1. Opposite angles in any quadrilateral inscribed in a circle are supplements of each other. (Their measures add up to 180 degrees)

there fore Angle A+Angle C =180

(2x+4)+4y-4=180

x+2y=90 --------i

Similary Angle B+Angle D =180

(x+10)+5y+5=180

x+5y=165-----ii

After solving i and ii

we will get y =25 and x=40

2. Angle A =Angle c=62 ( an Exterior angle of cyclic quadilatreral is equal to its opposite interior angle )

In Triangle ACE

Angle CEF=Angle A+Angle ACE (Sum of Interior two angles equal to opposite exterior angle )

Angle CEF=105

Now Angel c+ Angle CEF+Angle b=180 ( Sum of angles of triangle)

62+105+Angle b=180

Angle b=13

In Traingle ABF

Angle A +Angle a +Angle b =180(Sum of angles of triangle)

62+Angle a +13=180

Angle a=105

1. Opposite angles in any quadrilateral inscribed in a circle are supplements of each other. (Their measures add up to 180 degrees)

there fore Angle A+Angle C =180

(2x+4)+4y-4=180

x+2y=90 --------i

Similary Angle B+Angle D =180

(x+10)+5y+5=180

x+5y=165-----ii

After solving i and ii

we will get y =25 and x=402. Angle A =Angle c=62 ( an Exterior angle of cyclic quadilatreral is equal to its opposite interior angle )In Triangle ACEAngle CEF=Angle A+Angle ACE (Sum of Interior two angles equal to opposite exterior angle )Angle CEF=105Now Angel c+ Angle CEF+Angle b=180 ( Sum of angles of triangle)62+105+Angle b=180

Angle b=13

In Traingle ABF

Angle A +Angle a +Angle b =180(

Sum of angles of triangle)62+Angle a +13=180Angle a=105