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An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his eye to the top of the tower is 30º. The heights of the tower is:
Answer & Solution
Correct Answer:
Option
A
Let AB be the observer and CD be the tower.
Draw BE ⊥ CD
Then CE = AB = 1.6m
BE = AC = 20√3m
$$\frac{{DE}}{{BE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ \Rightarrow DE = \frac{{BE}}{{\sqrt 3 }} = \frac{{20\sqrt 3 }}{{\sqrt 3 }} = 20$$
∴ CD = CE + DE = (1.6 + 20) m = 21.6 m
Draw BE ⊥ CD
Then CE = AB = 1.6m
BE = AC = 20√3m
$$\frac{{DE}}{{BE}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ \Rightarrow DE = \frac{{BE}}{{\sqrt 3 }} = \frac{{20\sqrt 3 }}{{\sqrt 3 }} = 20$$
∴ CD = CE + DE = (1.6 + 20) m = 21.6 m
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