The angle of elevation of the top of a vertical tower from a point...

Let the height of the tower be $h = 50(\sqrt{3} + 1)$ m.

Let the initial point be $A$. The horizontal distance of $A$ from the foot of the tower is $x = h \cot 45^\circ = h$.

The observer walks a distance $d$ along a slope of $30^\circ$.

Horizontal distance covered = $d \cos 30^\circ = \dfrac{d\sqrt{3}}{2}$.

Vertical distance covered = $d \sin 30^\circ = \dfrac{d}{2}$.

From the new position, the horizontal distance to the tower is $h - \dfrac{d\sqrt{3}}{2}$ and the vertical distance to the top of the tower is $h - \dfrac{d}{2}$.

The new angle of elevation is $60^\circ$, so:

$\tan 60^\circ = \dfrac{h - \dfrac{d}{2}}{h - \dfrac{d\sqrt{3}}{2}}$

$\sqrt{3} \left( h - \dfrac{d\sqrt{3}}{2} \right) = h - \dfrac{d}{2}$

$h\sqrt{3} - \dfrac{3d}{2} = h - \dfrac{d}{2}$

$h(\sqrt{3} - 1) = d$

Substitute $h = 50(\sqrt{3} + 1)$:

$d = 50(\sqrt{3} + 1)(\sqrt{3} - 1)$

$d = 50(3 - 1) = 100$.

Answer: $100$