1. **State the problem:** We need to find the length $x$ of the side opposite the $55^\circ$ angle in a triangle where the other two sides are $19$ m and $17$ m. 2. **Formula used:** The Law of Cosines states: $$a^2 = b^2 + c^2 - 2bc \cos A$$ where $a$ is the side opposite angle $A$, and $b$ and $c$ are the other two sides. 3. **Assign values:** Here, $a = x$, $b = 19$, $c = 17$, and $A = 55^\circ$. 4. **Apply the formula:** $$x^2 = 19^2 + 17^2 - 2 \times 19 \times 17 \times \cos 55^\circ$$ 5. **Calculate squares:** $$x^2 = 361 + 289 - 2 \times 19 \times 17 \times \cos 55^\circ$$ 6. **Calculate product:** $$2 \times 19 \times 17 = 646$$ 7. **Substitute:** $$x^2 = 361 + 289 - 646 \times \cos 55^\circ$$ 8. **Calculate cosine:** $$\cos 55^\circ \approx 0.5736$$ 9. **Multiply:** $$646 \times 0.5736 \approx 370.8$$ 10. **Simplify:** $$x^2 = 361 + 289 - 370.8 = 650 - 370.8 = 279.2$$ 11. **Take square root:** $$x = \sqrt{279.2} \approx 16.7$$ **Final answer:** $$x \approx 16.7 \text{ meters}$$