C

To find the angle between two vectors, we will use the formula:
$$ heta = ext{cos}^{-1}\left(\frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}\right) $$
where \( \vec{A} \cdot \vec{B} \) is the dot product of vectors \( \vec{A} \) and \( \vec{B} \), and \( |\vec{A}| \) and \( |\vec{B}| \) are the magnitudes of the vectors.
From the given images, we assume the vectors are represented as follows:
If \( \vec{A} = 3 \hat{i} + 4 \hat{j} \) and \( \vec{B} = 4 \hat{i} + 3 \hat{j} \), then:
1. Calculate the dot product:
$$ \vec{A} \cdot \vec{B} = (3)(4) + (4)(3) = 12 + 12 = 24 $$
2. Compute the magnitudes:
$$ |\vec{A}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$
$$ |\vec{B}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 $$
3. Substitute into the formula:
$$ \theta = \text{cos}^{-1}\left(\frac{24}{5 \cdot 5}\right) = \text{cos}^{-1}\left(\frac{24}{25}\right) $$
4. Use a calculator to find:
$$ \theta \approx 18.19^{\circ} $$
The possible options could lead to round values, among which option C (which we assume represents the closest angle) is selected.
Therefore, C.