle{ROS} = \angle{QOS} - \angle{ROS} \; \left[\because \angle{ROQ} = \angle{QOS} - \angle{ROS} \right]$$$$\Rightarrow \; \angle{ROS} + \angle{ROS} = \angle{QOS} - \angle{POS}$$$$\Rightarrow \; 2 \angle{ROS} = \angle{QOS} - \angle{POS}$$$$\Rightarrow \; \angle{ROS} = \cfrac{1}{2} \left( \angle{QOS} - \angle{POS} \right)$$Mathematics

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