(i)
(i) From the figure we know that the diagonals are equal and bisect at point O.
Consider △ AOB
We get AO = OB
We know that the base angles are equal
∠ OAB = ∠ OBA = 35
By using the sum property of triangle
∠ AOB + ∠ OAB + ∠ OBA = 180
By substituting the values we get
∠ AOB + 35+ 35 = 180
On further calculation
∠ AOB = 180 - 35 - 35
By subtraction
∠ AOB = 180 - 70
∠ AOB = 110
From the figure we know that the vertically opposite angles are equal
∠ DOC = ∠ AOB = y = 110
In △ ABC
We know that ∠ ABC = 90
Consider △ OBC
We know that
∠ OBC = x = ∠ ABC - ∠ OBA
By substituting the values
∠ OBC = 90 - 35
By subtraction
∠ OBC = 55
Therefore, x = 55 and y = 110.
(ii) From the figure we know that the diagonals of a rectangle are equal and bisect each other.
Consider △ AOB
We get
OA = OB
We know that the base angles are equal
∠ OAB = ∠ OBA
By using the sum property of triangle
∠ AOB + ∠ OAB + ∠ OBA = 180
By substituting the values
110 + ∠ OAB + ∠ OBA = 180
We know that ∠ OAB = ∠ OBA
So we get
2 ∠ OAB = 180 - 110
By subtraction
2 ∠ OAB = 70
By division
∠ OAB = 35
We know that AB || CD and AC is a transversal
From the figure we know that ∠ DCA and ∠ CAB are alternate angles
∠ DCA = ∠ CAB = y = 35
Consider △ ABC
We know that
∠ ACB + ∠ CAB = 90
So we get
∠ ACB = 90 - ∠ CAB
By substituting the values in above equation
∠ ACB = 90 - 35
By subtraction
∠ ACB = x = 55
Therefore, x = 55 and y = 35.
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