A vertical pole stands at a corner O of a horizontal rhombus OABC ....

Let the height of the vertical pole at $O$ be $h$.

The horizontal distance to the adjacent corner $A$ is:
$OA = h \cot 45^\circ = h$

Since $OABC$ is a rhombus, all its sides are equal. Thus, the side length of the rhombus is $a = OA = h$.

The horizontal distance to the opposite corner $B$ (which is the length of the diagonal $OB$) is:
$OB = h \cot 30^\circ = h\sqrt{3}$

In $\triangle OAB$, the lengths of the sides are $OA = a$, $AB = a$, and $OB = a\sqrt{3}$.

Using the cosine rule in $\triangle OAB$ to find $\angle OAB$:
$\cos(\angle OAB) = \dfrac{OA^2 + AB^2 - OB^2}{2 \cdot OA \cdot AB}$

$\cos(\angle OAB) = \dfrac{a^2 + a^2 - (a\sqrt{3})^2}{2 \cdot a \cdot a}$

$\cos(\angle OAB) = \dfrac{2a^2 - 3a^2}{2a^2} = -\dfrac{1}{2}$

Therefore, $\angle OAB = 120^\circ$.

In a rhombus, adjacent angles are supplementary. The angle at $O$, which is $\angle AOC$, is:
$\angle AOC = 180^\circ - 120^\circ = 60^\circ$

Thus, the acute angle of the rhombus is $60^\circ$.

Answer: $60^\circ$