1. **State the problem:** We are given a circle with two lines from a common external point: one tangent and one secant. The angles formed are labeled as $(5x - 6)^\circ$, $3^\circ$, and $(4x + 8)^\circ$. We need to find the value of $x$. 2. **Recall the tangent-secant angle theorem:** The angle formed outside the circle by a tangent and a secant is half the difference of the intercepted arcs. 3. **Set up the equation:** The angle between the tangent and secant is $3^\circ$, so $$3 = \frac{1}{2} \left| (5x - 6) - (4x + 8) \right|$$ 4. **Simplify inside the absolute value:** $$\left| (5x - 6) - (4x + 8) \right| = \left| 5x - 6 - 4x - 8 \right| = \left| x - 14 \right|$$ 5. **Rewrite the equation:** $$3 = \frac{1}{2} |x - 14|$$ 6. **Multiply both sides by 2:** $$2 \times 3 = \cancel{2} \times \frac{1}{\cancel{2}} |x - 14|$$ $$6 = |x - 14|$$ 7. **Solve the absolute value equation:** $$x - 14 = 6 \quad \text{or} \quad x - 14 = -6$$ 8. **Find $x$ values:** $$x = 20 \quad \text{or} \quad x = 8$$ 9. **Check for validity:** Both values are possible; the problem does not restrict $x$, so both are solutions. **Final answer:** $$x = 8 \text{ or } x = 20$$