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-dfrac -1-k-angle QOS-angle POS- k- 1234

Question

Toppr Admin · Accepted Answer

-xA0-PORQOR-xA0-xA0-90-o-xA0-xA0-1- -ORPQ-xA0-at O-Now-xA0-QOS-QOR-ROSQOS-xA0-90o-xA0-xA0-ROS-xA0-2- -by -1-POS-ROS-PORPOS-PORROSPOS-xA0-90o-xA0-ROS-xA0-3- -by -1-Subtracting -3- from -2-QOSPOS-90o-xA0-xA0-ROS-90o-xA0-xA0-ROS-2ROS2ROS-QOSPOS-i-e-xA0-ROS-12-QOSPOS-

In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. If $∠ R O S = \frac{1}{k} (∠ Q O S - ∠ P O S)$ , find $k$ .

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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}

In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. If ∠ROS=1k(∠QOS−∠POS), find k. 1234

PORQOR= $ 90^o $ ....(1) [ORPQ at O]Now, QOS=QOR+ROSQOS= 90o +ROS ....(2) [by (1)]POS+ROS=PORPOS=PORROSPOS= 90o ROS ....(3) [by (1)]Subtracting (3) from (2),QOSPOS={90o +ROS}{90o ROS}=2ROS2ROS={QOSPOS}i.e, ROS=12{QOSPOS}

In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. If $∠ R O S = \frac{1}{k} (∠ Q O S - ∠ P O S)$ , find $k$ .

1
2
3
4

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}