We know that, 

The sum of the opposite angles of cyclic quadrilateral should be 180°. 

And, in the cyclic quadrilateral ABCD, 

Angles ∠A and ∠C & angles ∠B and ∠D are the pairs of opposite angles.

So, ∠A + ∠C = 180° and 

∠B + ∠D = 180° 

Substituting the values given to the above two equations, we have 

For ∠A + ∠C = 180° 

⇒ ∠A = (2x + 4)° and ∠C = (2y + 10)° 

2x + 4 + 2y + 10 = 180° 

2x + 2y + 14 = 180° 

2x + 2y = 180° – 14° 

2x + 2y = 166 —— (i) 

And for, ∠B + ∠D = 180°, we have 

⇒ ∠B = (y+3)° and ∠D = (4x – 5)° 

y + 3 + 4x – 5 = 180° 

4x + y – 5 + 3 = 180° 

4x + y – 2 = 180° 

4x + y = 180° + 2° 

4x + y = 182° ——- (ii) 

Now for solving (i) and (ii), we perform 

Multiplying equation (ii) by 2 to get, 

8x + 2y = 364 —— (iii) 

And now, subtract equation (iii) from (i) to get 

-6x = -198 

x = −198/−6 

⇒ x = 33° 

Now, substituting the value of x = 33° in equation (ii) to find y 

4x + y = 182 

132 + y = 182 

y = 182 – 132 

⇒ y = 50 

Thus, calculating the angles of a cyclic quadrilateral we get: 

∠A = 2x + 4 

= 66 + 4 

= 70°

∠B = y + 3 

= 50 + 3 

= 53° 

∠C = 2y + 10 

= 100 + 10 

= 110° 

∠D = 4x – 5 

= 132 – 5 

= 127° 

Therefore, the angles of the cyclic quadrilateral ABCD are 

∠A = 70°, ∠B = 53°, ∠C = 110° and ∠D = 127°.