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The Acute Angle Between the Medians Drawn from the Acute Angles of a Right Angled Isosceles Triangle is - Mathematics

MCQ

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

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Q 2 | Page 133
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