Step 1: Set up the two distances in terms of the tower height.
Let the height of the tower be $h$, let $C$ be the foot of the tower, $A$ the friend's first position and $B$ the friend's position 30 seconds later, with $B$ closer to the tower than $A$.
At $A$, the angle of depression is $30^{\circ}$, so:
\[ \tan 30^{\circ} = \frac{h}{AC} \implies AC = \frac{h}{\tan 30^{\circ}} = h\sqrt{3} \]
At $B$, the angle of depression is $60^{\circ}$, so:
\[ \tan 60^{\circ} = \frac{h}{BC} \implies BC = \frac{h}{\tan 60^{\circ}} = \frac{h}{\sqrt{3}} \]
Step 2: Compare the two distances as a ratio instead of computing each one as a number.
Since both $AC$ and $BC$ are written in terms of the same $h$, take their ratio so that $h$ cancels out:
\[ \frac{AC}{BC} = \frac{h\sqrt{3}}{h/\sqrt{3}} = \sqrt{3} \times \sqrt{3} = 3 \]
So $AC = 3 \times BC$.
Step 3: Use the fact that speed is uniform, so time is proportional to distance covered.
The distance covered while walking from $A$ to $B$ is:
\[ AB = AC - BC = 3(BC) - BC = 2(BC) \]
The friend walked this distance $AB = 2(BC)$ in 30 seconds, moving at a constant (uniform) speed. Since the speed never changes, the time taken to cover any distance is directly proportional to that distance: covering a longer stretch always takes proportionally longer.
The remaining distance to the foot of the tower is $BC$, which is exactly half of $AB = 2(BC)$.
Because time is proportional to distance at a constant speed, covering half the distance takes half the time:
\[ \text{Remaining time} = 30 \times \frac{BC}{AB} = 30 \times \frac{BC}{2(BC)} = 30 \times \frac{1}{2} = 15 \text{ seconds} \]
Notice that the actual values of $h$, $AC$ and $BC$ never had to be calculated as numbers, since only their ratio mattered.
Final Answer:
The friend takes 15 more seconds to reach the foot of the tower.
\[ \boxed{15 \text{ seconds}} \]