Question

In the given $ABC$ is a right triangle and right angled at $B$ such that $\angle BCA=2\angle BAC$ Show that hypotenuse $AC=2BC$.

Answer 1

Given Let,$\angle BCA=2\theta$

$\angle BAC=\theta$

$\sin \theta =\frac{BC}{AC},\cos \theta \frac{AB}{AC}$

$\sin 2\theta =\frac{AB}{AC}$

$\sin 2\theta \Rightarrow 2\sin \theta \cos \theta =\frac{AB}{AC}$

$2\frac{BC}{AC}\times \frac{AB}{AC}=\frac{AB}{AC}$

$\Rightarrow 2BC=AC$

$\Rightarrow AC=2BC$