#### Question

A boy standing at a distance of 48 meters from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.

#### Solution

Let AB be the building and C be the position of the boy from the building.

Suppose the height of the building be h m.

Here, BC = 48 m and ∠ACB = 30º.

In right ∆ABC,

\[\tan30^\circ = \frac{AB}{BC}\]

\[ \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{48}\]

\[ \Rightarrow h = \frac{48}{\sqrt{3}} = 16\sqrt{3} m\]

Thus, the height of the building is

\[16\sqrt{3}\]

Is there an error in this question or solution?

Solution A Boy Standing at a Distance of 48 Meters from a Building Observes the Top of the Building and Makes an Angle of Elevation of 30°. Find the Height of the Building. Concept: Heights and Distances.