In the given figure, $RT||SQ$. If $\angle QPS=110^{\circ}$, $\angle PQS=40^{\circ}$, $\angle PSR=75^{\circ}$ and $\angle QRS=65^{\circ}$, then find the value of $\angle QRT$ "
Given: In the given figure, $RT||SQ$. If $\angle QPS=110^{\circ}$, $\angle PQS=40^{\circ}$, $\angle PSR=75^{\circ}$ and $\angle QRS=65^{\circ}$.
To do: To find the value of $\angle QRT$
Solution:
From the figure
In $\vartriangle PQS$,
$\angle PQS+\angle QPS+\angle PSQ=180^o$
$\Rightarrow \angle PSQ=180^o-\angle QPS-\angle PQS=180^o-100^o-40^o=40^o$
Given $\angle PSR=85^o $
$\Rightarrow \angle PSQ+\angle QSR=85^o$
$\Rightarrow \angle QSR=85^o-\angle PSQ=85^o-40^o =45^o$
In $\vartriangle QSR$
$\angle QRS+\angle PSQ+\angle SQR=180^o$
$\angle SQR=180^o-\angle QRS-\angle QSR=180^o-70^o-45^o =65^o$
AS $RT||SQ$
$\angle QRT=\angle SQR=65^o$ [$\because$ alternative angles]
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