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$PQRS$ is a rectangle. Find $\angle OPS$, if $\angle QOR$ is given. $( a).\ 42^o$ $( b).\ 56^o$.
"
Given: $PQRS$ is a rectangle.
To do: To find $\angle OPS$, if $\angle QOR$ is:$( a).\ 42^o$ $( b).\ 56^o$.
Solution:
$( a).\ \because PQRS$ is a rectangle, therefore diagonals $PR$ and $QS$ are equal in length.
$\therefore \frac{1}{2}PR=\frac{1}{2}QS$
$\Rightarrow OS=OP\ .......\ ( i)$
In $\vartriangle OPS$
$\angle POS=\angle QOR=42^o$ [given]
$\angle OSP=\angle OPS$ [$\because OS=OP$]
$\therefore \angle POS+\angle OPS+\angle OPS=180^o$
$\Rightarrow 42^o+2\angle OPS=180^o$
$\Rightarrow 2\angle OPS=180^-42^o$
$\Rightarrow 2\angle OPS=138^o$
$\Rightarrow \angle OPS=\frac{138^o}{2}$
$\Rightarrow \angle OPS=69^o$
$( b)$ Similarly, When $\angle QOR=56^o$
$\Rightarrow 2\angle OPS=180^o-56^o$
$\Rightarrow 2\angle OPS=124^o$
$\Rightarrow \angle OPS=\frac{124^o}{2}$
$\Rightarrow \angle OPS=62^o$
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