1. **State the problem:** We have a triangle with vertices T, V, and U. Given: - Side TV = $u$ (unknown) - Side VU = 12 - Side TU = 13 - Angle at vertex U = 22° We need to find the length $u$. 2. **Identify the formula:** We can use the Law of Cosines, which relates the sides and the included angle in a triangle: $$c^2 = a^2 + b^2 - 2ab \cos(C)$$ where $c$ is the side opposite angle $C$. 3. **Assign sides and angle:** - Let side $TV = u$ be opposite angle $U = 22^\circ$. - The other two sides are $VU = 12$ and $TU = 13$. 4. **Apply the Law of Cosines:** $$u^2 = 12^2 + 13^2 - 2 \times 12 \times 13 \times \cos(22^\circ)$$ 5. **Calculate each term:** $$12^2 = 144$$ $$13^2 = 169$$ $$2 \times 12 \times 13 = 312$$ 6. **Calculate $\cos(22^\circ)$:** $$\cos(22^\circ) \approx 0.9272$$ 7. **Substitute values:** $$u^2 = 144 + 169 - 312 \times 0.9272$$ $$u^2 = 313 - 289.4784$$ $$u^2 = 23.5216$$ 8. **Find $u$ by taking the square root:** $$u = \sqrt{23.5216}$$ $$u \approx 4.85$$ **Final answer:** $$u \approx 4.9$$ (rounded to the nearest tenth)