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.

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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,

$\tan 30 ° = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{48} \Rightarrow h = \frac{48}{\sqrt{3}} = 16 \sqrt{3} m$
Thus, the height of the building is $16 \sqrt{3}$ m.

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A 1.5-m-tall boy is standing at some distance from a 30-m-tall building. The angle of elevation from his eyes to the top of the building increases from $30^{\circ}$ to $60^{\circ}$ as he walks towards the building. Find the distance he walked towards the building.

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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.

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°=ABBC⇒13=h48⇒h=483=163 m Thus, the height of the building is 163 m.