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The image displays a right-angled triangle $$\triangle ABM$$, with the right angle at B. We are given that $$\angle BAM = 30^{\circ}$$ and $$AM - MB = 7$$.

This problem involves trigonometry in a right-angled triangle. We can express the sides AM and MB in terms of AB using trigonometric ratios.

Let $$AB = x$$. Then, in $$\triangle ABM$$:

• $$MB = AB \tan(30^{\circ}) = x \cdot \frac{1}{\sqrt{3}} = \frac{x}{\sqrt{3}}$$
• $$AM = \frac{AB}{\cos(30^{\circ})} = \frac{x}{\frac{\sqrt{3}}{2}} = \frac{2x}{\sqrt{3}}$$

We are given $$AM - MB = 7$$. Substituting the expressions for AM and MB:

• $$\frac{2x}{\sqrt{3}} - \frac{x}{\sqrt{3}} = 7$$
• $$\frac{x}{\sqrt{3}} = 7$$
• $$x = 7\sqrt{3}$$

Therefore, $$AB = 7\sqrt{3}$$.

We can find the lengths of the other sides:

• $$MB = \frac{7\sqrt{3}}{\sqrt{3}} = 7$$
• $$AM = \frac{2 \cdot 7\sqrt{3}}{\sqrt{3}} = 14$$

Check: $$AM - MB = 14 - 7 = 7$$, which matches the given condition.

Final Answer: The lengths of the sides are $$AB = 7\sqrt{3}$$, $$MB = 7$$, and $$AM = 14$$.