The angle of elevation of the top of a building from a point P on the horizontal is $$30^\circ$$. If point P is 48 m away from the foot of the building, what is the height of the building?

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Accepted Answer

The problem involves a right triangle where:  
- The distance from point P to the foot of the building = 48 m (base).  
- The angle of elevation = $$30^\circ$$.  
- Let the height of the building = $$h$$.

Using the tangent function:  
$$
	an(30^\circ) = \frac{	ext{Height}}{	ext{Base}}.
$$  
Substitute values:  
$$
	an(30^\circ) = \frac{1}{\sqrt{3}}, \quad 	ext{Base} = 48.
$$  
So:  
$$
\frac{1}{\sqrt{3}} = \frac{h}{48}.
$$  
Solve for $$h$$:  
$$
h = \frac{48}{\sqrt{3}} = 16\sqrt{3}.
$$

Thus, the height of the building is **$$16\sqrt{3}$$** m.

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