Solution
We are given:
$$ XY = (x + 2)\,\text{cm}, \quad XZ = (2x + 4)\,\text{cm}, \quad YZ = (2x - 1)\,\text{cm}, \quad \angle YXZ = 60^\circ $$
Using the law of cosines:
$$ YZ^2 = XY^2 + XZ^2 - 2 \cdot XY \cdot XZ \cdot \cos(\angle YXZ) $$
Substitute the given values:
$$ (2x - 1)^2 = (x + 2)^2 + (2x + 4)^2 - 2 \cdot (x + 2) \cdot (2x + 4) \cdot \frac{1}{2} $$
Step 1: Expand each term
1. Expand $(2x - 1)^2$:
$$ (2x - 1)^2 = 4x^2 - 4x + 1 $$
2. Expand $(x + 2)^2$:
$$ (x + 2)^2 = x^2 + 4x + 4 $$
3. Expand $(2x + 4)^2$:
$$ (2x + 4)^2 = 4x^2 + 16x + 16 $$
4. Expand $2 \cdot (x + 2) \cdot (2x + 4) \cdot \frac{1}{2}$:
$$ 2 \cdot (x + 2) \cdot (2x + 4) \cdot \frac{1}{2} = (x + 2)(2x + 4) $$
$$ (x + 2)(2x + 4) = 2x^2 + 8x + 8 $$
Step 2: Substitute and simplify
Substitute all expanded terms into the law of cosines equation:
$$ 4x^2 - 4x + 1 = x^2 + 4x + 4 + 4x^2 + 16x + 16 - (2x^2 + 8x + 8) $$
Simplify the right-hand side:
$$ x^2 + 4x + 4 + 4x^2 + 16x + 16 - 2x^2 - 8x - 8 = 3x^2 + 12x + 12 $$
Equating both sides:
$$ 4x^2 - 4x + 1 = 3x^2 + 12x + 12 $$
Step 3: Rearrange into standard quadratic form
Rearrange all terms to one side:
$$ 4x^2 - 4x + 1 - 3x^2 - 12x - 12 = 0 $$
$$ x^2 - 16x - 11 = 0 $$
Step 4: Solve the quadratic equation
Use the quadratic formula:
$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 1,\; b = -16,\; c = -11 $$
Substitute the values:
$$ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(-11)}}{2(1)} $$
$$ x = \frac{16 \pm \sqrt{256 + 44}}{2} $$
$$ x = \frac{16 \pm \sqrt{300}}{2} $$
$$ x = \frac{16 \pm 10\sqrt{3}}{2} $$
$$ x = 8 \pm 5\sqrt{3} $$
Step 5: Choose the valid solution
Since $x$ represents a length, it must be positive. Thus:
$$ x = 8 + 5\sqrt{3} $$
Final Answer
$$ \boxed{x = 8 + 5\sqrt{3}} $$
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