[Geometry]
Problem: Triangle $A B C$ has vertices $A(0,5), B(3,0)$ and $C(8,3)$. Determine the measure of $\angle A C B$.

Solution: First, we calculate the side lengths of $\triangle A B C$ :

$$
\begin{aligned}
& A B=\sqrt{(0-3)^{2}+(5-0)^{2}}=\sqrt{34} \\
& B C=\sqrt{(3-8)^{2}+(0-3)^{2}}=\sqrt{34} \\
& A C=\sqrt{(0-8)^{2}+(5-3)^{2}}=\sqrt{68}
\end{aligned}
$$

Since $A B=B C$ and $A C=\sqrt{2} A B=\sqrt{2} B C$, then $\triangle A B C$ is an isosceles right-angled triangle, with the

<img_3758>
right angle at $B$.

Therefore, $\angle A C B=45^{\circ}$.
First, we calculate the side lengths of $\triangle A B C$ :

$$
\begin{aligned}
& A B=\sqrt{(0-3)^{2}+(5-0)^{2}}=\sqrt{34} \\
& B C=\sqrt{(3-8)^{2}+(0-3)^{2}}=\sqrt{34} \\
& A C=\sqrt{(0-8)^{2}+(5-3)^{2}}=\sqrt{68}
\end{aligned}
$$

Line segment $A B$ has slope $\frac{5-0}{0-3}=-\frac{5}{3}$.

Line segment $B C$ has slope $\frac{0-3}{3-8}=\frac{3}{5}$.

Since the product of these two slopes is -1 , then $A B$ and $B C$ are perpendicular.

Therefore, $\triangle A B C$ is right-angled at $B$.

Since $A B=B C$, then $\triangle A B C$ is an isosceles right-angled triangle, so $\angle A C B=45^{\circ}$.
First, we calculate the side lengths of $\triangle A B C$ :

$$
\begin{aligned}
& A B=\sqrt{(0-3)^{2}+(5-0)^{2}}=\sqrt{34} \\
& B C=\sqrt{(3-8)^{2}+(0-3)^{2}}=\sqrt{34} \\
& A C=\sqrt{(0-8)^{2}+(5-3)^{2}}=\sqrt{68}
\end{aligned}
$$

Using the cosine law,

$$
\begin{aligned}
A B^{2} & =A C^{2}+B C^{2}-2(A C)(B C) \cos (\angle A C B) \\
34 & =68+34-2(\sqrt{68})(\sqrt{34}) \cos (\angle A C B) \\
0 & =68-2(\sqrt{2} \sqrt{34})(\sqrt{34}) \cos (\angle A C B) \\
0 & =68-68 \sqrt{2} \cos (\angle A C B) \\
68 \sqrt{2} \cos (\angle A C B) & =68 \\
\cos (\angle A C B) & =\frac{1}{\sqrt{2}}
\end{aligned}
$$

Since $\cos (\angle A C B)=\frac{1}{\sqrt{2}}$ and $0^{\circ}<\angle A C B<180^{\circ}$, then $\angle A C B=45^{\circ}$.

Final answer: $45^{\circ}$