(i) Here, ABCD is rectangle.

We know that the diagonals of a rectangle are congruent and bisect each other.

∴ In ∆ AOB, we have OA = OB

This means that ∆ AOB is isosceles triangle.

We know that base angles of isosceles triangle are equal.

∴ ∠​OAB = ∠​OBA = 35°

∴ ∴​ x = 90° − 35° = 55°

Also, ∠AOB = 180° − (35° + 35°) = 110°

∴ y = ∠AOB​ = 110°​ …Vertically opposite angles

Hence, x = 55° and y = 110°​​

(ii) Here, ABCD is rectangle.

We know that the diagonals of a rectangle are congruent and bisect each other.

∴ In ∆ AOB, we have OA = OB

This means that ∆ AOB is isosceles triangle.

We know that base angles of isosceles triangle are equal.

∴ ∠​OAB = ∠​OBA = × (180° − 110°) = 35°

∴​ y = ∠BAC = 35° … alternate angles with transversal AC
Also, x = 90° – y … ​∵∠C = 90° = x + y
∴​ x = 90° − 35° = 55°
Hence, x = 55° and y = 35°​​