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Answer :
We have, \(∠Q = 130° and ∠R = 90^{\circ}\) and
\(\bar { SP } || \bar { RQ }\)
\(∠P + ∠Q = 180^{\circ}\quad\) [Adjacent angles]
\(\Rightarrow ∠P + 130^{\circ} = 180^{\circ}\)
\(\Rightarrow ∠P = 180^{\circ} – 130^{\circ} = 50^{\circ}\)
and also we have, \(∠S + ∠R = 180^{\circ}\quad\) [Adjacent angles]
\(\Rightarrow ∠S + 90^{\circ} = 180^{\circ}\)
\(\Rightarrow ∠S = 180^{\circ} – 90^{\circ} = 90^{\circ}\)
Hence, \(m∠P = 50^{\circ} \;and \; m∠S = 90^{\circ}\)
Alternate Method:
\(∠Q = 130^{\circ}, ∠R = 90^{\circ} and ∠S = 90^{\circ}\)
We know that
\(∠P + ∠Q + ∠R + ∠Q = 360^{\circ}\quad\) [Angle sum property of quadrilateral]
\(\Rightarrow ∠P + 130^{\circ}+ 90^{\circ} + 90^{\circ} = 360^{\circ}\)
\(\Rightarrow ∠P + 310^{\circ} = 360^{\circ}\)
\(\Rightarrow ∠P = 360^{\circ}– 310° = 50^{\circ}\)
Hence \(m∠P = 50^{\circ}\)