1. **State the problem:** We have triangle $\triangle ABC$ with sides $AB = x$ cm, $BC = x + 2$ cm, $AC = 5$ cm, and angle $\angle ABC = 60^\circ$. We need to find $x$. 2. **Apply the Law of Cosines:** The Law of Cosines states for any triangle: $$c^2 = a^2 + b^2 - 2ab \cos C$$ where $c$ is the side opposite angle $C$. 3. **Identify sides and angle:** Here, angle $B = 60^\circ$ is between sides $AB = x$ and $BC = x + 2$, and opposite side is $AC = 5$. 4. **Write the Law of Cosines for this triangle:** $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos 60^\circ$$ Substitute values: $$5^2 = x^2 + (x+2)^2 - 2 \cdot x \cdot (x+2) \cdot \cos 60^\circ$$ 5. **Simplify:** $$25 = x^2 + (x^2 + 4x + 4) - 2x(x+2) \cdot \frac{1}{2}$$ Since $\cos 60^\circ = \frac{1}{2}$, the last term simplifies: $$25 = x^2 + x^2 + 4x + 4 - x(x+2)$$ 6. **Expand and simplify further:** $$25 = 2x^2 + 4x + 4 - (x^2 + 2x)$$ $$25 = 2x^2 + 4x + 4 - x^2 - 2x$$ $$25 = x^2 + 2x + 4$$ 7. **Bring all terms to one side:** $$x^2 + 2x + 4 - 25 = 0$$ $$x^2 + 2x - 21 = 0$$ 8. **Solve quadratic equation:** Use quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=2$, $c=-21$. Calculate discriminant: $$\Delta = 2^2 - 4 \cdot 1 \cdot (-21) = 4 + 84 = 88$$ Calculate roots: $$x = \frac{-2 \pm \sqrt{88}}{2} = \frac{-2 \pm 2\sqrt{22}}{2} = -1 \pm \sqrt{22}$$ 9. **Select valid solution:** Since $x$ is a length, it must be positive. $ -1 - \sqrt{22} < 0 $ (discard) $ -1 + \sqrt{22} > 0 $ (valid) 10. **Final answer:** $$x = -1 + \sqrt{22} \approx 3.69$$ cm