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# The cosine of angle between the medians drawn from vertices of isosceles rt. angled triangle is?

$(A)\:\:1/5\:\:\:\qquad\:(B)\:\:2/5\:\:\:\qquad\:(C)\:\:3/5\:\:\:\qquad\:(D)\:\:4/5$

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• The Angle between $\overrightarrow {AC}$ and $\overrightarrow {BD}$ is given by $cos\theta=\bigg|\large\frac{\overrightarrow {AC}.\overrightarrow {BD}}{|\overrightarrow {AC}||\overrightarrow {BD}|}\bigg|$
Let the isosceles rt angled triangle be $OAB$ where $O$ be origin, $A$ be on $x\:axis$ and $B$ be on $y\:axis$.
then $\overrightarrow {OA}=a\hat i$ and $\overrightarrow {OB}=a\hat j$
Let $AC$ and $BD$ be the medians drawn from vertices.
$\Rightarrow$ $C$ is mid point of $OB$ and $D$ is mid point of $OA$
$\Rightarrow\:\overrightarrow {OC}=\large\frac{a}{2}$$\hat j and \overrightarrow {OD}=\large\frac{a}{2}$$\hat i$ (From section formula.)
$\Rightarrow\:\overrightarrow {AC}=\overrightarrow {OC}-\overrightarrow {OA}=-a\hat i+\large\frac{b}{2}$$\hat j \overrightarrow {BD}=\overrightarrow {OD}-\overrightarrow {OB}=\large\frac{a}{2}$$\hat i-a\hat j$
$\therefore$ The angle between the medians $\overrightarrow {AC}$ and $\overrightarrow {BD}$ is given by
$cos\theta=\bigg|\large\frac{\overrightarrow {AC}.\overrightarrow {BD}}{|\overrightarrow {AC}||\overrightarrow {BD}|}\bigg|=\large\frac{a^2}{5a^2/4}=\large\frac{4}{5}$

edited Jan 12, 2014