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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,

$$\angle{ROP} = \angle{ROQ}$$

$$\angle{POS} + \angle{ROS} = \angle{ROQ} \; \left[\because \angle{ROP} = \angle{POS} + \angle{ROS} \right]$$

$$\Rightarrow \; \angle{POS} + \angle{ROS} = \angle{QOS} - \angle{ROS} \; \left[\because \angle{ROQ} = \angle{QOS} - \angle{ROS} \right]$$

$$\Rightarrow \; \angle{ROS} + \angle{ROS} = \angle{QOS} - \angle{POS}$$

$$\Rightarrow \; 2 \angle{ROS} = \angle{QOS} - \angle{POS}$$

$$\Rightarrow \; \angle{ROS} = \cfrac{1}{2} \left( \angle{QOS} - \angle{POS} \right)$$

1017915_1063971_ans_df96fcef9f1d4069876453ce69398b25.png

Mathematics

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