Question

The acute angle between the medians drawn from the acute angles of an isosceles right angled triangle is :

• A. ${\cos }^{-1}\left(\frac{2}{3}\right)$
• B. ${\cos }^{-1}\left(\frac{3}{4}\right)$
• C. ${\cos }^{-1}\left(\frac{4}{5}\right)$
• D. $none$

Answer 1

The correct option is C ${\cos }^{-1}\left(\frac{4}{5}\right)$

Let ABC be an isosceles rt angled triangle be $OAB$ where O be origin. A be on x-axis and B be on y-axis.

then $\overrightarrow{OA}=a\hat{i}$ and $\overrightarrow{OB}=a\hat{j}$

Let AC and BD be the medians drawn from vertices.

C is mid point of OB and D is mid point of OA.

$\overrightarrow{OC}=\frac{a}{2}\overrightarrow{j},\overrightarrow{OD}=\frac{a}{2}\overrightarrow{i}$

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}$

$-a\overrightarrow{i}+\frac{b}{2}\overrightarrow{j}$

$\overrightarrow{BD}=\overrightarrow{OD}-\overrightarrow{OB}$

$\frac{a}{2}\overrightarrow{i}-a\overrightarrow{j}$

Angle between medians $\overrightarrow{AC}and\overrightarrow{BD}$ is given by

$\cos \theta =\left|\frac{\overrightarrow{AC}\overrightarrow{BD}}{|\overrightarrow{AC}||\overrightarrow{BD}|}\right|=\frac{a^2}{\frac{5a^2}{4}}=\frac{4}{5}$