First, determine the vectors \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\):

\(\overrightarrow{AC} = -6\mathbf{i} + 10\mathbf{k}\)

\(\overrightarrow{BC} = -8\mathbf{j} + 10\mathbf{k}\)

Calculate the dot product \(\overrightarrow{AC} \cdot \overrightarrow{BC}\):

\(\overrightarrow{AC} \cdot \overrightarrow{BC} = (-6)(0) + (0)(-8) + (10)(10) = 100\)

Find the magnitudes of \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\):

\(|\overrightarrow{AC}| = \sqrt{(-6)^2 + 0^2 + 10^2} = \sqrt{136}\)

\(|\overrightarrow{BC}| = \sqrt{0^2 + (-8)^2 + 10^2} = \sqrt{164}\)

Use the scalar product formula to find \(\cos \angle ACB\):

\(\overrightarrow{AC} \cdot \overrightarrow{BC} = |\overrightarrow{AC}| |\overrightarrow{BC}| \cos \angle ACB\)

\(100 = \sqrt{136} \sqrt{164} \cos \angle ACB\)

\(\cos \angle ACB = \frac{100}{\sqrt{136} \sqrt{164}}\)

Calculate \(\angle ACB\):

\(\angle ACB = \cos^{-1}\left(\frac{100}{\sqrt{136} \sqrt{164}}\right) \approx 48.0^{\circ}\)