We are given that $\angle WZY = 140^\circ$ and need to find $\angle WZX$. The problem shows that:

$\angle WZX = (8b + 49)^\circ \quad \text{and} \quad \angle XZY = (3b + 102)^\circ$

Using the Angle Addition Postulate, we know:

$\angle WZY = \angle WZX + \angle XZY$

Substitute the known expressions:

$140^\circ = (8b + 49)^\circ + (3b + 102)^\circ$

Now, simplify and solve for $b$:

$140 = (8b + 49) + (3b + 102)$ $140 = 11b + 151$

Subtract 151 from both sides:

$140 - 151 = 11b$ $-11 = 11b$

Now, solve for $b$:

$b = -1$

Now substitute $b = -1$ back into the expression for $\angle WZX$:

$\angle WZX = 8(-1) + 49 = -8 + 49 = 41^\circ$

Thus, $\angle WZX = 41^\circ$.

Would you like further details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How does the Angle Addition Postulate help solve angle problems?
2. What happens if the angle sum exceeds $180^\circ$ in this kind of diagram?
3. How would the problem change if $\angle WZY$ were given as obtuse?
4. What would happen if $b$ were positive? Does it affect the overall geometry?
5. Can you derive other equations for $\angle WZY$ if the angles were described differently?

Tip: Always check your final result by substituting values back into the original equations to verify your solution!