Mathematics

If $PQ\bot PS,\ PQ\parallel SR,\ \angle SOR = 28^{o}$ and $\angle QRT = 65^{o}$, then find $x$ & $y$.

SOLUTION
$\because PQ\parallel SR$
$\angle PQR = \angle QRT$ (Alternate interior angle)
$\Rightarrow x + {28^ \circ } = {65^ \circ }$
$\Rightarrow x = {65^ \circ } - {28^ \circ }$
$\Rightarrow x = {37^ \circ }$
$In\,\,\,\Delta PQS$
$\angle P + \angle Q + \angle S = {108^ \circ }$ (By angle sum property)
$\Rightarrow {90^ \circ } + x + y = {180^ \circ }$
$\Rightarrow {90^ \circ } + {37^ \circ }+y = {180^ \circ }$
$\Rightarrow y = {108^ \circ } - 127$
$\Rightarrow y = {53^ \circ }$

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Subjective Medium Published on 09th 09, 2020
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