Step 1: Let the height of the tower be $h$ and let the friend be at $C$ initially and at $D$ after $30$ s.
Step 2: At angle of depression $30^\circ$ (point $C$) :
$$\tan 30^\circ = \frac{h}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{BC} \Rightarrow BC = \sqrt{3}h$$
Step 3: At angle of depression $60^\circ$ (point $D$) :
$$\tan 60^\circ = \frac{h}{BD} \Rightarrow \sqrt{3} = \frac{h}{BD} \Rightarrow BD = \frac{h}{\sqrt{3}}$$
Step 4: Distance covered in $30$ seconds :
$$CD = BC - BD = \sqrt{3}h - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$$
Step 5: Remaining distance to the foot of the tower :
$$BD = \frac{h}{\sqrt{3}}$$
Step 6: Since the speed is uniform, time is proportional to distance :
$$\frac{\text{time for } BD}{\text{time for } CD} = \frac{BD}{CD} = \frac{\frac{h}{\sqrt{3}}}{\frac{2h}{\sqrt{3}}} = \frac{1}{2}$$
Step 7: $$\text{Required time} = \frac{1}{2} \times 30 = 15\ \text{seconds}$$
Hence the friend takes $15$ seconds to reach the foot of the tower.