Answer
$ \theta =\dfrac{\pi}{2}$ or $90 ^{\circ}$; $\overrightarrow{CA} \perp\overrightarrow{DB}$
Work Step by Step
We have two diagonals, let us say $\overrightarrow{CA}$ and $\overrightarrow{DB}$, of a rhombus ABCD .
The formula to calculate the angle between two diagonals $\overrightarrow{CA}$ and $\overrightarrow{DB}$ of a rhombus is:
$ \theta = \cos ^{-1} (\dfrac{\overrightarrow{CA} \cdot \overrightarrow{DB}}{|\overrightarrow{CA}||\overrightarrow{DB}|})=\cos ^{-1} (\dfrac{(\overrightarrow{DA})^2-(\overrightarrow{AB})^2}{ |\overrightarrow{CA}||\overrightarrow{DB}|})$
As ABCD is a rhombus, thus $|\overrightarrow{DA}|=|\overrightarrow{AB}|$
or, $\theta=\cos ^{-1} (0)$
or, $ \theta =\dfrac{\pi}{2}$ or $90 ^{\circ}$
Hence, $\overrightarrow{CA} \perp\overrightarrow{DB}$
