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

In $$\triangle ABC, \angle C = 90^\circ$$ and $$CD$$ is perpendicular to $$AB$$ at $$D$$.

Given : $$\frac{AD}{BD} = \sqrt{k}$$

To find : $$\frac{AC}{BC} = ?$$

Solution : Let $$AD=\sqrt{k}$$ and $$BD=1$$

We know that, $$(CD)^2=(AD)\times(BD)$$

=> $$(CD)^2=\sqrt{k}\times1=\sqrt{k}$$

In right $$\triangle$$ BCD,

=> $$(BC)^2=(CD)^2+(BD)^2$$

=> $$(BC)^2=\sqrt{k}+1$$ -----------(i)

Similarly, $$(AC)^2=\sqrt{k}+k=\sqrt{k}(\sqrt{k}+1)$$ ---------------(ii)

Dividing equation (ii) by (i), we get :

=> $$(\frac{AC}{BC})^2=\frac{\sqrt{k}(\sqrt{k}+1)}{\sqrt{k}+1}$$

=> $$\frac{AC}{BC}=\sqrt[4]{k}$$

=> Ans - (D)