1. **State the problem:** We have triangle XYZ with sides XY = 5 cm, XZ = y cm, YZ = x cm, and angle $\angle X = 60^\circ$. Given $x = y - 1$, find the perimeter $k$ of the triangle. 2. **Formula and rules:** Use the Law of Cosines to relate the sides and the included angle: $$x^2 = 5^2 + y^2 - 2 \times 5 \times y \times \cos 60^\circ$$ Recall $\cos 60^\circ = \frac{1}{2}$. 3. **Substitute $x = y - 1$ and $\cos 60^\circ = \frac{1}{2}$:** $$ (y - 1)^2 = 25 + y^2 - 2 \times 5 \times y \times \frac{1}{2} $$ 4. **Expand and simplify:** $$ y^2 - 2y + 1 = 25 + y^2 - 5y $$ 5. **Cancel $y^2$ from both sides:** $$ \cancel{y^2} - 2y + 1 = 25 + \cancel{y^2} - 5y $$ 6. **Bring all terms to one side:** $$ -2y + 1 - 25 + 5y = 0 $$ $$ 3y - 24 = 0 $$ 7. **Solve for $y$:** $$ 3y = 24 $$ $$ y = \frac{24}{3} $$ $$ y = 8 $$ 8. **Find $x$:** $$ x = y - 1 = 8 - 1 = 7 $$ 9. **Calculate perimeter $k$:** $$ k = 5 + y + x = 5 + 8 + 7 = 20 $$ **Final answer:** $$ k = 20 \text{ cm} $$