Answer :

(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°
​​


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