In $\triangle ABC, \angle C = 90^\circ$ and CD is perpendicular to AB at D. If $\frac{AD}{BD} = \sqrt{k}$, then $\frac{AC}{BC} = ?$

$\sqrt[4]{k}$

If ABC is a right angled triangle and CD is perpendicular AB we have,

= \(\frac{AD}{BD}\) = \( {(\frac{AC}{BC}) }^{ 2} \)

= \(\sqrt {k }\) = \( {(\frac{AC}{BC}) }^{ 2} \)

= \(\frac{AC}{BC}\) = \(\sqrt[4]{ k}\)

Therefore, the answer is \(\sqrt[4]{ k}\).