To solve the problem, let's analyze the given information in the triangle $ABC$:

• $AB = x$ cm
• $BC = (x + 2)$ cm
• $AC = 5$ cm
• $\angle ABC = 60^\circ$

This setup is ideal for applying the Cosine Rule, as we know two sides and the included angle, and we need to solve for $x$.

Using the Cosine Rule:

The Cosine Rule states: $c^2 = a^2 + b^2 - 2ab \cdot \cos C$ where $c$ is the side opposite angle $C$, and $a$ and $b$ are the other two sides.

In this case:

• $a = x$
• $b = x + 2$
• $c = 5$
• $\angle C = 60^\circ$

Substitute into the formula: $5^2 = x^2 + (x + 2)^2 - 2 \cdot x \cdot (x + 2) \cdot \cos 60^\circ$ Since $\cos 60^\circ = 0.5$, we have: $25 = x^2 + (x + 2)^2 - x(x + 2)$

Now, expand and simplify:

1. $(x + 2)^2 = x^2 + 4x + 4$
2. $x(x + 2) = x^2 + 2x$

So: $25 = x^2 + x^2 + 4x + 4 - x^2 - 2x$ Combine terms: $25 = x^2 + 2x + 4$ Subtract 25 from both sides: $0 = x^2 + 2x - 21$ This is a quadratic equation. We can solve it using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 2$, and $c = -21$: $x = \frac{-2 \pm \sqrt{2^2 + 4 \cdot 21}}{2 \cdot 1}$ [