A vertical pole stands inside a triangular park with side lengths...

Let the foot of the pole be $P$ and its height be $h$.

The angles of elevation of the top of the pole from the nearest point on each of the three sides are equal. The nearest point on a side is the foot of the perpendicular drawn from $P$ to that side. Since the angles of elevation are equal, the perpendicular distances from $P$ to the three sides must be equal.

This implies that $P$ is the incentre of the triangular park, and the horizontal distance to each side is the inradius $r$.

The sides of the park are $a = 13$ m, $b = 14$ m, and $c = 15$ m.
The semi-perimeter is $s = \dfrac{13 + 14 + 15}{2} = 21$ m.

Using Heron's formula, the area of the triangle is:
$\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21 \times 8 \times 7 \times 6} = 84$ m$^2$.

The inradius $r$ is given by:
$r = \dfrac{\Delta}{s} = \dfrac{84}{21} = 4$ m.

From the right-angled triangle formed by the pole and the inradius, we have:
$\tan(30^{\circ}) = \dfrac{h}{r}$

$h = r \tan(30^{\circ}) = 4 \times \dfrac{1}{\sqrt{3}} = \dfrac{4}{\sqrt{3}}$ m.

Answer: $\dfrac{4}{\sqrt{3}}$ m