Question

In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. If $\angle ROS=\frac{1}{k}(\angle QOS-\angle POS)$, find $k$.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

PORQOR= $ 90^o $ ....(1) [ORPQ at O]
Now, QOS=QOR+ROS
QOS= ${90}^o$ +ROS ....(2) [by (1)]
POS+ROS=POR
POS=PORROS
POS= ${90}^o$ ROS ....(3) [by (1)]
Subtracting (3) from (2),
QOSPOS={${90}^o$ +ROS}{${90}^o$ ROS}
=2ROS
2ROS={QOSPOS}
i.e, ROS=12{QOSPOS}