The problem presented in the image shows that ray $GH$ bisects $\angle FGI$, and the given angles are $(5x - 3)^\circ$ and $(6x - 10)^\circ$. Let's solve each part step-by-step.

Part a: Solve for $x$ and find $\angle FGH$.

Since $GH$ bisects $\angle FGI$, the two angle measures should be equal. Therefore, we can set the two expressions equal to each other:

$5x - 3 = 6x - 10$

Solving for $x$:

1. Subtract $5x$ from both sides: $-3 = x - 10$

2. Add 10 to both sides: $7 = x$

Thus, $x = 7$.

Finding $\angle FGH$:

Now substitute $x = 7$ into one of the angle expressions (for example, $5x - 3$) to find $\angle FGH$:

$\angle FGH = 5(7) - 3 = 35 - 3 = 32^\circ$

So, $\angle FGH = 32^\circ$.

Part b: Find $\angle HGI$.

Since $GH$ bisects the angle, $\angle HGI$ must be equal to $\angle FGH$. Therefore:

$\angle HGI = 32^\circ$

Part c: Find $\angle FGI$.

Since $\angle FGI$ is the sum of the two bisected angles, we have:

$\angle FGI = \angle FGH + \angle HGI = 32^\circ + 32^\circ = 64^\circ$

Summary of Solutions:

a. $x = 7$, $\angle FGH = 32^\circ$
b. $\angle HGI = 32^\circ$
c. $\angle FGI = 64^\circ$

Would you like further details or have any other questions?

Here are 5 related questions:

1. What is the significance of an angle bisector in geometry?
2. How do we verify that the bisected angles are equal?
3. What would happen if the given angles were unequal? Could $GH$ still be a bisector?
4. How would the problem change if additional angles or relationships were provided in the diagram?
5. Can you generalize the formula for bisecting angles involving variables?

Tip: Always remember that when a ray bisects an angle, the two resulting angles are equal by definition!