$\tan\alpha=\dfrac{5}{12}\\↔\dfrac{\sin\alpha}{\cos\alpha}=\tan\alpha\left(\dfrac{5}{12}\right)\\↔\sin\alpha=\tan\alpha.\cos\alpha\\\sin^\alpha+\cos^2\alpha=1\\↔(\tan\alpha.\cos\alpha)^2+\cos^2\alpha=1\\↔\tan^2\alpha.\cos^2\alpha+\cos^2\alpha=1\\↔\left(\dfrac{5}{12}\right)^2\cos^2\alpha+\cos^2\alpha=1\\↔\dfrac{25}{144}\cos^2\alpha+\cos^2\alpha=1\\↔\dfrac{169}{144}\cos^2\alpha=1\\↔\cos^2\alpha=\dfrac{144}{169}$
Xét $ΔABC$ vuông tại $A$:
$\cos\alpha=\dfrac{AB}{BC}>0$
mà $\cos^2\alpha=\dfrac{144}{169}$
$→\cos\alpha=\dfrac{12}{13}\\↔\dfrac{AB}{BC}=\dfrac{12}{13}\\↔\dfrac{AB}{13}=\dfrac{12}{13}\\↔AB=12cm$
Áp dụng định lý Pytago vào $ΔABC$ vuông tại $A$:
$→AC=\sqrt{BC^2-AB^2}=\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5cm$
Vậy $AB=12cm,\,AC=5cm$