Question

Let $ABC$ be a triangle such that $AB=BC$. Let $F$ be the midpoint of $AB$ and $X$ be a point on $BC$ such that $FX$ is perpendicular to $AB$. If $BX=3XC$ then the ratio $\frac{BC}{AC}$ equals

• A. $\sqrt{3}$
• B. $\sqrt{2}$
• C. $\sqrt{\frac{3}{2}}$
• D. $1$

Answer 1

Answer: (C)

$\sqrt{\frac{3}{2}}$

$\cos \theta =\frac{2}{3}$.

Now, $\cos \theta =1-2{\sin }^2\frac{\theta }{2}\to \frac{2}{3}=1-2{\sin }^2\frac{\theta }{2}$

${\sin }^2\frac{\theta }{2}=\frac{1}{6}\to sin\frac{\theta }{2}=\frac{1}{\sqrt{6}}$

$2\alpha =\pi -\theta \to \alpha =\frac{\pi }{2}-\frac{\theta }{2}$

$\cos \alpha =\cos (\frac{\pi }{2}-\frac{\theta }{2})=\sin \frac{\theta }{2}$

$\cos \alpha =\frac{1}{\sqrt{6}}$

$\cos \alpha =\frac{y}{2a}=\frac{1}{\sqrt{6}}$

$\frac{a}{y}=\sqrt{\frac{3}{2}}$

Hence, $\left(C\right)$