1. **State the problem:** We need to find the perimeter of triangle $\triangle XYZ$ where two sides are given as 34.4 and 36, and the third side $x$ is unknown. The angles given are $63^\circ$ at $Y$ and $57^\circ$ at $X$. 2. **Use the triangle angle sum rule:** The sum of angles in any triangle is $180^\circ$. So, $$x = 180^\circ - 63^\circ - 57^\circ = 60^\circ.$$ This means the angle at $Z$ is $60^\circ$. 3. **Use the Law of Cosines to find side $x$ opposite angle $Z$:** The Law of Cosines states: $$x^2 = a^2 + b^2 - 2ab \cos(C)$$ where $a=34.4$, $b=36$, and $C=60^\circ$. 4. **Calculate $x^2$:** $$x^2 = 34.4^2 + 36^2 - 2 \times 34.4 \times 36 \times \cos(60^\circ)$$ $$= 1183.36 + 1296 - 2 \times 34.4 \times 36 \times 0.5$$ $$= 2479.36 - 1238.4$$ $$= 1240.96$$ 5. **Find $x$ by taking the square root:** $$x = \sqrt{1240.96} \approx 35.23$$ 6. **Find the perimeter $P$ of $\triangle XYZ$:** $$P = 34.4 + 36 + 35.23 = 105.63$$ 7. **Round to the nearest tenth:** $$P \approx 105.6$$ **Final answer:** The perimeter of $\triangle XYZ$ is approximately $105.6$ units.