Question

In the given figure $ABC$ is a right triangle and right angled at $B$ such that $\angle BCA=2\angle BAC$. Find $\frac{AC}{BC}$.

Answer 1

$x+2x+90={180}^{\circ }$

$⟹3x={90}^{\circ }$

$⟹x={30}^{\circ }$

$\cos C=\frac{BC}{AC}$

$⟹\cos \left({60}^{\circ }\right)=\frac{BC}{AC}$

$⟹\frac{1}{2}=\frac{BC}{AC}$

$⟹\frac{AC}{BC}=\frac{2}{1}$