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Answer :
(i)Here, ABCD is a parallelogram.
\(∠B = ∠D\quad \) [Opposite angles of a parallelogram are equal]
\(∠D = 100^{\circ}\)
\(\Rightarrow y = 100^{\circ}\)
\(∠A + ∠B = 180^{\circ}\quad\) [Adjacent angles of a parallelogram are supplementary]
\(\Rightarrow z + 100^{\circ} = 180^{\circ}\)
\(\Rightarrow z = 180^{\circ} – 100^{\circ} = 80^{\circ}\)
\(∠A = ∠C\quad\) [Opposite angles of a parallelogram are equal]
\(x = 80^{\circ}\)
\(Hence, x = 80^{\circ}, y = 100^{\circ} and\; z = 80^{\circ}\)
(ii)We have PQRS as a parallelogram.
\(∠P + ∠S = 180^{\circ}\quad \) [Adjacent angles of parallelogram]
\(\Rightarrow x + 50^{\circ} = 180^{\circ}\)
\(\Rightarrow x = 180^{\circ} – 50^{\circ} = 130^{\circ}\)
\(Now, ∠P = ∠R\quad\)[Opposite angles are equal]
\(\Rightarrow x = y\)
\(\Rightarrow y = 130^{\circ}\)
\(Also, y = z\quad\) [Alternate angles]
\(\Rightarrow z = 130^{\circ}\)
\(Hence, x = 130^{\circ}, y = 130^{\circ} and z = 130^{\circ}\)
(iii) Here we have, ABCD as a rhombus because diagonals intersect at \(90^{\circ} \)
\(So, x = 90^{\circ}\)
So,now in ∆OCB we have,
\(x + y + 30^{\circ} = 180^{\circ}\quad \) [Angle sum property]
\(\Rightarrow 90^{\circ} + y + 30^{\circ} = 180^{\circ}\)
\(\Rightarrow y + 120^{\circ} = 180^{\circ}\)
\(\Rightarrow y = 180^{\circ} – 120^{\circ} = 60^{\circ}\)
\(y = z \quad \)[Alternate angles]
\(\Rightarrow z = 60^{\circ}\)
\(Hence, x = 90^{\circ}, y = 60^{\circ} and z = 60^{\circ}\)
(iv) ABCD is a parallelogram
∠A + ∠B = 180^{\circ} (Adjacent angles of a parallelogram are supplementary)
\(\Rightarrow x + 80^{\circ} = 180^{\circ}\)
\(\Rightarrow x = 180^{\circ} – 80^{\circ} = 100^{\circ}\)
\(Now, ∠D = ∠B\quad \) [Opposite angles of a parallelogram are equal]
\(\Rightarrow y = 80^{\circ}\)
\(Also, z = ∠B = 80^{\circ}\quad\) (Alternate angles)
\(So, x = 100^{\circ}, y = 80^{\circ} and z = 80^{\circ}\)
(v) In this case we have ABCD as a parallelogram.
\(∠D = ∠B \quad\)[Opposite angles of a parallelogram are equal]
\(y = 112^{\circ}\)
\(x + y + 40^{\circ} = 180^{\circ}\quad\) [Angle sum property]
\(\Rightarrow x + 112^{\circ} + 40^{\circ} = 180^{\circ}\)
\(\Rightarrow x + 152^{\circ} = 180^{\circ}\)
\(\Rightarrow x = 180^{\circ} – 152^{\circ} = 28^{\circ}\)
\(z = x = 28^{\circ} \quad \)[Alternate angles]
\(So\; we \;have, x = 28^{\circ}, y = 112^{\circ}, z = 28^{\circ}\).