Question : From the top of a hill 240 m high, the angles of depression of the top and bottom of a pole are 30° and 60°, respectively. The difference (in m) between the height of the pole and its distance from the hill is:
Option 1: $120(2-\sqrt3$)
Option 2: $120(\sqrt3-1)$
Option 3: $80(\sqrt3-1$)
Option 4: $80(2-\sqrt3)$
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Correct Answer: $80(2-\sqrt3)$
Solution :
As per the figure shown above,
AD = 240 m
In $\triangle ADE$,
$⇒\tan 60^{\circ}=\frac{AD}{DE}$
$⇒\sqrt3=\frac{240}{DE}$
$⇒DE=80\sqrt3$
In $\triangle ABC$,
$⇒\tan 30^{\circ}=\frac{AB}{BC}$
$⇒\frac{1}{\sqrt3}=\frac{AB}{BC}$
$⇒BC=AB{\sqrt3}$
From the figure, BC = DE.
$⇒80\sqrt3=AB{\sqrt3}$
$⇒AB=80$ m
The height of the pole BD = 240 – AB = 240 – 80 = 160 m
The difference between the height of the pole and its distance from the hill,
$=160-80\sqrt3=80(2-\sqrt3)$
Hence, the correct answer is $80(2-\sqrt3)$.
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