Question : A person 1.8 metres tall is $30 \sqrt{3}$ metres away from a tower. If the angle of elevation from his eye to the top of the tower is 30°, then what is the height (in m) of the tower?
Option 1: 32.5
Option 2: 37.8
Option 3: 30.5
Option 4: 31.8
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Correct Answer: 31.8
Solution :
Given: A person 1.8 metres tall is $30 \sqrt{3}$ metres away from a tower.
The angle of elevation from his eye to the top of the tower is $30^{\circ}$.
We know the formulas, $\tan 30^{\circ}=\frac{1}{\sqrt3}$ and $\tan \theta=\frac{\text{Perpendicular}}{\text{Base}}$.
Let the tower's height and the person's height be AB and DE, respectively.
In $\triangle ACD$,
$\frac{AC}{DC}=\tan 30^{\circ}$
⇒ $\frac{x}{30\sqrt3}=\frac{1}{\sqrt3}$
⇒ $x=30$ m
The height of the tower AB = 30 + 1.8 = 31.8 m
Hence, the correct answer is 31.8 m.
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