1. **State the problem:** We need to find the tangent of angle $B$ in a right triangle with vertices $A$, $B$, and $C$. The side lengths given are $AB=17$ units and $BC=8$ units, with the right angle at vertex $C$. 2. **Recall the definition of tangent in a right triangle:** $$\tan(B) = \frac{\text{opposite side to } B}{\text{adjacent side to } B}$$ 3. **Identify the sides relative to angle $B$:** - Opposite side to $B$ is $AC$ (unknown). - Adjacent side to $B$ is $BC=8$ units. 4. **Find the length of side $AC$ using the Pythagorean theorem:** $$AB^2 = BC^2 + AC^2$$ $$17^2 = 8^2 + AC^2$$ $$289 = 64 + AC^2$$ $$AC^2 = 289 - 64 = 225$$ $$AC = \sqrt{225} = 15$$ 5. **Calculate $\tan(B)$:** $$\tan(B) = \frac{AC}{BC} = \frac{15}{8}$$ 6. **Simplify the fraction:** The fraction $\frac{15}{8}$ is already in simplest form. **Final answer:** $$\boxed{\tan(B) = \frac{15}{8}}$$