Question

The acute angle between the medians drawn from the acute angles of a right angled isosceles 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) ${\cos }^{-1}\left(\frac{5}{6}\right)$

Answer 1

(c) ${\cos }^{-1}\left(\frac{4}{5}\right)$

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.



∴ Slope of BD = m₁
= $\frac{0-a}{{\displaystyle \frac{a}{2}-0}}$
= $-2$

Slope of AE = m₂
= $\frac{{\displaystyle \frac{a}{2}}-0}{0-a}$
$=-\frac{1}{2}$

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

$$\begin{gathered} \tan \theta =\left|\frac{-2+{\displaystyle \frac{1}{2}}}{1+1}\right| \\ =\frac{3}{4} \\ \Rightarrow \cos \theta =\frac{4}{\sqrt{3^2+4^2}} \\ \Rightarrow \cos \theta =\frac{4}{5} \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{4}{5}\right) \end{gathered}$$

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