Question

The cosine of the obtuse angle formed by medians drawn from the vertices of the acute angles of an isosceles right angled triangle is

• A. $-\frac{1}{5}$
• B. $-\frac{2}{5}$
• C. $-\frac{3}{5}$
• D. $-\frac{4}{5}$

Answer 1

Answer: (D)

$-\frac{4}{5}$

Let $AB=BC=\lambda$
$\therefore AC=\lambda \sqrt{2}$ and $BC=\frac{\lambda }{2}$
$\because AD$ and $CE$ are medians $G$ is centroid.
Then $GA=GC=\frac{2}{3}CE$
$=\frac{2}{3}\sqrt{{BC}^2+{CE}^2}$
$=\frac{2}{3}\sqrt{{\lambda }^2+\frac{{\lambda }^2}{4}}$
$=\frac{2}{3}\sqrt{{\lambda }^2(1+\frac{1}{4})}$
$=\frac{2\lambda }{3}\sqrt{\frac{5}{4}}=\frac{\lambda \sqrt{5}}{3}$
$\therefore \cos \theta =\frac{{GA}^2+{GB}^2-{AC}^2}{2.AG.GC}$
$=\frac{{\left(\frac{\lambda \sqrt{5}}{3}\right)}^2+{\left(\frac{\lambda \sqrt{5}}{3}\right)}^2-{\left(\lambda \sqrt{2}\right)}^2}{2\times \frac{\lambda \sqrt{5}}{3}\times \frac{\lambda \sqrt{5}}{3}}$
$=\frac{{\lambda }^2[\frac{5}{9}+\frac{5}{9}-2]}{2\times {\lambda }^2\times \frac{5}{9}}$
Like terms get cancelled in numerator and denominator
$=\frac{\frac{10-18}{9}}{\frac{10}{9}}=-\frac{8}{10}=-\frac{4}{5}$