From the top of a vertical cliff of 40m high, the angle of depression of an object that is at the base level of the cliff is $$30^{\circ}$$. If the object is at a distance $$x$$ units from the base of the cliff, then $$\frac{1}{x + 40}$$

A

$$\frac{\sqrt{3} -1}{80}$$

B

$$\frac{4\sqrt{3}}{40}$$

C

$$\frac{1}{40\sqrt{3}}$$

D

$$\frac{1}{40 - \sqrt{3}}$$