A vertical pole of height 15 m stands at the centre of a circular...

Let the height of the pole be $h = 15\text{ m}$ and the centre of the park be $O$.

Point $A$ lies on the boundary of the circular park. The horizontal distance $OA$ is the radius $R$ of the park.
Since the angle of elevation from $A$ is $30^\circ$:
$R = h \cot 30^\circ = 15\sqrt{3}\text{ m}$

The angle of elevation of the top of the pole is maximum when the observer is at the minimum horizontal distance from the pole. This occurs at the foot of the perpendicular from $O$ to the chord $AB$. Let this minimum distance be $d$.
Since the maximum angle of elevation is $45^\circ$:
$d = h \cot 45^\circ = 15(1) = 15\text{ m}$

The length of the chord $AB$ can be found using the right-angled triangle formed by the radius, the perpendicular $d$, and half of the chord:
$\dfrac{AB}{2} = \sqrt{R^2 - d^2}$
$\dfrac{AB}{2} = \sqrt{(15\sqrt{3})^2 - 15^2} = \sqrt{675 - 225} = \sqrt{450} = 15\sqrt{2}\text{ m}$

The full length of the path is $AB = 2 \times 15\sqrt{2} = 30\sqrt{2}\text{ m}$.

The square of the length of the path is:
$AB^2 = (30\sqrt{2})^2 = 1800$

Answer: $1800$