Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

As it is given that, ∠AOD = 90°

And also given that, ∠AEC = 90°.

But these are corresponding angles.

[tex]\red{\rm\implies \:OD \: \parallel \: BC \: } \\ [/tex]

Now, OC is transversal.

So, it means ∠DOC = ∠OCB = 30°

[ Alternate interior angles ]

We know,

Angle subtended at the centre by an arc is double the angle subtended on the circumference of a circle by same arc.

So, using this theorem, we have

∠DOC = 2∠DBC

[tex]\rm \: 30 \degree \: = \: 2y \\ [/tex]

[tex]\red{\rm\implies \:\boxed{ \rm{y \: = \: 15 \degree \: \: }}}\\ [/tex]

Now,

∠AOC = ∠AOD + ∠DOC

∠AOC = 90° + 30° = 120°

Now,

[tex]\rm \: \angle \: AOC \: = \: 2\angle ABC \: \\ [/tex]

[tex]\rm \: \: 2\angle ABC \: = \: 120\degree \\ [/tex]

[tex]\rm\implies \:\rm \: \: \angle ABC \: = \: 60\degree \\ [/tex]

[tex]\rm\implies \:\rm \: \: \angle ABE \: = \: 60\degree \\ [/tex]

Now, In triangle ABE

∠ABE + ∠AEB + ∠BAE = 180°

[tex]\rm \: 60\degree + 90\degree + x = 180\degree \\ [/tex]

[tex]\rm \: 150\degree + x = 180\degree \\ [/tex]

[tex]\rm \: x = 180\degree - 150\degree \\ [/tex]

[tex]\red{\rm\implies \:\boxed{ \rm{ \:x \: = \: 30\degree \: \: }}} \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional information :-

1. Angle in same segments are equal.

2. Angle in semi-circle is right angle.

3. The sum of opposite angles of a cyclic quadrilateral is supplementary.

4. Exterior angle of a cyclic quadrilateral is equals to interior opposite angle.