Let's analyze the given problem carefully.

Given information:

• Points $B, D, E, F,$ and $G$ are on the circumference of the circle.
• $AC$ is a tangent to the circle at $B$.
• Given angles:
  • $\angle DFG = 75^\circ$
  • $\angle ABG = 48^\circ$

We need to find: (a) $\angle DEG$
(b) $\angle DOG$
(c) $\angle DBC$

Step 1: Finding $\angle DEG$

Since $D, E, F, G$ are points on the circle, $\angle DFG$ and $\angle DEG$ are opposite angles in a cyclic quadrilateral.
Using the opposite angle property of a cyclic quadrilateral:

$\angle DEG = \angle DFG = 75^\circ$

Step 2: Finding $\angle DOG$

Angle $DOG$ is the central angle subtended by arc $DG$, while $\angle DEG$ is the inscribed angle subtended by the same arc.
Using the central angle theorem:

$\angle DOG = 2 \times \angle DEG = 2 \times 75^\circ = 150^\circ$

Step 3: Finding $\angle DBC$

Since $AC$ is a tangent at $B$, we use the tangent-secant angle theorem, which states:

$\angle DBC = \angle ABG = 48^\circ$

Final Answers:

(a) $\angle DEG = 75^\circ$
(b) $\angle DOG = 150^\circ$
(c) $\angle DBC = 48^\circ$

Would you like me to explain any step in more detail?

Related Questions:

1. What is the sum of opposite angles in a cyclic quadrilateral?
2. How do you prove that the angle at the center is twice the angle at the circumference?
3. What is the alternate segment theorem?
4. How do you find an angle in a cyclic quadrilateral given two angles?
5. What is the relationship between a tangent and a chord in a circle?

Tip: In any cyclic quadrilateral, the sum of opposite angles is always $180^\circ$.