A 1.8 m tall man is standing at some distance from a 31.8 m tall building. The a

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A 1.8 m tall man is standing at some distance from a 31.8 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.

Options:

A . 30 m

B .

20 \sqrt{3}

m

C .

20

m

D .

30 \sqrt{3}

m

Answer: Option B
:
B

The situation can be represented by the figure above,
$t a n (∠ A C B) = \frac{A B}{B C} \Rightarrow t a n 60^{\circ} = \frac{30}{B C} \Rightarrow B C = \frac{30}{t a n 60^{\circ}} = 10 \sqrt{3} m \Rightarrow t a n (∠ A D B) = \frac{A B}{B D} \Rightarrow t a n 30^{\circ} = \frac{30}{B D} \Rightarrow B D = \frac{30}{\frac{1}{\sqrt{3}}} = 30 \sqrt{3} m \Rightarrow D i s t a n c e w a l k e d = D C = B D - B C = 30 \sqrt{3} - 10 \sqrt{3} = 20 \sqrt{3} m$

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