Question 11 Mark
In the figure, $POQ$ is a line. Ray $OR$ is perpendicular to line $PQ. OS$ is another ray lying between rays $OP$ and $OR.$ Prove that $\angle ROS={1\over2}(\angle QOS-\angle POS)$
Answer
Ray $OR$ is perpendicular to line $PQ$
$\therefore \angle QOR = \angle POR = 90^\circ . . . .(1)$
$\angle QOS = \angle QOR + \angle ROS. . . .(2)$
$\angle POS = \angle POR – \angle ROS. . . .(3)$
From $(2)$ and $(3),$
$\therefore \angle QOS – \angle POS = (\angle QOR – \angle POR) + 2\angle ROS = 2\angle ROS . . . [$Using $(1)]$
$\therefore \angle ROS={1\over2}(\angle QOS-\angle POS)$
View full question & answer→Ray $OR$ is perpendicular to line $PQ$
$\therefore \angle QOR = \angle POR = 90^\circ . . . .(1)$
$\angle QOS = \angle QOR + \angle ROS. . . .(2)$
$\angle POS = \angle POR – \angle ROS. . . .(3)$
From $(2)$ and $(3),$
$\therefore \angle QOS – \angle POS = (\angle QOR – \angle POR) + 2\angle ROS = 2\angle ROS . . . [$Using $(1)]$
$\therefore \angle ROS={1\over2}(\angle QOS-\angle POS)$
