Mathematics

# $POQ$ 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,

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

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

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

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

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

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

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Subjective Medium Published on 09th 09, 2020
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