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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