1. **Problem statement:** Solve for the angle $x$ in a triangle with sides 2, 3, and 9, where $x$ is the angle opposite the side of length 9. 2. **Formula used:** We use the Law of Cosines, which relates the sides and angles of a triangle: $$\cos(x) = \frac{a^2 + b^2 - c^2}{2ab}$$ where $c$ is the side opposite angle $x$, and $a$, $b$ are the other two sides. 3. **Apply the formula:** Here, $a=2$, $b=3$, and $c=9$. $$\cos(x) = \frac{2^2 + 3^2 - 9^2}{2 \times 2 \times 3} = \frac{4 + 9 - 81}{12} = \frac{-68}{12} = -\frac{17}{3}$$ 4. **Interpretation:** Since $\cos(x)$ must be between $-1$ and $1$, and $-\frac{17}{3} < -1$, this means such a triangle cannot exist with sides 2, 3, and 9. 5. **Conclusion:** No valid angle $x$ exists for these side lengths; the triangle is not possible. **Answer:** None of the options (a. 45, b. 5, c. 6, d. 2) are valid because the triangle cannot exist with these side lengths.