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Question
The acute angle between the medians drawn from the acute angles of a right angled isosceles triangle is
Options
\[\cos^{- 1} \left( \frac{2}{3} \right)\]
\[\cos^{- 1} \left( \frac{3}{4} \right)\]
\[\cos^{- 1} \left( \frac{4}{5} \right)\]
\[\cos^{- 1} \left( \frac{5}{6} \right)\]
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Solution
\[\cos^{- 1} \left( \frac{4}{5} \right)\]
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
=\[\frac{0 - a}{\frac{a}{2} - 0}\]
= -2
Slope of AE = m2
= \[\frac{\frac{a}{2} - 0}{0 - a}\]
\[= - \frac{1}{2}\]
Let \[\theta\] be the angle between BD and AE.
\[\tan \theta = \left| \frac{- 2 + \frac{1}{2}}{1 + 1} \right|\]
\[ = \frac{3}{4}\]
\[ \Rightarrow \cos \theta = \frac{4}{\sqrt{3^2 + 4^2}}\]
\[ \Rightarrow \cos \theta = \frac{4}{5}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{4}{5} \right)\]
Hence, the acute angle between the medians is \[\cos^{- 1} \left( \frac{4}{5} \right)\].
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