In the picture below, ABCD is a square with vertices on a circle and XYZ is such an equilateral triangle. P and Q are points on the circules:

i) How much is □ APB?

ii) How much is □ XQZ?

(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°