The angle of elevation of a flying drone from a point on the ground is $$60$$ degrees. After flying away from the point for 5 seconds the angle of elevation drops to $$30$$ degrees. If the drone is flying horizontally at a constant height of $$1000\surd3$$ m, the distance travelled by the drone is:

Initially, the drone was PD away from the point of observation (P). 
Angle APD is $$60$$ degrees. 
And in right-angled triangle ADP, AD $$=1000\surd3$$ m.

$$\Rightarrow$$ Tan (APD) $$=\dfrac{\text{AD}}{\text{DP}}$$

$$\Rightarrow$$ Tan $$60$$ $$=\dfrac{1000\sqrt{\ 3}}{\text{PD}}$$

$$\Rightarrow$$ $$\sqrt{\ 3}=\dfrac{1000\sqrt{\ 3}}{\text{PD}}$$

$$\Rightarrow$$ PD $$=1000$$ m.

Similarly, in triangle BCP, we are given that angle BPC is $$30$$ degrees, as the angle of elevation changed.

$$\Rightarrow$$ Tan $$30$$ $$=\dfrac{1000\sqrt{\ 3}}{\text{PC}}$$

$$\Rightarrow$$ $$\dfrac{1}{\sqrt{\ 3}}=\dfrac{1000\sqrt{\ 3}}{\text{PC}}$$

$$\Rightarrow$$ PC $$=3000$$ m.

The distance travelled by the drone is CD, i.e. $$3000-1000=2000$$ m.

Hence, Option A is correct.