Question : From a tower 125 metres high, the angle of depression of two objects, which are in a horizontal line through the base of the tower, are $45^{\circ}$ and $30^{\circ}$ and they are on the same side of the tower. The distance (in metres) between the objects is:
Option 1: $125\sqrt{3}$
Option 2: $125\left ( {\sqrt3-1} \right )$
Option 3: $\frac{125}{\left ( {\sqrt3-1} \right )}$
Option 4: $125\left ({\sqrt3+1} \right)$
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Correct Answer: $125\left ( {\sqrt3-1} \right )$
Solution :
We have,
Height of the tower ($AB$) = 125 m
Angle of depression = 45° and 30°
In $\Delta ABC$,
$\tan 45^{\circ} = \frac{AB}{BC}$
$⇒1 = \frac{125}{BC}$
$⇒BC=125$.............(i)
In $\Delta ADB$,
$\tan 30^{\circ} = \frac{AB}{BD}$
$⇒\frac{1}{\sqrt3} = \frac{125}{CD+BC}$
Substituting the value of $AB$ from equation (i), we get,
$⇒\frac{1}{\sqrt3} = \frac{125}{CD+125}$
$⇒CD=125\sqrt3-125$
$\therefore CD=125(\sqrt3-1)$ m
Hence, the correct answer is $125(\sqrt3-1)$.
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