On the same side of a tower, two objects are located. When observed from the top of the tower, their angles of depression are $45^o$ and $60^o$. If the height of the tower is $50\sqrt{3}$, then the distance between the objects is 

Let AB be the tower which is $50\sqrt{3}m$ high and C and D be the positions of the two towers such that 

 $\angle ADB=45^0$ and $\angle ACB=60^0$

Now, in triangle ABC,

 $\tan 60^0=\frac{AB}{BC}$

   $\Rightarrow$$\sqrt{3}=\frac{}{50\sqrt{3}}BC$

  $\Rightarrow$$BC=50m$

Also, in triangle ABD,

$\tan 45^0=\frac{AB}{BD}$ 

 $\Rightarrow \tan 45^0=\frac{AB}{BC+CD}$

           $\Rightarrow$$1=\frac{}{50\sqrt{3}}50+CD$

$\Rightarrow$$50+CD=50\sqrt{3}$

         $\Rightarrow$$CD=50\sqrt{3}-50$

          $\Rightarrow$$CD=50\left(\sqrt{3}-1\right)m$.

Hence option C is correct.