In Fig. 6.17, $\mathrm{POQ}$ is a line. Ray $\mathrm{OR}$ is perpendicular to line $\mathrm{PQ}$. OS is another ray lying between rays $O P$ and OR. Prove that$\angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})$"

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Given:

$POQ$ is a line, Ray $OR$ is perpendicular to line $PQ$ and $OS$ is another ray lying between rays $OP$ and $OR$.

To do:

We have to prove that $\angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$.

Solution:

Ray $OR \perp POQ$.

This implies,

$\angle POR = 90^o$

$\angle POS + \angle ROS = 90^o$.....…(i)

$\angle ROS = 90^o - \angle POS$

$\angle POS + \angle QOS = 180^o$          (Linear pair)

$= 2(∠POS + ∠ROS)$             [From (i)]

$\angle POS + \angle QOS = 2\angle ROS + 2\angle POS$

$2\angle ROS = \angle POS + \angle QOS - 2\angle POS$

$2\angle ROS =\angle QOS - \angle POS$

$\angle ROS = \frac{1}{2}(\angle QOS - \angle POS)$

Hence proved.

Updated on 10-Oct-2022 13:40:30