Question

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

विकल्प

• \[\cos^{- 1} \left( \frac{2}{3} \right)\]

• \[\cos^{- 1} \left( \frac{3}{4} \right)\]

• \[\cos^{- 1} \left( \frac{4}{5} \right)\]

• \[\cos^{- 1} \left( \frac{5}{6} \right)\]

Answer 1

\[\cos^{- 1} \left( \frac{4}{5} \right)\]

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 \[\cos^{- 1} \left( \frac{4}{5} \right)\].