From the top of a lamp post of height x metres, two objects on the ground on the sameside of it (and in line with the foot of the lamp post) are observed at angles of depression of $$30^\circ$$ and $$60^\circ$$, respectively. The distance between the objects is $$32\sqrt{3}$$ m. The value of x is:

from the question we draw the diagram 

Let BD =$$y $$

In the $$\triangle ABC $$ 

$$ \tan 30^\circ = \frac{x} {32\sqrt {3} +y} $$ 

$$ \frac {1} {\sqrt {3} } = \frac {x} {32 \sqrt {3} +y} $$ ...... Equestion (1)

then $$\triangle ABC $$

$$ \tan 60^\circ = \frac{x} {y} $$

$$ \sqrt {3} = \frac{x} {y} $$

$$ y = \frac {x} {\sqrt {3} } $$

Put the value $$ y $$ in Equation  (1)

$$ \frac {1} {\sqrt {3}} = \frac {x} { 32 \sqrt {3} + \frac {x} {\sqrt {3}}} $$

$$\Rightarrow 32 + \frac{x}{3} = x $$

$$ \Rightarrow 2 \frac {x}{3} = 32 $$

$$\Rightarrow x = 48  $$ Ans