Given the value of one of the 4 central angles, we can calculate the other central angles:

We are given that ∠BOD=40°

∠AOC=∠BOD (vertical angles are equal), therefore, ∠AOC=40°.

∠BOD+∠AOD=180° (the sum of supplementary angles is 180°), therefore ∠AOD=180°-40°=140°.

Calculating the angles in triangle ACO:

We are given that O the is center of the circle and the chords AB and CD intersect at point O, therefore AB and CD are diameters and create 4 radii CO=BO=AO=DO.

Since CO=AO the triangle ACO is isosceles and the angles ∠ACO and ∠CAO are equal (mark them as x).

The sum of angles in a triangle is equal to 180°, therefore ACO+CAO+AOC=180°, plugging the values we get an equation that we can solve:

x+x+40=180

2x=140

x=70

The angle ∠ACO is equal to 70 degrees.

Calculating the angles in triangle ADO:

We are given that O the is center of the circle and the chords AB and CD intersect at point O, therefore AB and CD are diameters and create 4 radii CO=BO=AO=DO.

Since AO=DO the triangle ADO is isosceles and the angles ∠DAO and ∠ADO are equal (mark them as x).

The sum of angles in a triangle is equal to 180°, therefore DAO+ADO+AOD=180°.

We found that the angle ∠ACO is equal to 140 degrees.

Plugging the values, we get an equation that we can solve:

x+x+140=180

2x=40

x=20

The angle ∠ADO is equal to 20 degrees.

The ratio between the angles ∠ACO and ∠ADO is 70/20=7/2=3.5