Answer :

(i)



Join AC.


Now, APBC is a cyclic quadrilateral, as it lies on a circle


Therefore,


APB + ACB = 180° [Sum of opposite angles of a cyclic quadrilateral is 180°]


Now, Diagonals of the square bisects the angles of a square




APB + 45° = 180°


APB = 135°


(ii)



As, XYZQ is a cyclic quadrilateral


XYZ + XQZ = 180° [Sum of opposite angles of a cyclic quadrilateral is 180°]


Now, XYZ = 60°, as XYZ is an equilateral triangle


XQZ = 180° – 60° = 120°


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