Q. 8

In the picture be

(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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