Question

Mark the correct alternative in each of the following :

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

• A. $co{s}^{-1}\left(\frac{2}{3}\right)$
• B. $co{s}^{-1}\left(\frac{3}{4}\right)$
• C. $co{s}^{-1}\left(\frac{4}{5}\right)$
• D. $co{s}^{-1}\left(\frac{5}{6}\right)$

Answer 1

The correct option is C

$co{s}^{-1}\left(\frac{4}{5}\right)$

Let the coordinates of the right-angled isosceles triangle be $O(0,0),A(a,0)$ and $B(0,a)$

Here, BD and AE are the medians drawn from the acute angles B and A, respectively.

$\therefore$ Slope of BD = $m_1$

$=\frac{0-\alpha }{\frac{\alpha }{2}-0}$

$=-2$

Slope of AE = $m_2$

$=\frac{\frac{\alpha }{2}-0}{0-\alpha }$

$=-\frac{1}{2}$

Let $\theta$ be the angle between BD and AE

$tan\theta =\left|\frac{-2+\frac{1}{2}}{1+1}\right|$

$=\frac{3}{4}$

$\Rightarrow cos\theta =\frac{4}{\sqrt{3^2+4^4}}$

$\Rightarrow cos\theta =\frac{4}{5}$

$\Rightarrow \theta =co{s}^{-1}\left(\frac{4}{5}\right)$

Hence, the acute angle between the medians is $co{s}^{-1}\left(\frac{4}{5}\right)$