Let the first tower be \(AB\) (top \(A\), base \(B\)) and the second tower be \(CD\) (top \(C\), base \(D\)). The bases \(B\) and \(D\) are 25 m apart.

From the top \(A\): angle of depression to base \(D\) is \(60^\circ\) and to top \(C\) is \(30^\circ\).

1. From right \(\triangle ABD\): \[ \tan 60^\circ=\frac{AB}{BD}=\frac{AB}{25}\;\Rightarrow\; AB=25\sqrt{3}. \]
2. From right \(\triangle ACD\): vertical difference \(=AB-CD\) and horizontal \(=25\). \[ \tan 30^\circ=\frac{AB-CD}{25}=\frac{1}{\sqrt{3}} \;\Rightarrow\; AB-CD=\frac{25}{\sqrt{3}}. \]
3. Substitute \(AB=25\sqrt{3}\): \[ 25\sqrt{3}-CD=\frac{25}{\sqrt{3}} \;\Rightarrow\; CD=25\!\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) =\frac{50}{\sqrt{3}}\;\text{m}. \]