# 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

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

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

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

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

#### 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?

#### APPEARS IN

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