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Prove that \(\angle{ROS}\) = \(\frac{1}{2} \) (\(\angle{QOS}\) - \(\angle{POS}\)).

Answer :

Since it is given that, OR is perpendicular to PQ, we have,\(\angle{PQR}\) = \(\angle{ROQ}\) = \(90^\circ\)

Also, we can say that,

\(\angle{POS}\) + \(\angle{ROS}\) = \(90^\circ\)

\(\Rightarrow \) \(\angle{ROS}\) = \(90^\circ\) - \(\angle{POS}\) ...(i)

Now, by adding, \(\angle{ROS}\) on both the side, we get,

2\(\angle{ROS}\) = \(90^\circ\) - \(\angle{POS}\) + \(\angle{ROS}\)

\(\Rightarrow \) 2\(\angle{ROS}\) = (\(90^\circ\) + \(\angle{ROS}\)) - \(\angle{POS}\)

\(\Rightarrow \) 2\(\angle{ROS}\) = \(\angle{QOS}\) - \(\angle{POS}\)

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

Hence, proved.

- In figure, lines AB and CD intersect at O. If \(\angle{AOC}\) + \(\angle{BOE}\) = \(70^\circ\) and \(\angle{BOD}\) = \(40^\circ\), find \(\angle{BOE}\) and reflex \(\angle{COE}\).
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