Question : From a point 12 m above the water level, the angle of elevation of the top of a hill is 60° and the angle of depression of the base of the hill is 30°. What is the height (in m) of the hill?
Option 1: $48 \sqrt{3}$
Option 2: $36$
Option 3: $36 \sqrt{3}$
Option 4: $48$
Correct Answer: $48$
Solution :
Given :
Point 12 m above the water level
The angle of elevation of the top of a hill = $60^\circ$
The angle of depression of the base of the hill = $30^\circ$
Now, In triangle ABE
$\Rightarrow \tan 30^\circ\ =\ \frac{AB}{BE}$
$\Rightarrow \frac{1}{\sqrt{3}}\ =\ \frac{12}{BE}$
$\Rightarrow BE\ = 12\sqrt3$
As we know,
BE = AD and AB = DE
In triangle ACD
$ \tan60^\circ\ =\ \frac{CD}{AD}$
$\Rightarrow \sqrt{3}\ =\ \frac{CD}{12\sqrt{3}}$
$\Rightarrow CD = 36$
$\Rightarrow CE\ =\ CD\ +\ DE$
$\Rightarrow CE\ =\ 36\ +\ 12\ =\ 48$ m
Hence, the correct answer is $48$.
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