A tower stands vertically on the ground. From a point on the ground which is 18 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be $$30^\circ$$. Find the height of the tower.

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The problem involves calculating the height of the tower using trigonometry. Let:  
- $$AB$$ represent the height of the tower.  
- $$AC = 18 \, 	ext{m}$$, which is the horizontal distance from the point to the base of the tower.  
- The angle of elevation is $$30^\circ$$.

Using the tangent function:  
$$
	an 30^\circ = \frac{	ext{Height of Tower (AB)}}{	ext{Distance (AC)}} = \frac{AB}{18}
$$
We know that $$	an 30^\circ = \frac{1}{\sqrt{3}}$$. Substituting this value:  
$$
\frac{1}{\sqrt{3}} = \frac{AB}{18}
$$  
Rearranging to solve for $$AB$$:  
$$
AB = \frac{18}{\sqrt{3}}
$$  
Rationalizing the denominator:  
$$
AB = \frac{18 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = 6\sqrt{3} \, 	ext{m}
$$

Hence, the height of the tower is **$$6\sqrt{3} \, 	ext{m}$$**.

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