Finding tangent for angle of right triangle

Given : Right $\mathrm{\u25b3}ABC$ CH is altitude and $AH=2$, $BH=8$. Here is drawing

Find the $tan\mathrm{\angle}CBA$

Here is how I solved the exercise: From that fact that ABC is right triangle $\Rightarrow A{C}^{2}=AH\cdot AB$ (Altitude on hypotenuse theorem) \Rightarrow AC=\sqrt{2 \cdot 10}=\sqrt{20} and $tan\mathrm{\angle}CBA=\frac{AC}{AB}=\frac{2\ast \sqrt{5}}{10}=\frac{\sqrt{5}}{5}$.

However the answer in test is 12. I can't find my mistake.

Given : Right $\mathrm{\u25b3}ABC$ CH is altitude and $AH=2$, $BH=8$. Here is drawing

Find the $tan\mathrm{\angle}CBA$

Here is how I solved the exercise: From that fact that ABC is right triangle $\Rightarrow A{C}^{2}=AH\cdot AB$ (Altitude on hypotenuse theorem) \Rightarrow AC=\sqrt{2 \cdot 10}=\sqrt{20} and $tan\mathrm{\angle}CBA=\frac{AC}{AB}=\frac{2\ast \sqrt{5}}{10}=\frac{\sqrt{5}}{5}$.

However the answer in test is 12. I can't find my mistake.