We are given a circle with center O, and the points B, C, and D lie on the circumference. Two tangents, ABE and ADF, touch the circle at B and D respectively. The angles provided are:

• ∠DAB = 40°
• ∠CBE = 75°

We need to calculate the angle ∠ODC.

Steps:

1. Identify key angles:

   • ∠DAB = 40° (given).
   • ∠CBE = 75° (given).

2. Use the property of tangents:

   • Since ABE is tangent to the circle at B, and ADF is tangent at D, the angles between the tangents and the radii drawn to the points of tangency are 90°. This means ∠OBA and ∠ODA are 90°.

3. Identify cyclic quadrilateral:

   • B, C, D, O form a cyclic quadrilateral. In any cyclic quadrilateral, opposite angles sum to 180°. Therefore: $\angle BOC + \angle BDC = 180^\circ$

4. Calculate central angle ∠BOC:

   • Since ∠CBE = 75°, this angle subtends the arc BC. The central angle subtended by the same arc is twice the inscribed angle (by the circle theorems). Therefore: $\angle BOC = 2 \times 75^\circ = 150^\circ$

5. Find angle ∠BDC:

   • From the cyclic quadrilateral property: $\angle BOC + \angle BDC = 180^\circ$ Substituting the value of ∠BOC: $150^\circ + \angle BDC = 180^\circ$ Solving for ∠BDC: $\angle BDC = 180^\circ - 150^\circ = 30^\circ$

Thus, the size of angle ∠ODC is 30°.

Would you like further explanation or any details about the method? Here are a few related questions to expand on this:

1. How can you verify the properties of cyclic quadrilaterals?
2. What are the key circle theorems that apply to tangents and radii?
3. Why is the central angle double the inscribed angle subtended by the same arc?
4. How do you calculate angles when tangents are involved in a circle diagram?
5. How do opposite angles in a cyclic quadrilateral relate to each other?

Tip: Remember that tangents to a circle form a right angle with the radius at the point of tangency.