Step 1: Understanding the Concept:
We are given the vertices of a parallelogram. We need to find the equations or vectors representing its diagonals and then find the angle between them.
The diagonals are the line segments connecting opposite vertices, AC and BD.
Step 2: Key Formula or Approach:
The angle $\theta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by $\cos \theta = \frac{|\vec{u} \cdot \vec{v}|}{|\vec{u}| |\vec{v}|}$.
Alternatively, if the lines have slopes $m_1$ and $m_2$, the acute angle is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Step 3: Detailed Explanation:
Let's find the vectors representing the diagonals AC and BD.
Vector $\vec{AC} = (x_c - x_a)\hat{i} + (y_c - y_a)\hat{j} = (2 - 2)\hat{i} + (3 - (-1))\hat{j} = 0\hat{i} + 4\hat{j}$.
Vector $\vec{BD} = (x_d - x_b)\hat{i} + (y_d - y_b)\hat{j} = (4 - 0)\hat{i} + (0 - 2)\hat{j} = 4\hat{i} - 2\hat{j}$.
Now, let's find the acute angle $\theta$ between them using the dot product:
\[ \cos \theta = \frac{|\vec{AC} \cdot \vec{BD}|}{|\vec{AC}| |\vec{BD}|} \]
\[ \vec{AC} \cdot \vec{BD} = (0)(4) + (4)(-2) = 0 - 8 = -8 \]
\[ |\vec{AC}| = \sqrt{0^2 + 4^2} = 4 \]
\[ |\vec{BD}| = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]
\[ \cos \theta = \frac{|-8|}{4 \cdot 2\sqrt{5}} = \frac{8}{8\sqrt{5}} = \frac{1}{\sqrt{5}} \]
Since $\cos \theta = \frac{1}{\sqrt{5}}$, we can find $\tan \theta$ using a right-angled triangle where adjacent side is 1 and hypotenuse is $\sqrt{5}$.
Opposite side $= \sqrt{(\sqrt{5})^2 - 1^2} = \sqrt{5 - 1} = \sqrt{4} = 2$.
Therefore, $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{2}{1} = 2$.
This implies $\theta = \tan^{-1} 2$.
Step 4: Final Answer:
The acute angle between the diagonals is $\tan^{-1} 2$.