Two towers face each other separated by a distance of 25 meters. As seen from the top of the first tower, the angle of depression of the second tower’s base is 60° and that of the top is 30°. The height (in meters) of the second tower is

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 \(AB=25\sqrt{3}\): \[ 25\sqrt{3}-CD=\frac{25}{\sqrt{3}} \;\Rightarrow\; CD=25\!\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) =\frac{50}{\sqrt{3}}\;\text{m}. \]