In Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that $∠$ ROS = $\frac{1}{2}$ ( $∠$ QOS – $∠$ POS)

Question

In Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that  $∠$ ROS =  $\frac{1}{2}$  ( $∠$ QOS –  $∠$ POS)

cdquestions Admin · Accepted Answer

It is given that (OR ⊥ PQ) and  $∠$ POQ = 180° We can rewrite it as  $∠$ ROP =  $∠$ ROQ = 90 0 Since,  $∠$ ROP = $∠$ ROQ It can be written as $∠$ POS +  $∠$ ROS =  $∠$ ROQ $∠$ POS +  $∠$ ROS =  $∠$ QOS –  $∠$ ROS $∠$ SOR +  $∠$ ROS =  $∠$ QOS –  $∠$ POS $⇒$  2 $∠$ ROS =  $∠$ QOS –  $∠$ POS i.e,  $∠$ ROS =  $\frac{1}{2}$  ( $∠$ QOS –  $∠$ POS)

In Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that \(∠\)ROS = \(\frac{1}{2}\) (\(∠\)QOS – \(∠\)POS)

Solution and Explanation

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