A 1.8 m tall man is standing at some distance from a building. The angle of elev

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A 1.8 m tall man is standing at some distance from a building. The angle of elevation from his eyes to the top of the building increases from

30^{\circ}

to

60^{\circ}

as he walks 20 m towards the building. Find the height of the building (in m).

Options:

A . 10 m

B . 20√3 m

C . 20 m

D . (10√3 + 1.8) m

Answer: Option D
:
D

A 1.8 M Tall Man Is Standing At Some Distance From A Buildin...

The situation can be represented by the figure above.

t a n (∠ A C B) = \frac{A B}{B C} \Rightarrow B C = \frac{A B}{t a n 60^{\circ}} . . . . . (1) t a n (∠ A D B) = \frac{A B}{B D} \Rightarrow t a n 30^{\circ} = \frac{A B}{B C + C D} \Rightarrow B C + C D = \frac{A B}{t a n 30^{\circ}} . . . . . (2) on subtracting equation(1) from (2) \Rightarrow C D = \frac{A B}{t a n 30^{\circ}} - \frac{A B}{t a n 60^{\circ}} \Rightarrow A B = 10 \sqrt{3}

Hence, the height of the building is
= AB + BG =

(10 \sqrt{3} + 1.8) m .

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