Question : In $\triangle$ABC, $\angle$C = 90° and CD is perpendicular to AB at D. If $\frac{\text{AD}}{\text{BD}}=\sqrt{k}$, then $\frac{\text{AC}}{\text{BC}}$=?
Option 1: $\sqrt{k}$
Option 2: $\frac{1}{\sqrt{k}}$
Option 3: $\sqrt[4]{k}$
Option 4: $k$
Correct Answer: $\sqrt[4]{k}$
Solution :
In $\triangle$ABC is a right-angled triangle.
The CD is perpendicular to AB
⇒ $\frac{AD}{BD}$ = $(\frac{AC}{BC})^{2}$
Also,
⇒ $\sqrt{k}$ = $(\frac{AC}{BC})^2$
⇒ $\frac{AC}{BC}$ = $\sqrt{k}^{\frac{1}{2}}$ = $\sqrt[4]{k}$
Hence, the correct answer is $\sqrt[4]{k}$.
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