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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
\[16\sqrt{3}\]
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