Given: ∠OAC = 50°

Consider ΔAOC

∠OAC = ∠OCA = 50° ( OA = OC = radius, angles opposite to equal sides are equal)

Now, by angle sum property

∠OAC + ∠OCA + ∠AOC = 180°

50° + 50° + ∠AOC = 180°

∠AOC = 180° – 50° – 50°

∠AOC = 80°

Now angle ∠BOD = ∠AOC = 80° (vertically opposite angles)

Now, consider ΔBOD

Here,

OB = OD (radius)

∠OBD = ∠ODB (angles opposite to equal angles are equal)

Let ∠ODB = x

By angle sum property

∠ODB + ∠OBD + ∠BOD = 180°

x + x + 80° = 180°

2x = 180° – 80°

2x = 100°

x = 50°

∴ ∠ODB = 50°